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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s107289826
|
p02853
|
u209631375
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 837
|
We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.
|
input_number = input()
input_list = input_number.split()
coding_contest_rank = int(input_list[0])
robot_maneuver_rank = int(input_list[1])
if coding_contest_rank == 1:
coding_contest_money = 300000
if robot_maneuver_rank == 1:
robot_maneuver_money = 300000
if coding_contest_rank == 2:
coding_contest_money = 200000
if robot_maneuver_rank == 2:
robot_maneuver_money = 200000
if coding_contest_rank == 3:
coding_contest_money = 100000
if robot_maneuver_rank == 3:
robot_maneuver_money = 100000
if coding_contest_rank > 3:
coding_contest_money = 0
if robot_maneuver_rank > 3:
robot_maneuver_money = 0
elif coding_contest_rank == 1 and robot_maneuver_rank == 1:
total_money = coding_contest_money + robot_maneuver_money + 400000
print(total_money)
print(coding_contest_money + robot_maneuver_money)
|
s115009337
|
Accepted
| 17
| 3,064
| 848
|
input_number = input()
input_list = input_number.split()
coding_contest_rank = int(input_list[0])
robot_maneuver_rank = int(input_list[1])
if coding_contest_rank == 1:
coding_contest_money = 300000
if robot_maneuver_rank == 1:
robot_maneuver_money = 300000
if coding_contest_rank == 2:
coding_contest_money = 200000
if robot_maneuver_rank == 2:
robot_maneuver_money = 200000
if coding_contest_rank == 3:
coding_contest_money = 100000
if robot_maneuver_rank == 3:
robot_maneuver_money = 100000
if coding_contest_rank > 3:
coding_contest_money = 0
if robot_maneuver_rank > 3:
robot_maneuver_money = 0
if coding_contest_rank == 1 and robot_maneuver_rank == 1:
total_money = coding_contest_money + robot_maneuver_money + 400000
print(total_money)
else:
print(coding_contest_money + robot_maneuver_money)
|
s774425603
|
p03023
|
u225388820
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 122
|
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.
|
s=input()
b=len(s)
a=7-b
for i in range(b):
if s[i]=="o":
a+=1
if a>=0:
print('YES')
else:
print('NO')
|
s703822477
|
Accepted
| 17
| 2,940
| 27
|
print(180*(int(input())-2))
|
s288057261
|
p02398
|
u956226421
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,352
| 133
|
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
|
I = input().split()
a = I[0]
b = I[1]
c = I[2]
cnt = 0
while a <= b:
if c % a == 0:
cnt += 1
a += 1
print(str(cnt))
|
s924487049
|
Accepted
| 30
| 7,708
| 148
|
I = input().split()
a = int(I[0])
b = int(I[1])
c = int(I[2])
cnt = 0
while a <= b:
if c % a == 0:
cnt += 1
a += 1
print(str(cnt))
|
s606201478
|
p02612
|
u127025777
| 2,000
| 1,048,576
|
Wrong Answer
| 35
| 9,076
| 30
|
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.
