wrong_submission_id
stringlengths 10
10
| problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 131k
1.05M
| wrong_status
stringclasses 2
values | wrong_cpu_time
float64 10
40k
| wrong_memory
float64 2.94k
3.37M
| wrong_code_size
int64 1
15.5k
| problem_description
stringlengths 1
4.75k
| wrong_code
stringlengths 1
6.92k
| acc_submission_id
stringlengths 10
10
| acc_status
stringclasses 1
value | acc_cpu_time
float64 10
27.8k
| acc_memory
float64 2.94k
960k
| acc_code_size
int64 19
14.9k
| acc_code
stringlengths 19
14.9k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s311877673
|
p03719
|
u558764629
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 60
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A,B,C = map(int,input().split())
print('YNeos'[A<=C<=B::2])
|
s578619845
|
Accepted
| 17
| 2,940
| 61
|
A,B,C = map(int,input().split())
print(' YNeos'[A<=C<=B::2])
|
s556599509
|
p04029
|
u475966842
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 63
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N=int(input())
s=0
for i in range(1,N):
s=s+i
print(s)
|
s278850526
|
Accepted
| 17
| 2,940
| 65
|
N=int(input())
s=0
for i in range(1,N+1):
s=s+i
print(s)
|
s218592362
|
p03556
|
u518042385
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 54
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
n=int(input())
num=int((n-1)**(1/2))
print((num+1)**2)
|
s850500968
|
Accepted
| 17
| 2,940
| 38
|
print((int((int(input()))**(1/2)))**2)
|
s965947956
|
p02388
|
u117053676
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,628
| 85
|
Write a program which calculates the cube of a given integer x.
|
import math
print(1)
x = int(input("x="))
print(2)
y = int(math.pow(x,3))
print(y)
|
s013081289
|
Accepted
| 20
| 5,572
| 31
|
x = int(input())
print(x*x*x)
|
s892141312
|
p02409
|
u566311709
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,616
| 252
|
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
n = int(input())
l = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
for _ in range(n):
b, f, r, v = map(int, input().split())
l[b - 1][f - 1][r - 1] += v
for b in range(4):
for f in range(3):
print(*l[b][f])
print("#" * 20)
|
s901682446
|
Accepted
| 20
| 5,620
| 327
|
n = int(input())
l = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
for _ in range(n):
b, f, r, v = map(int, input().split())
l[b - 1][f - 1][r - 1] += v
for b in range(4):
for f in range(3):
for r in range(10):
print(" %d" % l[b][f][r], end = "")
print("")
if b != 3:
print("#" * 20)
|
s401600011
|
p03387
|
u029234056
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 140
|
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
T=list(map(int,input().split()))
T.sort()
L1=T[0]
L2=T[1]
L3=T[2]
ans=0
ans+=L3-L2
print(ans)
ans+=(L3-ans-L1+1)//2+(L3-L1-ans)%2
print(ans)
|
s542793941
|
Accepted
| 19
| 3,060
| 129
|
T=list(map(int,input().split()))
T.sort()
L1=T[0]
L2=T[1]
L3=T[2]
ans=0
ans+=L3-L2
ans+=(L3-ans-L1+1)//2+(L3-L1-ans)%2
print(ans)
|
s160455403
|
p03644
|
u370429695
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 175
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
split_num = 0
flag = 0
while flag == 0:
if n % 2 == 1:
flag += 1
break
else:
split_num += 1
n = n / 2
print(split_num)
|
s011781973
|
Accepted
| 18
| 3,060
| 190
|
n = int(input())
cnt = 0
num = 1
for i in range(1,n+1):
a = 0
c = i
while c % 2 == 0:
c = c // 2
a += 1
if a > cnt:
num = i
cnt = a
print(num)
|
s498061886
|
p02400
|
u003684951
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,640
| 100
|
Write a program which calculates the area and circumference of a circle for given radius r.
|
from math import pi
r = float(input())
area=r*r*pi
cir =(r+r)*pi
print(f'{(area):.6f}{(cir):.6f}')
|
s286829310
|
Accepted
| 20
| 5,640
| 109
|
from math import pi
r = float(input())
area = r * r * pi
cir =(r+r) * pi
print(f'{(area):.6f} {(cir):.6f}')
|
s164102828
|
p02742
|
u620238824
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 213
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
R,C = list(map(int, input().split()))
if R % 2 == 0:
print(R / 2 * C)
else:
if C % 2 == 0:
print(((R // 2 + 1) + (R // 2)) * C / 2)
else:
print((R // 2 + 1)*(C//2+1) + (R // 2)*(C//2))
|
s801016738
|
Accepted
| 17
| 3,060
| 151
|
h, w = map(int,input().split())
if h == 1 or w == 1:
print(1)
elif h % 2 == 1 and w % 2 == 1:
print(h * w// 2 + 1)
else:
print(h * w // 2)
|
s806185354
|
p03448
|
u860855494
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 373
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
result = 0
A = int(input()) #500
B = int(input()) #100
C = int(input()) #50
X = int(input())
for a in range(A):
if a * 500 <= X:
tmp = a * 500
for b in range(B):
if tmp + b * 100 <= X:
tmp += b * 100
for c in range(C):
if tmp + c * 50 <= X:
tmp += c * 50
if tmp == X:
result += 1
print(result)
|
s778543473
|
Accepted
| 51
| 3,060
| 251
|
result = 0
A = int(input()) #500
B = int(input()) #100
C = int(input()) #50
X = int(input())
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
if 500*a + 100*b + 50*c == X:
result += 1
print(result)
|
s088578472
|
p03448
|
u273038590
| 2,000
| 262,144
|
Wrong Answer
| 199
| 4,852
| 369
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a=["","",""]
for i in range(3):
a[i]=int(input())
x=int(input())
cnt=0
for i in range(a[0]+1):
print("----i="+str(i))
for j in range(a[1]+1):
print("----j="+str(j))
for k in range(a[2]+1):
ans = x-500*i - 100*j - 50*k
print("----k="+str(k) + str(ans))
if ans==0:
cnt+=1
print(cnt)
|
s870421251
|
Accepted
| 54
| 3,060
| 256
|
a=["","",""]
for i in range(3):
a[i]=int(input())
x=int(input())
cnt=0
for i in range(a[0]+1):
for j in range(a[1]+1):
for k in range(a[2]+1):
ans = x-500*i - 100*j - 50*k
if ans==0:
cnt+=1
print(cnt)
|
s593036912
|
p03574
|
u118147328
| 2,000
| 262,144
|
Wrong Answer
| 31
| 3,444
| 476
|
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
H,W = map(int, input().split())
print(H,W)
S = [input() for i in range(H)]
count = []
for i in range(H):
for j in range(W):
if S[i][j] == ".":
cnt = 0
for x in [-1, 0, 1]:
for y in [-1, 0, 1]:
if 0 <= i+x < H and 0 <= j+y < W:
if S[i+x][j+y] == "#":
cnt += 1
print(cnt,end="")
else:
print("#",end="")
print()
|
s881184626
|
Accepted
| 31
| 3,444
| 460
|
H,W = map(int, input().split())
S = [input() for i in range(H)]
count = []
for i in range(H):
for j in range(W):
if S[i][j] == ".":
cnt = 0
for x in [-1, 0, 1]:
for y in [-1, 0, 1]:
if 0 <= i+x < H and 0 <= j+y < W:
if S[i+x][j+y] == "#":
cnt += 1
print(cnt,end="")
else:
print("#",end="")
print()
|
s744652179
|
p04043
|
u264681142
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 106
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a, b, c = map(int, input().split())
if sorted((a,b,c)) == [5,5,7]:
print("Yes")
else:
print("No")
|
s557683872
|
Accepted
| 16
| 2,940
| 106
|
a, b, c = map(int, input().split())
if sorted((a,b,c)) == [5,5,7]:
print("YES")
else:
print("NO")
|
s120544705
|
p02613
|
u514687406
| 2,000
| 1,048,576
|
Wrong Answer
| 150
| 9,412
| 252
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
# import sys;input = lambda : sys.stdin.readline()
import collections
d=collections.defaultdict(lambda:0)
for _ in range(int(input())):
s=input()
d[s]+=1
print("AC X",d['AC'])
print("WA X",d['WA'])
print("TLE X",d["TLE"])
print("RE X",d['RE'])
|
s994077948
|
Accepted
| 146
| 9,476
| 252
|
# import sys;input = lambda : sys.stdin.readline()
import collections
d=collections.defaultdict(lambda:0)
for _ in range(int(input())):
s=input()
d[s]+=1
print("AC x",d['AC'])
print("WA x",d['WA'])
print("TLE x",d["TLE"])
print("RE x",d['RE'])
|
s767226843
|
p03712
|
u721970149
| 2,000
| 262,144
|
Wrong Answer
| 31
| 3,856
| 325
|
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
H, W = map(int,input().split())
S = [list(input()) for i in range(H)]
Ans = []
for i in range(H+2) :
Ansi = ["#" for x in range(W+2)]
if i != 0 and i != H+1 :
for j in range(W) :
print(i, j)
Ansi[j+1] = S[i-1][j]
Ans.append(Ansi)
for i in range(H+2) :
print("".join(Ans[i]))
|
s187829943
|
Accepted
| 20
| 3,188
| 301
|
H, W = map(int,input().split())
S = [list(input()) for i in range(H)]
Ans = []
for i in range(H+2) :
Ansi = ["#" for x in range(W+2)]
if i != 0 and i != H+1 :
for j in range(W) :
Ansi[j+1] = S[i-1][j]
Ans.append(Ansi)
for i in range(H+2) :
print("".join(Ans[i]))
|
s670281848
|
p03044
|
u795630164
| 2,000
| 1,048,576
|
Wrong Answer
| 685
| 30,932
| 511
|
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
N = int(input())
data = []
for i in range(N-1):
a, b, c = map(int, input().split())
data.append((a,b,c))
data.append((b,a,c))
data.sort()
ans = [0] * (N+1)
went = [False] * (N+1)
went[0] = True
went[1] = True
before = 1
for i in range(2*(N-1)):
a, b, c = data[i]
if went[a]:
if not went[b]:
went[b] = True
if c % 2 == 0:
ans[b] = ans[a]
else:
ans[b] = 1 - ans[a]
for i in range(1, N+1):
print(ans[i])
print()
|
s256701431
|
Accepted
| 653
| 39,596
| 561
|
N = int(input())
data = [[] for _ in range(N+1)]
for i in range(N-1):
a, b, c = map(int, input().split())
data[a].append((b,c))
data[b].append((a,c))
ans = [0] * (N+1)
went = [False] * (N+1)
went[0] = True
went[1] = True
go = [1]
while len(go) > 0:
before = go.pop()
for i, c in data[before]:
if not went[i]:
went[i] = True
if c % 2 == 0:
ans[i] = ans[before]
else:
ans[i] = 1 - ans[before]
go.append(i)
for i in range(1, N+1):
print(ans[i])
|
s335112251
|
p02927
|
u814986259
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 3,060
| 198
|
Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have?
|
M,D = map(int, input().split())
ans = 0
for i in range(1, M+1):
for j in range(1, D+1):
tmp = 1
while(j > 0):
tmp *= j%10
j = j // 10
if tmp == i:
ans += 1
print(ans)
|
s167514220
|
Accepted
| 19
| 2,940
| 221
|
M,D = map(int, input().split())
ans = 0
for i in range(1,M+1):
for j in range(2, D//10 + 1):
for k in range(2, 10):
if j*10 + k > D:
break
else:
if j*k == i:
ans += 1
print(ans)
|
s869328917
|
p03671
|
u437638594
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 69
|
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
a, b, c = map(int, input().split())
print(sum(sorted([a, b, c])[:1]))
|
s951672007
|
Accepted
| 17
| 2,940
| 69
|
a, b, c = map(int, input().split())
print(a + b + c - max([a, b, c]))
|
s246534377
|
p00718
|
u798803522
| 1,000
| 131,072
|
Wrong Answer
| 130
| 8,332
| 2,381
|
Prof. Hachioji has devised a new numeral system of integral numbers with four lowercase letters "m", "c", "x", "i" and with eight digits "2", "3", "4", "5", "6", "7", "8", "9". He doesn't use digit "0" nor digit "1" in this system. The letters "m", "c", "x" and "i" correspond to 1000, 100, 10 and 1, respectively, and the digits "2", ...,"9" correspond to 2, ..., 9, respectively. This system has nothing to do with the Roman numeral system. For example, character strings > "5m2c3x4i", "m2c4i" and "5m2c3x" correspond to the integral numbers 5234 (=5*1000+2*100+3*10+4*1), 1204 (=1000+2*100+4*1), and 5230 (=5*1000+2*100+3*10), respectively. The parts of strings in the above example, "5m", "2c", "3x" and "4i" represent 5000 (=5*1000), 200 (=2*100), 30 (=3*10) and 4 (=4*1), respectively. Each of the letters "m", "c", "x" and "i" may be prefixed by one of the digits "2", "3", ..., "9". In that case, the prefix digit and the letter are regarded as a pair. A pair that consists of a prefix digit and a letter corresponds to an integer that is equal to the original value of the letter multiplied by the value of the prefix digit. For each letter "m", "c", "x" and "i", the number of its occurrence in a string is at most one. When it has a prefix digit, it should appear together with the prefix digit. The letters "m", "c", "x" and "i" must appear in this order, from left to right. Moreover, when a digit exists in a string, it should appear as the prefix digit of the following letter. Each letter may be omitted in a string, but the whole string must not be empty. A string made in this manner is called an _MCXI-string_. An MCXI-string corresponds to a positive integer that is the sum of the values of the letters and those of the pairs contained in it as mentioned above. The positive integer corresponding to an MCXI-string is called its MCXI-value. Moreover, given an integer from 1 to 9999, there is a unique MCXI-string whose MCXI-value is equal to the given integer. For example, the MCXI-value of an MCXI-string "m2c4i" is 1204 that is equal to `1000 + 2*100 + 4*1`. There are no MCXI-strings but "m2c4i" that correspond to 1204. Note that strings "1m2c4i", "mcc4i", "m2c0x4i", and "2cm4i" are not valid MCXI-strings. The reasons are use of "1", multiple occurrences of "c", use of "0", and the wrong order of "c" and "m", respectively. Your job is to write a program for Prof. Hachioji that reads two MCXI-strings, computes the sum of their MCXI-values, and prints the MCXI-string corresponding to the result.
|
trial = int(input())
numlist = [str(n) for n in range(2,10)]
cases = []
for t in range(trial):
cases.append(input().split(" "))
print(cases)
for case in range(len(cases)):
num = 0
for ba in range(2):
for n in range(len(cases[case][ba])):
if cases[case][ba][n] == "m":
if n - 1 == 0:
num += int(cases[case][ba][n-1]) * 1000
else:
num += 1000
elif cases[case][ba][n] == "c":
if (n - 1 == 0 and cases[case][ba][n-1] != "m") or cases[case][ba][n-1] in numlist:
num += int(cases[case][ba][n-1]) * 100
else:
num += 100
elif cases[case][ba][n] == "x":
if (n - 1 == 0 and cases[case][ba][n-1] not in ["m","c"]) in numlist:
num += int(cases[case][ba][n-1]) * 10
else:
num += 10
elif cases[case][ba][n] == "i":
if (n - 1 == 0 and cases[case][ba][n-1] not in ["m","c","x"]) or cases[case][ba][n-1] in numlist:
num += int(cases[case][ba][n-1]) * 1
else:
num += 1
else:
num = str(num)
answer = ""
print(num)
for n in range(len(num)):
if len(num) - 1 - n == 3:
if num[n] == "1":
answer += "m"
elif num[n] == "0":
pass
else:
answer += str(num[n]) + "m"
if len(num) - 1 - n == 2:
if num[n] == "1":
answer += "c"
elif num[n] == "0":
pass
else:
answer += str(num[n]) + "c"
if len(num) - 1 - n == 1:
if num[n] == "1":
answer += "x"
elif num[n] == "0":
pass
else:
answer += str(num[n]) + "x"
if len(num) - 1 - n == 0:
if num[n] == "1":
answer += "i"
elif num[n] == "0":
pass
else:
answer += str(num[n]) + "i"
else:
print(answer)
|
s414359065
|
Accepted
| 110
| 8,368
| 2,393
|
trial = int(input())
numlist = [str(n) for n in range(2,10)]
cases = []
for t in range(trial):
cases.append(input().split(" "))
for case in range(len(cases)):
num = 0
for ba in range(2):
#print(num)
for n in range(len(cases[case][ba])):
if cases[case][ba][n] == "m":
if n - 1 == 0:
num += int(cases[case][ba][n-1]) * 1000
else:
num += 1000
elif cases[case][ba][n] == "c":
if (n - 1 == 0 and cases[case][ba][n-1] != "m") or cases[case][ba][n-1] in numlist:
num += int(cases[case][ba][n-1]) * 100
else:
num += 100
elif cases[case][ba][n] == "x":
if (n - 1 == 0 and cases[case][ba][n-1] not in ["m","c"]) or cases[case][ba][n-1] in numlist:
num += int(cases[case][ba][n-1]) * 10
else:
num += 10
elif cases[case][ba][n] == "i":
if (n - 1 == 0 and cases[case][ba][n-1] not in ["m","c","x"]) or cases[case][ba][n-1] in numlist:
num += int(cases[case][ba][n-1]) * 1
else:
num += 1
else:
num = str(num)
answer = ""
for n in range(len(num)):
if len(num) - 1 - n == 3:
if num[n] == "1":
answer += "m"
elif num[n] == "0":
pass
else:
answer += str(num[n]) + "m"
if len(num) - 1 - n == 2:
if num[n] == "1":
answer += "c"
elif num[n] == "0":
pass
else:
answer += str(num[n]) + "c"
if len(num) - 1 - n == 1:
if num[n] == "1":
answer += "x"
elif num[n] == "0":
pass
else:
answer += str(num[n]) + "x"
if len(num) - 1 - n == 0:
if num[n] == "1":
answer += "i"
elif num[n] == "0":
pass
else:
answer += str(num[n]) + "i"
else:
print(answer)
|
s445920066
|
p03409
|
u929618357
| 2,000
| 262,144
|
Wrong Answer
| 52
| 3,064
| 400
|
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point. At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.