|
a = int(input())
print(a%1000)
|
s230270794
|
Accepted
| 29
| 9,152
| 83
|
a = int(input())
ans = a % 1000
if ans == 0 :
print(0)
else :
print(1000 - ans)
|
s450642370
|
p00049
|
u024715419
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,568
| 305
|
ある学級の生徒の出席番号と ABO 血液型を保存したデータを読み込んで、おのおのの血液型の人数を出力するプログラムを作成してください。なお、ABO 血液型には、A 型、B 型、AB 型、O 型の４種類の血液型があります。
|
a = 0
b = 0
o = 0
ab = 0
while True:
try:
n, bt = input().split(",")
if bt =="A":
a += 1
elif bt =="B":
b += 1
elif bt =="O":
o += 1
elif bt =="AB":
ab += 1
except:
break
print(a, b, o, ab, sep="\n")
|
s869883679
|
Accepted
| 20
| 5,560
| 305
|
a = 0
b = 0
o = 0
ab = 0
while True:
try:
n, bt = input().split(",")
if bt =="A":
a += 1
elif bt =="B":
b += 1
elif bt =="O":
o += 1
elif bt =="AB":
ab += 1
except:
break
print(a, b, ab, o, sep="\n")
|
s312183658
|
p02380
|
u814278309
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,724
| 200
|
For given two sides of a triangle _a_ and _b_ and the angle _C_ between them, calculate the following properties: * S: Area of the triangle * L: The length of the circumference of the triangle * h: The height of the triangle with side a as a bottom edge
|
import math
pi=math.pi
a,b,C=map(int, input().split())
S=(1/2)*a*b*math.sin((pi/180)*C)
h=math.sqrt(a**2+b**2-2*a*b*math.cos((pi/180)*C))
L=a+b+h
print(f"{S:.8f}")
print(f"{L:.8f}")
print(f"{h:.8f}")
|
s189283832
|
Accepted
| 30
| 5,716
| 255
|
import math
a,b,c = map(int,input().split())
th = (math.pi/180)*c
s = 1/2 * a * b * math.sin(th)
c2 = math.sqrt(a**2 + b**2 - 2*a*b*math.cos(th))
l = a + b + c2
h = 2 * s / a
print('{:.8f}'.format(s))
print('{:.8f}'.format(l))
print('{:.8f}'.format(h))
|
s176308474
|
p03494
|
u931209993
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 227
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
num = int(input())
a = [int(e) for e in input().split()]
cnt = 0
while 1:
cnt = 0
for i in range(num):
if a[i]%2 != 0:
cnt += 1
if cnt != 0:
break
for i in range(num):
a[i] = int(a[i]/2)
print(a)
|
s529537173
|
Accepted
| 19
| 3,060
| 251
|
num = int(input())
a = [int(e) for e in input().split()]
cnt = 0
cnt2 = 0
while 1:
cnt = 0
for i in range(num):
if a[i]%2 != 0:
cnt += 1
if cnt != 0:
break
for i in range(num):
a[i] = int(a[i]/2)
cnt2 += 1
print(cnt2)
|
s283974140
|
p03407
|
u374146618
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 83
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
a,b,c=map(int, input().split())
if c>=(a+b):
print("Yes")
else:
print("No")
|
s259312310
|
Accepted
| 17
| 2,940
| 83
|
a,b,c=map(int, input().split())
if c<=(a+b):
print("Yes")
else:
print("No")
|
s881911776
|
p02608
|
u693694535
| 2,000
| 1,048,576
|
Wrong Answer
| 884
| 9,276
| 231
|
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).
|
N=int(input())
Z=[0]*10005
for i in range(1,102):
for j in range(1,102):
for k in range(1,102):
n=i**2+j**2+k**2+i*j+j*k+k*i
if n<=N:
Z[n]+=1
for i in range(N+1):
print(Z[i])
|
s519013459
|
Accepted
| 871
| 9,280
| 231
|
N=int(input())
Z=[0]*10005
for i in range(1,102):
for j in range(1,102):
for k in range(1,102):
n=i**2+j**2+k**2+i*j+j*k+k*i
if n<=N:
Z[n-1]+=1
for i in range(N):
print(Z[i])
|
s673253789
|
p03815
|
u612975321
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,136
| 107
|
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())
ans = 2 * x // 11
m = x % 11
if m >= 7:
ans += 2
elif m >= 1:
ans += 1
print(ans)
|
s136253648
|
Accepted
| 30
| 9,060
| 112
|
x = int(input())
ans = x // 11
ans *= 2
m = x % 11
if m >= 7:
ans += 2
elif m >= 1:
ans += 1
print(ans)
|
s731375979
|
p02694
|
u764401543
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,228
| 155
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X = int(input())
base = 100
rate = 1.01
year = 1
while True:
base = base * rate // 1
if base > X:
print(year)
break
year += 1
|
s917500895
|
Accepted
| 23
| 9,164
| 175
|
import math
X = int(input())
base = 100
rate = 1.01
year = 1
while True:
base = math.floor(base) * rate
if base > X:
print(year)
break
year += 1
|
s802302383
|
p03997
|
u357751375
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 203
|
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.
|
a = input()
b = input()
h = input()
print((int(a) + int(b)) * int(h) / 2)
|
s172490902
|
Accepted
| 25
| 9,036
| 68
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s710309405
|
p02239
|
u007270338
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,616
| 643
|
Write a program which reads an directed graph $G = (V, E)$, and finds the shortest distance from vertex $1$ to each vertex (the number of edges in the shortest path). Vertices are identified by IDs $1, 2, ... n$.