|
N = int(input())
red = [3*N] * (2*N)
blue = [0] * (2*N)
for i in range(N):
a, b = (int(j) for j in input().split())
red[a] = b
for i in range(N):
a, b = (int(j) for j in input().split())
blue[a] = b
ans = 0
for i in range(1, 2*N):
for j in range(blue[i]-1, -1, -1):
if j in red:
ans += 1
blue[i] = 0
red[red.index(j)] = 3 * N
print(ans)
|
s630718603
|
Accepted
| 62
| 3,064
| 441
|
N = int(input())
red = [3*N] * (2*N)
blue = [-1] * (2*N)
for i in range(N):
a, b = (int(j) for j in input().split())
red[a] = b
for i in range(N):
a, b = (int(j) for j in input().split())
blue[a] = b
ans = 0
for i in range(1, 2*N):
for j in range(blue[i]-1, -1, -1):
if j in red and red.index(j) < i:
ans += 1
blue[i] = 0
red[red.index(j)] = 3 * N
break
print(ans)
|
s681357574
|
p03556
|
u556610039
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 53
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
import math
n = int(input())
print(int(math.sqrt(n)))
|
s410491252
|
Accepted
| 17
| 2,940
| 63
|
import math
n = int(input())
r = int(math.sqrt(n))
print(r * r)
|
s820381184
|
p02975
|
u286486951
| 2,000
| 1,048,576
|
Wrong Answer
| 42
| 14,212
| 55
|
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
N = int(input())
nums = list(map(int, input().split()))
|
s118543836
|
Accepted
| 78
| 14,212
| 263
|
N = int(input())
nums = list(map(int, input().split()))
sums = 0
for i in range(len(nums)):
sums ^= nums[i]
ans = True
for i in range(len(nums)):
if nums[i] == sums ^ nums[i]:
pass
else:
ans = False
if ans:
print("Yes")
else:
print("No")
|
s711897055
|
p02442
|
u126478680
| 1,000
| 262,144
|
Wrong Answer
| 20
| 5,592
| 326
|
Compare given two sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$ and $B = \\{b_0, b_1, ..., b_{m-1}$ lexicographically.
|
n = int(input())
a = list(map(int, input().split(' ')))
m = int(input())
b = list(map(int, input().split(' ')))
loop_num = n if n <= m else m
for i in range(loop_num):
if a[i] < b[i]:
print(1)
break
elif a[i] > b[i]:
print(0)
break
if len(a) < len(b):
print(1)
else:
print(0)
|
s595027674
|
Accepted
| 20
| 5,620
| 376
|
n = int(input())
a = list(map(int, input().split(' ')))
m = int(input())
b = list(map(int, input().split(' ')))
rst = None
loop_num = n if n <= m else m
for i in range(loop_num):
if a[i] < b[i]:
rst = 1
elif a[i] > b[i]:
rst = 0
if rst != None: break
if rst == None:
if len(a) < len(b):
rst = 1
else:
rst = 0
print(rst)
|
s294587731
|
p03385
|
u835482198
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 74
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s = sorted(input())
if s == 'abc':
print("Yes")
else:
print("No")
|
s896372696
|
Accepted
| 17
| 2,940
| 83
|
s = "".join(sorted(input()))
if s == 'abc':
print("Yes")
else:
print("No")
|
s451209588
|
p04012
|
u999669171
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 126
|
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = input()
w_set = set( w )
for each_ch in w_set:
if w.count( each_ch ) % 2 == 1:
print( "NO" )
exit()
print( "YES" )
|
s144838550
|
Accepted
| 17
| 2,940
| 124
|
w = input()
w_set = set( w )
for each_ch in w_set:
if w.count( each_ch ) % 2 == 1:
print( "No" )
exit()
print( "Yes" )
|
s427527694
|
p03068
|
u684695949
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 179
|
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
N = int(input())
S = input().rstrip()
K = int(input())
val = list(S)[K-1]
print(N,S,K,list(S),val)
output=[tmp if tmp == val else '*' for tmp in list(S) ]
print("".join(output))
|
s339172279
|
Accepted
| 17
| 2,940
| 156
|
N = int(input())
S = input().rstrip()
K = int(input())
val = list(S)[K-1]
output=[tmp if tmp == val else '*' for tmp in list(S) ]
print("".join(output))
|
s234875527
|
p02612
|
u312814337
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,180
| 142
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
yen = N // 100 +1
shoki = yen * 1000
oturi = shoki - N
if oturi == 1000:
print(int(oturi - 1000))
else:
print(int(oturi))
|
s635684451
|
Accepted
| 29
| 9,168
| 143
|
N = int(input())
yen = N // 1000 +1
shoki = yen * 1000
oturi = shoki - N
if oturi == 1000:
print(int(oturi - 1000))
else:
print(int(oturi))
|
s145522266
|
p03997
|
u886581995
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 158
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
# -*- coding: utf-8 -*-
"""
Created on Fri Nov 29 16:01:33 2019
@author: virfi
"""
a = int(input())
b = int(input())
h = int(input())
S = (a+b)*h/2
print(S)
|
s488318519
|
Accepted
| 17
| 2,940
| 80
|
a = int(input())
b = int(input())
h = int(input())
S = int((a+b)*h/2)
print(S)
|
s549396492
|
p02612
|
u735891571
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,064
| 31
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N%1000)
|
s046917760
|
Accepted
| 31
| 9,148
| 77
|
N = int(input())
if N%1000 == 0:
print(0)
else:
print(1000-N%1000)
|
s321201248
|
p02842
|
u970082363
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 115
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
import math
n = int(input())
m = (n+1)//1.08
l = math.floor(m*1.08)
if m==l:
print(m)
else:
print(":(")
|
s949932134
|
Accepted
| 17
| 2,940
| 125
|
import math
n = int(input())
m = math.floor((n+1)/1.08)
l = math.floor(m*1.08)
if n==l:
print(m)
else:
print(":(")
|
s116415751
|
p03110
|
u823044869
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 275
|
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
n = int(input())
otoshidama = [float(i[0]) if i[1] == "JPY" else float(i[0])*380000 for j in range(n) for i in [input().split()]]
print(otoshidama)
print(sum(otoshidama))
|
s016954905
|
Accepted
| 17
| 2,940
| 257
|
n = int(input())
otoshidama = [float(i[0]) if i[1] == "JPY" else float(i[0])*380000 for j in range(n) for i in [input().split()]]
print(sum(otoshidama))
|
s107424386
|
p00004
|
u536089081
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,580
| 185
|
Write a program which solve a simultaneous equation: ax + by = c dx + ey = f The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
|
from sys import stdin
for line in stdin:
a, b, c, d, e, f = map(float, line.split())
print('%lf %lf' % ((c * e - f * b) / (a * e - b * d), (c * d - a * f) / (b * d - a * e)) )
|
s917209378
|
Accepted
| 20
| 5,580
| 320
|
from sys import stdin
for line in stdin:
a, b, c, d, e, f = map(float, line.split())
anss = ['%.3f' % ((c * e - f * b) / (a * e - b * d)), '%.3f' % ((c * d - a * f) / (b * d - a * e))]
for index, ans in enumerate(anss):
if ans == '-0.000':
anss[index] = '0.000'
print(' '.join(anss))
|
s179236513
|
p03416
|
u742729271
| 2,000
| 262,144
|
Wrong Answer
| 121
| 3,064
| 201
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
A, B = map(int, input().split())
count=0
for i in range(10001, 100000, 1):
s = str(i)
a = int(s[4])
b = int(s[3])
c = int(s[1])
d = int(s[0])
if a==d and b==c:
count+=1
print(count)
|
s042632859
|
Accepted
| 118
| 3,064
| 194
|
A, B = map(int, input().split())
count=0
for i in range(A, B+1, 1):
s = str(i)
a = int(s[4])
b = int(s[3])
c = int(s[1])
d = int(s[0])
if a==d and b==c:
count+=1
print(count)
|
s917615143
|
p03455
|
u905582793
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 73
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=map(int,input().split())
if a*b%2:
print("Yes")
else:
print("No")
|
s932433388
|
Accepted
| 17
| 2,940
| 75
|
a,b=map(int,input().split())
if a*b%2:
print("Odd")
else:
print("Even")
|
s498004156
|
p03351
|
u518042385
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 132
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,x=map(int,input().split())
if abs(a-c)<=x:
print("Yes")
elif abs(a-b)<=x and abs(a-c)<=x:
print("Yes")
else:
print("No")
|
s707171532
|
Accepted
| 17
| 2,940
| 133
|
a,b,c,x=map(int,input().split())
if abs(a-c)<=x:
print("Yes")
elif abs(a-b)<=x and abs(b-c)<=x:
print("Yes")
else:
print("No")
|
s915290263
|
p02856
|
u163320134
| 2,000
| 1,048,576
|
Wrong Answer
| 616
| 3,060
| 144
|
N programmers are going to participate in the preliminary stage of DDCC 20XX. Due to the size of the venue, however, at most 9 contestants can participate in the finals. The preliminary stage consists of several rounds, which will take place as follows: * All the N contestants will participate in the first round. * When X contestants participate in some round, the number of contestants advancing to the next round will be decided as follows: * The organizer will choose two consecutive digits in the decimal notation of X, and replace them with the sum of these digits. The number resulted will be the number of contestants advancing to the next round. For example, when X = 2378, the number of contestants advancing to the next round will be 578 (if 2 and 3 are chosen), 2108 (if 3 and 7 are chosen), or 2315 (if 7 and 8 are chosen). When X = 100, the number of contestants advancing to the next round will be 10, no matter which two digits are chosen. * The preliminary stage ends when 9 or fewer contestants remain. Ringo, the chief organizer, wants to hold as many rounds as possible. Find the maximum possible number of rounds in the preliminary stage. Since the number of contestants, N, can be enormous, it is given to you as two integer sequences d_1, \ldots, d_M and c_1, \ldots, c_M, which means the following: the decimal notation of N consists of c_1 + c_2 + \ldots + c_M digits, whose first c_1 digits are all d_1, the following c_2 digits are all d_2, \ldots, and the last c_M digits are all d_M.
|
n=int(input())
sums=0
digits=0
for _ in range(n):
a,b=map(int,input().split())
sums+=a*b
digits+=b
ans=(digits-1)+(sums+9-1)//9
print(ans)
|
s099971816
|
Accepted
| 595
| 3,188
| 142
|
n=int(input())
sums=0
digits=0
for _ in range(n):
a,b=map(int,input().split())
sums+=a*b
digits+=b
ans=(digits-1)+(sums-1)//9
print(ans)
|
s483663504
|
p03485
|
u565204025
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 95
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
# -*- coding: utf-8 -*-
import math
a,b = map(int,input().split())
print(math.floor(a*b/2))
|
s353050993
|
Accepted
| 19
| 2,940
| 96
|
# -*- coding: utf-8 -*-
import math
a,b = map(int,input().split())
print(math.ceil((a+b)/2))
|
s387619774
|
p04029
|
u914330401
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 34
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n = int(input())
print(n*(n+1)/2)
|
s449073018
|
Accepted
| 17
| 2,940
| 35
|
n = int(input())
print(n*(n+1)//2)
|
s100941704
|
p03493
|
u393512980
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 29
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
s=input()
print(s.count('0'))
|
s945727502
|
Accepted
| 17
| 2,940
| 30
|
s=input()
print(s.count('1'))
|
s089434087
|
p03448
|
u957799665
| 2,000
| 262,144
|
Wrong Answer
| 110
| 9,264
| 292
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
cnt = 0
for a in range(A+1):
print("a=",a)
for b in range(B+1):
print("b=",b)
for c in range(C+1):
print("c=",c)
if 500*a+100*b+50*c == X:
cnt += 1
print(cnt)
|
s528013651
|
Accepted
| 56
| 9,096
| 226
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
cnt = 0
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
if 500*a+100*b+50*c == X:
cnt += 1
print(cnt)
|
s451930794
|
p02613
|
u699944218
| 2,000
| 1,048,576
|
Wrong Answer
| 147
| 16,272
| 244
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
S = [None for _ in range(N)]
for i in range(N):
S[i] = input()
a = S.count('AC')
b = S.count('WA')
c = S.count('TLE')
d = S.count('RE')
print('AC × %d' % a)
print('WA × %d' %b)
print('TLE × %d' %c)
print('RE × %d' %d)
|
s201984536
|
Accepted
| 146
| 16,152
| 239
|
N = int(input())
S = [None for _ in range(N)]
for i in range(N):
S[i] = input()
a = S.count('AC')
b = S.count('WA')
c = S.count('TLE')
d = S.count('RE')
print('AC x %d' %a)
print('WA x %d' %b)
print('TLE x %d' %c)
print('RE x %d' %d)
|
s026029502
|
p03478
|
u133936772
| 2,000
| 262,144
|
Wrong Answer
| 34
| 2,940
| 122
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b = map(int,input().split())
ans = 0
for i in range(n):
if a<=sum(map(int,list(str(i))))<=b:
ans += i
print(ans)
|
s340015304
|
Accepted
| 34
| 2,940
| 124
|
n,a,b = map(int,input().split())
ans = 0
for i in range(n+1):
if a<=sum(map(int,list(str(i))))<=b:
ans += i
print(ans)
|
s815225079
|
p03852
|
u600402037
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 53
|
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
print('vowel' if input in ['aeiou'] else 'consonant')
|
s917469002
|
Accepted
| 17
| 2,940
| 54
|
print('vowel' if input() in 'aeiou' else 'consonant')
|
s337973574
|
p03555
|
u252210202
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 114
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
A = input()
D = input()
if A[0] == D[2] and A[1] == D[1] and A[2] == D[0]:
print("Yes")
else:
print("No")
|
s423885820
|
Accepted
| 17
| 2,940
| 113
|
A = input()
D = input()
if A[0] == D[2] and A[1] == D[1] and A[2] == D[0]:
print("YES")
else:
print("NO")
|
s303998817
|
p03457
|
u759412327
| 2,000
| 262,144
|
Wrong Answer
| 432
| 11,792
| 262
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
T = [0]
X = [0]
Y = [0]
f = True
for i in range(N-1):
t,x,y = list(map(int,input().split()))
T.append(t)
X.append(x)
Y.append(y)
if T[-2]-T[-1]<abs(X[-2]-X[-1])+abs(Y[-2]-Y[-1]):
f = False
if f:
print("Yes")
else:
print("No")
|
s617502557
|
Accepted
| 290
| 27,840
| 260
|
N = int(input())
P = [[0,0,0]]+[list(map(int,input().split())) for n in range(N)]
ans = "Yes"
for n in range(N):
dt = P[n+1][0]-P[n][0]
dx = abs(P[n+1][1]-P[n][1])
dy = abs(P[n+1][2]-P[n][2])
if dt<dx+dy or dt%2!=(dx+dy)%2:
ans = "No"
print(ans)
|
s380932066
|
p03795
|
u708211626
| 2,000
| 262,144
|
Wrong Answer
| 23
| 9,132
| 66
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
a=int(input())
import math
b=math.factorial(a)
print(b%(10**9+7))
|
s808919276
|
Accepted
| 25
| 9,000
| 44
|
a = int(input())
print((a*800)-(a//15*200))
|
s888581508
|
p02608
|
u946517952
| 2,000
| 1,048,576
|
Wrong Answer
| 271
| 9,440
| 383
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
import math
n = int(input())
nlist = [0]*n
for x in range(1,math.ceil(math.sqrt(n-1)-1)):
for y in range(1,math.ceil(math.sqrt(n-1)-1)):
for z in range(math.ceil(math.sqrt(n-1)-1)):
temp = x**2 + y**2 + z**2 + x*y + y*z + z*x
if temp > n:
break
else:
nlist[temp-1] +=1
for num in nlist:
print(num)
|
s976580636
|
Accepted
| 274
| 9,368
| 385
|
import math
n = int(input())
nlist = [0]*n
for x in range(1,math.ceil(math.sqrt(n-1)-1)):
for y in range(1,math.ceil(math.sqrt(n-1)-1)):
for z in range(1,math.ceil(math.sqrt(n-1)-1)):
temp = x**2 + y**2 + z**2 + x*y + y*z + z*x
if temp > n:
break
else:
nlist[temp-1] +=1
for num in nlist:
print(num)
|
s681351997
|
p02415
|
u024715419
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,424
| 39
|
Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string.