|
#coding:utf-8
n = int(input())
Q = [1]
M = [[False for i in range(n)] for j in range(n)]
dataList = []
for i in range(n):
data = list(map(int,input().split()))
dataList.append(data)
for j in range(data[1]):
v = data[j+2] - 1
M[i][v] = True
color = ["white" for i in range(n)]
d = [0 for i in range(n)]
while Q != []:
u = Q.pop() - 1
color[u] = "gray"
data = dataList[u]
for i in range(data[1]):
v = data[i+2] - 1
if M[u][v] == True and color[v] == "white":
d[v] = d[u] + 1
Q.append(v+1)
color[u] = "black"
print(" ".join([str(num) for num in d]))
|
s907031770
|
Accepted
| 1,880
| 6,000
| 509
|
#coding: utf-8
inf = 100000000
n = int(input())
List = [0] + [list(map(int,input().split())) for i in range(n)]
d = [inf for i in range(n+1)]
color = ["white" for i in range(n+1)]
s = 1
Q = [s]
d[s] = 0
color[s] = "gray"
while Q != []:
q = Q.pop(0)
k = List[q][1]
for i in List[q][2:2+k]:
if color[i] == "white":
Q.append(i)
d[i] = min(d[q] + 1, d[i])
color[q] = "gray"
for i, dis in enumerate(d[1:], 1):
if dis == inf:
dis = -1
print(i, dis)
|
s829801580
|
p03251
|
u785578220
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 219
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
x = max(b)
y = min(c)
p = max(a[2],x)
q = min(a[3],y)
if q-p > 0:
print("No war")
else:
print("War")
|
s155885702
|
Accepted
| 18
| 3,064
| 219
|
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
x = max(b)
y = min(c)
p = max(a[2],x)
q = min(a[3],y)
if q-p > 0:
print("No War")
else:
print("War")
|
s422919218
|
p02612
|
u095403885
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,148
| 88
|
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())
if n % 1000 == 0:
print(0)
else:
s = n // 1000
print(s + 1)
|
s464400980
|
Accepted
| 30
| 9,152
| 97
|
n = int(input())
if n % 1000 == 0:
print(0)
else:
s = n // 1000 + 1
print(1000*s - n)
|
s024881644
|
p03721
|
u845620905
| 2,000
| 262,144
|
Wrong Answer
| 477
| 20,804
| 242
|
There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3.
|
n, k = map(int, input().split())
d = [[0] * 2 for i in range(n)]
for i in range(n):
d[i][0], d[i][1] = map(int, input().split())
d = sorted(d)
s = 0
ans = 0
for a, b in d:
s += b
if (s > k):
break
ans = a
print(ans)
|
s304434444
|
Accepted
| 495
| 20,804
| 244
|
n, k = map(int, input().split())
d = [[0] * 2 for i in range(n)]
for i in range(n):
d[i][0], d[i][1] = map(int, input().split())
d = sorted(d)
s = 0
ans = 0
for a, b in d:
s += b
ans = a
if (s >= k):
break
print(ans)
|
s535175289
|
p04029
|
u650245944
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 639
|
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?
|
s = input()
tc = [ chr(i)*2 for i in range(ord('a'), ord('z')+1) ] + [ chr(i)+chr(j)+chr(i) for i in range(ord('a'), ord('z')+1) for j in range(ord('a'), ord('z')+1)]
if len(s) == 2:
if s[0] == s[1]:
print(1, 2)
else:
print(-1, -1)
else:
for i in tc:
if i in s:
t = i
break
else:
t = 0
if t == 0:
print(-1, -1)
else:
for i in range(len(s)-len(t)+1):
if s[i:i+len(t)] == t:
if len(t) == 2:
if i+len(t)+1 > len(s):
print(i, i+len(t))
else:
print(i+1, i+len(t)+1)
elif len(t) == 3:
print(i+1, i+len(t))
|
s669419894
|
Accepted
| 17
| 2,940
| 34
|
N = int(input())
print((N+1)*N//2)
|
s530285778
|
p03495
|
u489108157
| 2,000
| 262,144
|
Wrong Answer
| 182
| 33,312
| 275
|
Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them.