|
inp = input()
inp.swapcase()
print(inp)
|
s027927534
|
Accepted
| 20
| 7,396
| 35
|
inp = input()
print(inp.swapcase())
|
s455234194
|
p00006
|
u328733599
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,552
| 88
|
Write a program which reverses a given string str.
|
txt = input("")
for i in range(0,len(txt)):
print(txt[(len(txt)-i-1)],end="")
print()
|
s288933385
|
Accepted
| 20
| 5,552
| 87
|
txt = input("")
for i in range(0,len(txt)):
print(txt[(len(txt)-i-1)],end="")
print()
|
s880663445
|
p03547
|
u867848444
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 112
|
In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?
|
ori=list(input().split())
syu=sorted(ori)
if ori==syu:
print('<')
else:
print('>')
print(syu)
print(ori)
|
s688235182
|
Accepted
| 17
| 2,940
| 158
|
ori=list(input().split())
syu=sorted(ori)
if ori[0]==ori[1]:
print('=')
elif ori!=syu:
print('>')
else:
print('<')
|
s603361548
|
p03970
|
u363610900
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 130
|
CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.
|
s = sorted(list(input()))
t = 'CODEFESTIVAL2016'
cnt = 0
for i in range(len(s)):
if s[i] != t[i]:
cnt += 1
print(cnt)
|
s985689460
|
Accepted
| 17
| 2,940
| 111
|
S = input()
r = 'CODEFESTIVAL2016'
cnt = 0
for i in range(16):
if S[i] != r[i]:
cnt += 1
print(cnt)
|
s210629234
|
p03854
|
u597626771
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 170
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
l = input()
l = l.replace('eraser','')
l = l.replace('erase','')
l = l.replace('dream','')
l = l.replace('dreamer','')
if l == '':
print('Yes')
else:
print('No')
|
s741954266
|
Accepted
| 19
| 3,188
| 170
|
l = input()
l = l.replace('eraser','')
l = l.replace('erase','')
l = l.replace('dreamer','')
l = l.replace('dream','')
if l == '':
print('YES')
else:
print('NO')
|
s045591069
|
p03377
|
u392361133
| 2,000
| 262,144
|
Wrong Answer
| 16
| 2,940
| 77
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a, b, x = map(int, input().split())
print('Yes' if a <= x <= a + b else 'No')
|
s409066007
|
Accepted
| 26
| 9,172
| 77
|
a, b, x = map(int, input().split())
print("YES" if a <= x <= a + b else "NO")
|
s288010361
|
p03544
|
u781152930
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 151
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
# -*- coding: utf-8 -*-
N = str, input()
num = int(N[1])
L0 = 2
L1 = 1
x = 0
for i in range(2, num):
x = L1 + L0
L0 = L1
L1 = x
print(x)
|
s763355220
|
Accepted
| 17
| 2,940
| 151
|
# -*- coding: utf-8 -*-
N = str, input()
num = int(N[1])
L0 = 2
L1 = 1
x = 1
for i in range(1, num):
x = L1 + L0
L0 = L1
L1 = x
print(x)
|
s594011563
|
p03795
|
u904075088
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,020
| 112
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
#!/usr/bin/env python
# -*- coding: utf-8 -*-
num=int(input('N'))
quo=num//15
x=800*num
y=200*quo
print(x-y)
|
s665287986
|
Accepted
| 31
| 9,032
| 109
|
#!/usr/bin/env python
# -*- coding: utf-8 -*-
num=int(input())
quo=num//15
x=800*num
y=200*quo
print(x-y)
|
s384548805
|
p02603
|
u916242112
| 2,000
| 1,048,576
|
Wrong Answer
| 34
| 9,140
| 215
|
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
N = int(input())
a = list(map(int,input().split()))
S = 1000
K = 0
for i in range(N):
if i != N-1:
if a[i]<=a[i+1]:
K = S // a[i]
S -= K*a[i]
if a[i]>a[i+1]:
S += a[i]*K
else:
S += a[i]*K
print(S)
|
s307027040
|
Accepted
| 29
| 9,084
| 249
|
N = int(input())
a = list(map(int,input().split()))
S = 1000
K = 0
b = 1
last = 0
for i in range(N):
if i != N-1:
if a[i]<=a[i+1]:
S += a[i]*K
K = S // a[i]
S -= K*a[i]
else:
S += a[i]*K
K = 0
else:
S += a[i]*K
print(S)
|
s006312579
|
p02607
|
u060012100
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,092
| 161
|
We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd.
|
n = int(input("Enter your limit : "))
s = list(map(int,input().split()[:n]))
a = 0
for i in range(1,len(s)):
if((i%2 != 0) and (s[i]%2 !=0)):
a+=1
print(a)
|
s303388198
|
Accepted
| 28
| 9,012
| 124
|
n = int(input())
s = list(map(int,input().split()))
a = 0
for i in range(n):
if i%2 == 0 and s[i]%2 ==1:
a+=1
print(a)
|
s877762528
|
p03779
|
u318127926
| 2,000
| 262,144
|
Wrong Answer
| 40
| 9,160
| 94
|
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
x = int(input())
for i in range(100000):
if x<i*(i+1)//2:
print(i)
exit(0)
|
s663502993
|
Accepted
| 38
| 9,156
| 95
|
x = int(input())
for i in range(100000):
if x<=i*(i+1)//2:
print(i)
exit(0)
|
s086114376
|
p03378
|
u098994567
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,172
| 705
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
n, m, x = map(int, input().split())
a = list(map(int, input().split()))
goto_goal_0 = 0
goto_goal_n = 0
for i in range(x - 1, 0, -1):
if i in a:
goto_goal_0 += 1
# print("0 is {}".format(goto_goal_0))
for i in range(x, n + 1):
if i in a:
goto_goal_n += 1
print("n is {}".format(goto_goal_n))
if goto_goal_0 < goto_goal_n:
print(goto_goal_0)
else:
print(goto_goal_n)
|
s419112556
|
Accepted
| 30
| 9,168
| 706
|
n, m, x = map(int, input().split())
a = list(map(int, input().split()))
goto_goal_0 = 0
goto_goal_n = 0
for i in range(x - 1, 0, -1):
if i in a:
goto_goal_0 += 1
# print("0 is {}".format(goto_goal_0))
for i in range(x, n + 1):
if i in a:
goto_goal_n += 1
# print("n is {}".format(goto_goal_n))
if goto_goal_0 < goto_goal_n:
print(goto_goal_0)
else:
print(goto_goal_n)
|
s064596945
|
p02402
|
u684306364
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,580
| 127
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
n = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
print("{0} {1} {2}".format(max(a), min(a), sum(a)))
|
s367970695
|
Accepted
| 40
| 8,620
| 127
|
n = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
print("{0} {1} {2}".format(min(a), max(a), sum(a)))
|
s689501269
|
p03449
|
u042338793
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,064
| 313
|
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
N = int(input())
A1 = input().split()
A2 = input().split()
max_point = 0
point = int(A1[0])
for i in range(N):
for j in range(i):
point += int(A1[j])
for j in range(N-i):
point += int(A2[i+j])
if(point > max_point):
max_point = point
point = int(A1[0])
print(max_point)
|
s908167363
|
Accepted
| 45
| 3,064
| 315
|
N = int(input())
A1 = input().split()
A2 = input().split()
max_point = 0
point = int(A1[0])
for i in range(N):
for j in range(i):
point += int(A1[1+j])
for j in range(N-i):
point += int(A2[i+j])
if(point > max_point):
max_point = point
point = int(A1[0])
print(max_point)
|
s812596877
|
p03796
|
u403385724
| 2,000
| 262,144
|
Wrong Answer
| 38
| 9,040
| 99
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
N = int(input())
power = 1
for i in range(N):
power *= i
answer = power % (10^9 +7)
print(answer)
|
s652723747
|
Accepted
| 43
| 9,088
| 94
|
N = int(input())
answer = 1
for i in range(1,N+1):
answer = answer*i%(10**9+7)
print(answer)
|
s849503855
|
p00081
|
u136916346
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,608
| 294
|
平面上の異なる 3 点 P1(x1,y1), P2(x2,y2), Q(xq,yq) の座標の組を読み込んで、点 P1 点P2 を通る直線を対称軸として点 Q と線対称の位置にある点 R(x,y) を出力するプログラムを作成してください。なお、点 Q は、その対称軸上にないものとします。
|
while 1:
try:
x1,y1,x2,y2,xq,yq=list(map(float,input().split(",")))
except:
break
X=((2*x1-xq)*((y1-y2)**2)+xq*((x1-x2)**2)+2*(y1+yq)*(x1-x2)*(y1-y2))/((x1-x2)**2+(y1-y2)**2)
try:
Y=(y1-y2)/(x1-x2)*(X+xq-2*x1)+2*y1-yq
except:
Y=-(x1-x2)/(y1-y2)*(X-xq)+yq
print(X,Y)
|
s166653161
|
Accepted
| 30
| 6,388
| 317
|
import sys
from decimal import Decimal
for l in sys.stdin:
x1,y1,x2,y2,xq,yq=list(map(Decimal,l.split(",")))
a=x1-x2
b=y1-y2
c=xq-2*x1
d=yq-2*y1
e=yq-y1
X=(xq*a**2-c*b**2+2*e*a*b)/(a**2+b**2)
try:
Y=b/a*(X+c)-d
except:
Y=-a/b*(X-xq)+yq
print(" ".join(["{:.6f}".format(i) for i in [X,Y]]))
|
s151255090
|
p03993
|
u729836751
| 2,000
| 262,144
|
Wrong Answer
| 70
| 14,008
| 161
|
There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs.
|
N = int(input())
*a, = map(int, input().split())
count = 0
for i in range(N):
# print(a[a[i]-1])
if i + 1 == a[a[i]-1]:
count += 1
print(count)
|
s148311255
|
Accepted
| 67
| 14,008
| 164
|
N = int(input())
*a, = map(int, input().split())
count = 0
for i in range(N):
# print(a[a[i]-1])
if i + 1 == a[a[i]-1]:
count += 1
print(count//2)
|
s019845581
|
p03434
|
u962127640
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 205
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n = list(map(int , input().split()))
a = list(map(int , input().split()))
a = sorted(a)
al = 0
bo = 0
for i, aa in enumerate(a):
if i %2 == 0:
al += aa
else:
bo += aa
print(al -bo)
|
s557576901
|
Accepted
| 17
| 3,064
| 219
|
n = list(map(int , input().split()))
a = list(map(int , input().split()))
a = sorted(a, reverse=True)
al = 0
bo = 0
for i, aa in enumerate(a):
if i %2 == 0:
al += aa
else:
bo += aa
print(al -bo)
|
s582066032
|
p03999
|
u674052742
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,060
| 329
|
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.
|
# -*- coding: utf-8 -*-
"""
Created on Thu Apr 23 02:13:44 2020
@author: Kanaru Sato
"""
s = input()
n = len(s)
count = 0
for state in range(2**(n-1)):
nums = s
for i in range(n-1):
if state and (1<<i):
nums = nums[:n-1-i]+"+"+nums[n-1-i:]
count += sum(map(int, nums.split("+")))
print(count)
|
s334379949
|
Accepted
| 20
| 3,060
| 327
|
# -*- coding: utf-8 -*-
"""
Created on Thu Apr 23 02:13:44 2020
@author: Kanaru Sato
"""
s = input()
n = len(s)
count = 0
for state in range(2**(n-1)):
nums = s
for i in range(n-1):
if state & (1<<i):
nums = nums[:n-1-i]+"+"+nums[n-1-i:]
count += sum(map(int, nums.split("+")))
print(count)
|
s383670523
|
p02469
|
u067972379
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 246
|
Find the least common multiple (LCM) of given n integers.
|
def gcd(a, b):
if a < b:
a, b = b, a
while b:
a, b = b, a%b
return a
def lcm(a, b):
return a * b // gcd (a, b)
n = int(input())
A = list(map(int, input().split()))
ans = 1
for a in A:
ans = (ans, a)
print(ans)
|
s850594665
|
Accepted
| 20
| 5,600
| 249
|
def gcd(a, b):
if a < b:
a, b = b, a
while b:
a, b = b, a%b
return a
def lcm(a, b):
return a * b // gcd (a, b)
n = int(input())
A = list(map(int, input().split()))
ans = 1
for a in A:
ans = lcm(ans, a)
print(ans)
|
s380219078
|
p03448
|
u680004123
| 2,000
| 262,144
|
Wrong Answer
| 164
| 4,080
| 363
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
# -*- coding:utf-8 -*-
a = int(input())
b = int(input())
c = int(input())
x = int(input())
count = 0
for i in range(0,(a+1)):
for j in range(0,(b+1)):
for k in range(0,(c+1)):
print(500*i + 100*j +50*k)
if (500*i + 100*j +50*k) == x:
count += 1
print(count)
|
s223626751
|
Accepted
| 54
| 3,064
| 316
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
count = 0
for i in range(0,a+1):
for j in range(0,b+1):
for k in range(0,c+1):
cost = 500*i + 100*j + 50*k
if cost == x:
count += 1
print(count)
|
s660527803
|
p03149
|
u871596687
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 399
|
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
n = list(map(int,input().split()))
for i in range(len(n)):
if n[i]=="1":
for j in range(n):
if n[j]=="9":
for j in range(n):
if n[i]=="7":
for j in range(n):
if n[i]=="4":
print("YES")
exit()
print("NO")
|
s996755247
|
Accepted
| 17
| 2,940
| 112
|
s = input().split()
s.sort()
if ("".join(s)=="1479"):
print("YES")
else:
print("NO")
|
s833491523
|
p03433
|
u445380615
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 94
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
en = int(input())
ichien = int(input())
dst = "YES" if en % 500 <= ichien else "NO"
print(dst)
|
s777132421
|
Accepted
| 17
| 2,940
| 95
|
en = int(input())
ichien = int(input())
dst = "Yes" if en % 500 <= ichien else "No"
print(dst)
|
s208187397
|
p03477
|
u268792407
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 115
|
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a,b,c,d=map(int,input().split())
if a+b>c+d:
print('Left')
if a+b<c+d:
print('Right')
else:
print('Balanced')
|
s821200855
|
Accepted
| 17
| 2,940
| 117
|
a,b,c,d=map(int,input().split())
if a+b>c+d:
print('Left')
elif a+b<c+d:
print('Right')
else:
print('Balanced')
|
s283094482
|
p03970
|
u477320129
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 88
|
CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.