|
n,k=map(lambda x: int(x), input().split())
a=list(map(lambda x: int(x), input().split()))
dic={}
for i in range(n):
dic[a[i]]=dic.get(a[i],0)+1
count=list(dic.values())
count.sort(reverse=True)
print(count)
if len(count) <= k:
print(0)
else:
print(sum(count[k:]))
|
s523002371
|
Accepted
| 171
| 32,184
| 262
|
n,k=map(lambda x: int(x), input().split())
a=list(map(lambda x: int(x), input().split()))
dic={}
for i in range(n):
dic[a[i]]=dic.get(a[i],0)+1
count=list(dic.values())
count.sort(reverse=True)
if len(count) <= k:
print(0)
else:
print(sum(count[k:]))
|
s816582262
|
p02398
|
u825994660
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,532
| 117
|
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
|
a = list(map(int, input().split()))
cnt = 0
for i in range(a[0], a[1] + 1, 1):
if a[2] % i == 0:
cnt += 1
|
s094004617
|
Accepted
| 50
| 7,700
| 134
|
z = input()
a = list(map(int, z.split()))
cnt = 0
for i in range(a[0], a[1] + 1, 1):
if a[2] % i == 0:
cnt += 1
print(cnt)
|
s341765864
|
p03592
|
u747703115
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 14
|
We have a grid with N rows and M columns of squares. Initially, all the squares are white. There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted. Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.
|
print('Later')
|
s027241562
|
Accepted
| 236
| 9,096
| 177
|
n, m, k = map(int, input().split())
ans = 'No'
for i in range(n+1):
for j in range(m+1):
if i*(m-j)+j*(n-i)==k:
ans = 'Yes'
break
print(ans)
|
s079951910
|
p03494
|
u565204025
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 3,060
| 379
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
# -*- coding: utf-8 -*-
n = int(input())
a = list(map(int,input().split()))
def check(ar):
b = True
for i in range(len(ar)):
if ar[i] % 2 != 0:
b = False
break
return b
answer = 0
for i in range(1000000000):
if check(a) == True:
for j in range(len(a)):
a[j] = int(a[j]/2)
answer += 1
print(answer)
|
s086099219
|
Accepted
| 19
| 3,060
| 341
|
# -*- coding: utf-8 -*-
n = int(input())
a = list(map(int,input().split()))
flag = False
cnt = 0
for i in range(10000000000):
for j in range(n):
if a[j] % 2 == 0:
a[j] = int(a[j] / 2)
else:
flag = True
break
if flag == True:
break
else:
cnt += 1
print(cnt)
|
s751440720
|
p02608
|
u629540524
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,206
| 8,884
| 275
|
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).
|
n = 10**4
k = [0]*10**4
for x in range(1,10):
for y in range(1,10):
for z in range(1,10):
for i in range(n+1):
if x**2 + y** 2 + z**2 + x*y + y*z + z*x == i:
k[i-1] += 1
for i in range(int(input())):
print(k[i])
|
s606735026
|
Accepted
| 501
| 9,436
| 249
|
n = 10**4
ans = [0]*10**4
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
a = (x+y+z)**2 -(x*y+y*z+z*x)
if a <= n:
ans[a-1] += 1
for i in range(int(input())):
print(ans[i])
|
s288079709
|
p00036
|
u766477342
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,460
| 906
|
縦 8、横 8 のマスからなる図 1 のような平面があります。 □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ 図１ --- この平面上に、以下の A から G の図形のどれかが一つだけ置かれています。 | A --- ■| ■| | ---|---|---|--- ■| ■| | | | | | | | | B --- | ■| | ---|---|---|--- | ■| | | ■| | | ■| | | C --- ■| ■| ■| ■ ---|---|---|--- | | | | | | | | | | D --- | ■| | ---|---|---|--- ■| ■| | ■| | | | | | | E --- ■| ■| | ---|---|---|--- | ■| ■| | | | | | | | F --- ■| | | ---|---|---|--- ■| ■| | | ■| | | | | | G --- | ■| ■| ---|---|---|--- ■| ■| | | | | | | | たとえば、次の図 2 の例では E の図形が置かれています。 | □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| ■| ■| □| □| □| □| □ □| □| ■| ■| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ 図２ --- 平面の中で図形が占めているマスを 1、占めていないマスを 0 で表現した数字の列を読み込んで、置かれている図形の種類（A〜G）を出力するプログラムを作成してください。 ただし、ひとつの平面に置かれている図形は必ず１つで、複数の図形が置かれていることはありません。また、A〜G で表される図形以外のものが置かれていることはありません。