|
A = "CODEFESTIVAL2016"
S = input()
print(sum(1 if s == a else 0 for s, a in zip(A, S)))
|
s605171489
|
Accepted
| 23
| 3,064
| 88
|
A = "CODEFESTIVAL2016"
S = input()
print(sum(0 if s == a else 1 for s, a in zip(A, S)))
|
s908318730
|
p02659
|
u822130495
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 8,992
| 144
|
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
import math
A,B = map(float,input().split())
b = str(B).split(".")[1]
a1 = A * math.floor(B)
a2 = int(A * float(b)) / 100
print(int(a1 + a2))
|
s506421408
|
Accepted
| 31
| 10,488
| 123
|
from math import floor
from fractions import Fraction
a, b = input().split()
a = int(a)
b = Fraction(b)
print(floor(a * b))
|
s083184784
|
p04029
|
u027675217
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 61
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n = int(input())
sum=0
for i in range(n):
sum+=i
print(sum)
|
s301572239
|
Accepted
| 16
| 2,940
| 64
|
n = int(input())
sum=0
for i in range(1,n+1):
sum+=i
print(sum)
|
s518290629
|
p02390
|
u001166815
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,580
| 110
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
#coding utf-8
S = int(input())
s = S % 60
m = (s // 60) % 60
h = S // 3600
print('{0}:{1}:{2}'.format(h,m,s))
|
s962833838
|
Accepted
| 20
| 5,584
| 118
|
#coding utf-8
S = int(input())
s = S % 60
m = ((S // 60) % 60)
h = (S // 60) // 60
print('{0}:{1}:{2}'.format(h,m,s))
|
s981469154
|
p03943
|
u453642820
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 83
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
A=sorted(list(map(int,input().split())))
print("Yes" if sum(A[:1])==A[2] else "No")
|
s953978002
|
Accepted
| 17
| 2,940
| 83
|
A=sorted(list(map(int,input().split())))
print("Yes" if sum(A[:2])==A[2] else "No")
|
s491519202
|
p03448
|
u869919400
| 2,000
| 262,144
|
Wrong Answer
| 49
| 3,060
| 228
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
count = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if 500 * i + 100 * j + 50 * k == x:
count += 1
|
s919018400
|
Accepted
| 53
| 3,060
| 241
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
count = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if 500 * i + 100 * j + 50 * k == x:
count += 1
print(count)
|
s311098213
|
p03469
|
u331672674
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 108
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
# -*- coding: utf-8 -*-
year, month, day = input().split("/")
print("{}/{}/{}".format("2017", month, day))
|
s802773648
|
Accepted
| 17
| 2,940
| 108
|
# -*- coding: utf-8 -*-
year, month, day = input().split("/")
print("{}/{}/{}".format("2018", month, day))
|
s231873757
|
p03400
|
u595375942
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 196
|
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
n=int(input())
D,X=map(int,input().split())
cnt=0
for _ in range(n):
A=int(input())
if A>D:cnt+=1;continue
for i in range(D):
if i%A==0 or i==1:
cnt+=1
print(cnt+X)
|
s622218900
|
Accepted
| 19
| 3,188
| 216
|
n=int(input())
D,X=map(int,input().split())
cnt=0
for _ in range(n):
A=int(input())
cnt+=1
t=1
if A>D: continue
for i in range(2,D+1):
if i%(t*A+1)==0:
cnt+=1;t+=1
print(cnt+X)
|
s977011189
|
p03090
|
u505830998
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 3,996
| 850
|
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
import sys
bbn=1000000007
#+++++
def mk_path(a_i, max_n, ret):
#a_i = 1 ... max_n
#7 => 7,(6,1),(5,2),(4,3)
#8 => (8,1),(7,2),(6,3),(5,4)
off=0 if max_n % 2 == 1 else 1
for i in range(a_i+1, max_n+1):
if i == max_n - a_i + off:
continue
else:
ret.append((a_i, i))
return
def check_constraint(graph, n):
cc=[0]*n
for f,t in graph:
cc[f-1] += t
cc[t-1] += f
pa(('max:',max(cc),', min:', min(cc) ))
return
def main():
n = int(input())
ret=[]
for i in range(1,n+1):
mk_path(i,n,ret)
#check_constraint(ret, n)
for f,t in ret:
pass
print(f,t)
#+++++
isTest=False
def pa(v):
if isTest:
print(v)
if __name__ == "__main__":
if sys.platform =='ios':
sys.stdin=open('inputFile.txt')
isTest=True
else:
pass
#input = sys.stdin.readline
ret = main()
if ret is not None:
print(ret)
|
s353189546
|
Accepted
| 24
| 3,996
| 867
|
import sys
bbn=1000000007
#+++++
def mk_path(a_i, max_n, ret):
#a_i = 1 ... max_n
#7 => 7,(6,1),(5,2),(4,3)
#8 => (8,1),(7,2),(6,3),(5,4)
off=0 if max_n % 2 == 1 else 1
for i in range(a_i+1, max_n+1):
if i == max_n - a_i + off:
continue
else:
ret.append((a_i, i))
return
def check_constraint(graph, n):
cc=[0]*n
for f,t in graph:
cc[f-1] += t
cc[t-1] += f
pa(('max:',max(cc),', min:', min(cc) ))
return
def main():
n = int(input())
ret=[]
for i in range(1,n+1):
mk_path(i,n,ret)
#check_constraint(ret, n)
print(len(ret))
for f,t in ret:
pass
print(f,t)
#+++++
isTest=False
def pa(v):
if isTest:
print(v)
if __name__ == "__main__":
if sys.platform =='ios':
sys.stdin=open('inputFile.txt')
isTest=True
else:
pass
#input = sys.stdin.readline
ret = main()
if ret is not None:
print(ret)
|
s681845198
|
p02394
|
u027634846
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,652
| 127
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
w, h, x, y, r = map(int, input().split())
xr = x + r
yr = y + r
if w >= xr and h >= yr:
print('YES')
else:
print('NO')
|
s887395255
|
Accepted
| 20
| 7,672
| 293
|
w, h, x, y, r = map(int, input().split())
xr = x + r
xl = x + r
yu = y + r
yd = y + r
if x < 0 or y < 0:
print('No')
elif xr in range(0,w+1) and xl in range(0,w+1):
if yu in range(0,h+1) and yd in range(0,h+1):
print('Yes')
else:
print('No')
else:
print('No')
|
s407170976
|
p03493
|
u743272507
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 33
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
sum(list(map(int,list(input()))))
|
s163837621
|
Accepted
| 18
| 2,940
| 40
|
print(sum(list(map(int,list(input())))))
|
s713178326
|
p03472
|
u850664243
| 2,000
| 262,144
|
Wrong Answer
| 352
| 11,316
| 430
|
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
import math
N,H = map(int,input().split())
aList = []
bList = []
for count in range(0,N):
a,b = map(int, input().split())
aList.append(a)
bList.append(b)
bList.sort()
bList.reverse()
count = 0
damage = 0
for b in bList:
damage+=b
count+=1
if damage > H:
break
if damage < H:
remainingHP = H - sum(bList)
maxA = max(aList)
count = count + math.floor(remainingHP / maxA)
print(count)
|
s869694137
|
Accepted
| 355
| 11,264
| 448
|
import math
N,H = map(int,input().split())
aList = []
bList = []
for count in range(0,N):
a,b = map(int, input().split())
aList.append(a)
bList.append(b)
bList.sort()
bList.reverse()
maxA = max(aList)
count = 0
damage = 0
for b in bList:
if maxA < b:
damage+=b
count+=1
if damage >= H:
break
if damage < H:
remainingHP = H - damage
count = count + math.ceil(remainingHP / maxA)
print(count)
|
s728931654
|
p03719
|
u084324798
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 97
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a, b, c = map(int, input().split())
if a <= c and c <= b:
print("YES")
else:
print("NO")
|
s347829629
|
Accepted
| 17
| 2,940
| 97
|
a, b, c = map(int, input().split())
if a <= c and c <= b:
print("Yes")
else:
print("No")
|
s096869360
|
p04043
|
u168416324
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,136
| 85
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
if list(map(int,input().split())).sort()==[5,5,7]:
print("YES")
else:
print("NO")
|
s018961543
|
Accepted
| 29
| 8,976
| 86
|
if sorted(list(map(int,input().split())))==[5,5,7]:
print("YES")
else:
print("NO")
|
s292134646
|
p03359
|
u239375815
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 55
|
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
m,d = map(int,input().split())
print(m if d>m else m-1)
|
s443099630
|
Accepted
| 17
| 2,940
| 56
|
m,d = map(int,input().split())
print(m if d>=m else m-1)
|
s718322582
|
p03434
|
u061545295
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 162
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
N = int(input())
cards = map(int, input().split())
sorted_cards = sorted(cards)
Alice = sum(sorted_cards[0::2])
Bob = sum(sorted_cards[1::2])
print(Alice-Bob)
|
s284379403
|
Accepted
| 17
| 2,940
| 223
|
def solve():
N = int(input())
cards = map(int, input().split())
sorted_cards = sorted(cards, reverse=True)
Alice = sum(sorted_cards[0::2])
Bob = sum(sorted_cards[1::2])
print(Alice-Bob)
solve()
|
s963269715
|
p02613
|
u882869256
| 2,000
| 1,048,576
|
Wrong Answer
| 155
| 9,208
| 265
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
a=int(input())
b=0
d=0
e=0
f=0
g=0
while b<a:
c=input()
if c=='AC':
d=d+1
if c=='WA':
e=e+1
if c=='TLE':
f=f+1
if c=='RE':
g=g+1
b=b+1
print('AC','x',d)
print('WA','x',e)
print('TLE','x',f)
print('RE','x',f)
|
s781286392
|
Accepted
| 156
| 9,204
| 264
|
a=int(input())
b=0
d=0
e=0
f=0
g=0
while b<a:
c=input()
if c=='AC':
d=d+1
if c=='WA':
e=e+1
if c=='TLE':
f=f+1
if c=='RE':
g=g+1
b=b+1
print('AC','x',d)
print('WA','x',e)
print('TLE','x',f)
print('RE','x',g)
|
s103876448
|
p03477
|
u506302470
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 131
|
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
A,B,C,D=map(int,input().split())
if(A+B<C+D):
print("Left")
if(A+B>C+D):
print("Right")
if(A+B==C+D):
print("Balanced")
|
s809023470
|
Accepted
| 17
| 3,060
| 131
|
A,B,C,D=map(int,input().split())
if(A+B>C+D):
print("Left")
if(A+B<C+D):
print("Right")
if(A+B==C+D):
print("Balanced")
|
s492190435
|
p02612
|
u711909813
| 2,000
| 1,048,576
|
Wrong Answer
| 37
| 9,140
| 50
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
m = n // 1000
print(n - m * 1000)
|
s712721766
|
Accepted
| 32
| 9,052
| 159
|
n = int(input())
if n <= 1000:
print(1000 - n)
else:
if n % 1000 == 0:
print(0)
else:
m = n // 1000 + 1
print(m * 1000 - n)
|
s949458788
|
p03815
|
u085334230
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 147
|
Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7. Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation: * Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward. For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure. If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward. Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back. Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.
|
x = int(input())
print(x % 11)
if x % 11 == 0:
print(x // 11 * 2)
elif x % 11 <= 6:
print(x // 11 * 2 + 1)
else:
print(x // 11 * 2 + 2)
|
s465669520
|
Accepted
| 17
| 2,940
| 133
|
x = int(input())
if x % 11 == 0:
print(x // 11 * 2)
elif x % 11 <= 6:
print(x // 11 * 2 + 1)
else:
print(x // 11 * 2 + 2)
|
s272077015
|
p03140
|
u091051505
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,088
| 216
|
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
n = int(input())
a = input()
b = input()
c = input()
ans = 0
for i in range(n):
if (a[i] == b[i]) & (b[i] == c[i]):
continue
if (a[i] == b[i]) | (b[i] == c[i]):
ans += 1
continue
ans += 2
print(ans)
|
s698825790
|
Accepted
| 25
| 9,104
| 234
|
n = int(input())
a = input()
b = input()
c = input()
ans = 0
for i in range(n):
if (a[i] == b[i]) & (b[i] == c[i]):
continue
if (a[i] == b[i]) | (b[i] == c[i]) | (c[i] == a[i]):
ans += 1
continue
ans += 2
print(ans)
|
s870615698
|
p03624
|
u498401785
| 2,000
| 262,144
|
Wrong Answer
| 405
| 3,188
| 304
|
You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.
|
alp = "a b c d e f g h i j k l m n o p q r s t u v w x y z"
alp = list(map(str, alp.split()))
s = input()
for i in range(len(s)):
for j in range(26):
if(s[i] == alp[j]):
alp[j] = "?"
while("?" in alp):
alp.remove("?")
if(len(alp) == 0):
print("None")
else:
print(alp)
|
s546732574
|
Accepted
| 399
| 3,188
| 306
|
alp = "a b c d e f g h i j k l m n o p q r s t u v w x y z"
alp = list(map(str, alp.split()))
s = input()
for i in range(len(s)):
for j in range(26):
if(s[i] == alp[j]):
alp[j] = "?"
while("?" in alp):
alp.remove("?")
if(len(alp) == 0):
print("None")
else:
print(alp[0])
|
s393569779
|
p03023
|
u218843509
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 2,940
| 25
|
Given an integer N not less than 3, find the sum of the interior angles of a regular polygon with N sides. Print the answer in degrees, but do not print units.
|
print(180*int(input())-2)
|
s358967610
|
Accepted
| 17
| 2,940
| 27
|
print(180*(int(input())-2))
|
s577012086
|
p03695
|
u228759454
| 2,000
| 262,144
|
Wrong Answer
| 149
| 14,436
| 360
|
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
import numpy as np
N = int(input())
a_list = list(map(int, input().split()))
class_list = []
free_list = []
for a in a_list:
if a < 3200:
amod = a // 400
class_list.append(amod)
else:
free_list.append(a)
print(class_list)
print(free_list)
set_list = list(set(class_list))
print(len(set_list), len(set_list) + len(free_list))
|
s989183246
|
Accepted
| 152
| 12,484
| 396
|
import numpy as np
N = int(input())
a_list = list(map(int, input().split()))
class_list = []
free_list = []
for a in a_list:
if a < 3200:
amod = a // 400
class_list.append(amod)
else:
free_list.append(a)
set_list = list(set(class_list))
min_list = len(set_list)
max_list = len(set_list) + len(free_list)
if min_list == 0:
min_list = 1
print(min_list, max_list)
|
s681560383
|
p03477
|
u017415492
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 122
|
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a,b,c,d=map(int,input().split())
if (a+b)>(c+d):
print("Right")
elif a+b==c+d:
print("Balanced")
else:
print("Left")
|
s908123223
|
Accepted
| 18
| 3,060
| 122
|
a,b,c,d=map(int,input().split())
if (a+b)<(c+d):
print("Right")
elif a+b==c+d:
print("Balanced")
else:
print("Left")
|
s431968725
|
p03224
|
u777923818
| 2,000
| 1,048,576
|
Wrong Answer
| 140
| 16,556
| 611
|
You are given an integer N. Determine if there exists a tuple of subsets of \\{1,2,...N\\}, (S_1,S_2,...,S_k), that satisfies the following conditions: * Each of the integers 1,2,...,N is contained in exactly two of the sets S_1,S_2,...,S_k. * Any two of the sets S_1,S_2,...,S_k have exactly one element in common. If such a tuple exists, construct one such tuple.
|
# -*- coding: utf-8 -*-
OK = []
i = 2
while True:
OK.append(i*(i+1)//2)
if OK[-1] > 10**5:
break
i += 1
N = int(input())
if N in OK:
print("Yes")
W = OK.index(N) + 2
S = [[] for _ in range(W+1)]
j = 0
L = list(range(1, N+1))
l = 1
for i, w in enumerate(range(W, 0, -1)):
for s in range(l, l+w):
S[i].append(s)
l = l+w
l = 1
for i, w in enumerate(range(W, 0, -1)):
for j, s in enumerate(range(l, l+w)):
S[i+1+j].append(s)
l = l+w
for s in S:
print(len(s), *s)
else:
print("No")
|
s557069551
|
Accepted
| 151
| 16,556
| 624
|
# -*- coding: utf-8 -*-
OK = []
i = 1
while True:
OK.append(i*(i+1)//2)
if OK[-1] > 10**5:
break
i += 1
N = int(input())
if N in OK:
print("Yes")
W = OK.index(N) + 1
S = [[] for _ in range(W+1)]
j = 0
L = list(range(1, N+1))
l = 1
for i, w in enumerate(range(W, 0, -1)):
for s in range(l, l+w):
S[i].append(s)
l = l+w
l = 1
for i, w in enumerate(range(W, 0, -1)):
for j, s in enumerate(range(l, l+w)):
S[i+1+j].append(s)
l = l+w
print(len(S))
for s in S:
print(len(s), *s)
else:
print("No")
|
s297486251
|
p02807
|
u984276646
| 2,525
| 1,048,576
|
Wrong Answer
| 2,340
| 14,484
| 420
|
There are N slimes standing on a number line. The i-th slime from the left is at position x_i. It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}. Niwango will perform N-1 operations. The i-th operation consists of the following procedures: * Choose an integer k between 1 and N-i (inclusive) with equal probability. * Move the k-th slime from the left, to the position of the neighboring slime to the right. * Fuse the two slimes at the same position into one slime. Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
|
N = int(input())
A = list(map(int, input().split()))
mod = 1e+9 + 7
p = mod - 2
S = []
while p != 0:
S = [p%2] + S[:]
p //= 2
frac = 1
for i in range(N - 1):
frac *= i+1
frac %= mod
T = 0
for i in range(N - 1):
k = 1
for j in range(len(S)):
if S[j] == 1:
k *= i+1
k %= mod
if j != len(S) - 1:
k *= k
k %= mod
T += (frac * k * (A[N - 1] - A[i])) % mod
T %= mod
print(T%mod)
|
s665041538
|
Accepted
| 541
| 14,484
| 406
|
N = int(input())
A = list(map(int, input().split()))
mod = int(1e+9 + 7)
def inved(a):
x, y, u, v, k, l = 1, 0, 0, 1, a, mod
while l != 0:
x, y, u, v = u, v, x - u * (k // l), y - v * (k // l)
k, l = l, k % l
return x
frac = 1
for i in range(N - 1):
frac *= i+1
frac %= mod
T = 0
for i in range(N - 1):
k = inved(i+1)
T += (k * (A[N - 1] - A[i])) % mod
T %= mod
print((T*frac)%mod)
|
s435751128
|
p03760
|
u209918867
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 121
|
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
odd=input()
even=input()
passwd=''
for i in range(len(odd)):
passwd+=odd[i]
if len(even)-2 >= i:
passwd+=even[i]
|
s582718059
|
Accepted
| 18
| 2,940
| 103
|
o=input()
e=input()
s=''
for i in range(len(e)):
s+=o[i]+e[i]
print(s if len(o)==len(e) else s+o[-1])
|
s612689196
|
p02646
|
u871841829
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 9,160
| 149
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if A + (V*T) >= B + (W*T):
print("Yes")
else:
print("No")
|
s763559490
|
Accepted
| 23
| 9,016
| 217
|
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if V<=W:
print("NO")
import sys
sys.exit()
# import math
if abs(B-A)/(V-W) <= T:
print("YES")
else:
print("NO")
|
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