|
maxl = 8
offsetV = ('A', 'B', 'C', 'D', 'E', 'F', 'G')
offset = (((0, 1), (1, 0), (1, 1)),
((0, 1), (0, 2), (0, 3)),
((1, 0), (2, 0), (3, 0)),
((0, 1), (-1, 1), (-1, 2)),
((1, 0), (1, 1), (2, 1)),
((0, 1), (1, 1), (1, 2)),
((1, 0), (-1, 1), (0, 1)),
)
def match(l, x, y, oi):
for offX, offY in [a for a in offset[oi]]:
if 0 <= x + offX <= maxl and 0 <= y + offY < maxl:
pass
else:
return False
if l[y + offY][x + offX] == '0':
return False
return True
def solve():
l = [input() for i in range(9)]
for y in range(7):
for x in range(7):
if l[y][x] == '1':
for o in range(7):
if match(l, x, y, o):
return offsetV[o]
try:
while 1:
print(solve())
except:
pass
|
s383600691
|
Accepted
| 40
| 7,512
| 924
|
maxl = 8
offsetV = ('A', 'B', 'C', 'D', 'E', 'F', 'G')
offset = (((0, 1), (1, 0), (1, 1)),
((0, 1), (0, 2), (0, 3)),
((1, 0), (2, 0), (3, 0)),
((0, 1), (-1, 1), (-1, 2)),
((1, 0), (1, 1), (2, 1)),
((0, 1), (1, 1), (1, 2)),
((1, 0), (-1, 1), (0, 1)),
)
def match(l, x, y, oi):
for offX, offY in [a for a in offset[oi]]:
if 0 <= x + offX < maxl and 0 <= y + offY < maxl:
pass
else:
return False
if l[y + offY][x + offX] == '0':
return False
return True
def solve():
l = [input() for i in range(8)]
for y in range(8):
for x in range(8):
if l[y][x] == '1':
for o in range(7):
if match(l, x, y, o):
return offsetV[o]
try:
while 1:
print(solve())
l = input()
except:
pass
|
s996041397
|
p03408
|
u558242240
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,316
| 219
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
n = int(input())
from collections import defaultdict
s = defaultdict(int)
for i in range(n):
si = input()
s[si] += 1
m = int(input())
for i in range(m):
ti = input()
s[ti] -= 1
print(max(s[max(s)], 0))
|
s299998011
|
Accepted
| 21
| 3,316
| 225
|
n = int(input())
from collections import defaultdict
s = defaultdict(int)
for i in range(n):
si = input()
s[si] += 1
m = int(input())
for i in range(m):
ti = input()
s[ti] -= 1
print(max(0,max(s.values())))
|
s318656813
|
p02402
|
u586792237
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,584
| 97
|
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(input())
ns = list(map(int, input().split()))
print(n)
print(min(ns), max(ns), sum(ns))
|
s011851623
|
Accepted
| 20
| 6,584
| 89
|
n = int(input())
ns = list(map(int, input().split()))
print(min(ns), max(ns), sum(ns))
|
s878854274
|
p03573
|
u843318346
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 93
|
You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers.
|
arr = list(map(int,input().split()))
if arr[0]==arr[1]:
print(arr[0])
else:
print(arr[2])
|
s192593961
|
Accepted
| 17
| 2,940
| 67
|
a = list(map(int,input().split()))
a.sort()
print(a[0]*a[2]//a[1])
|
s172956022
|
p03944
|
u013408661
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 269
|
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
w,h,n=map(int,input().split())
w_d=0
h_d=0
for i in range(n):
x,y,a=map(int,input().split())
if a==0:
w_d=max(w_d,x)
if a==1:
w=min(w,x)
if a==2:
h_d=max(h_d,y)
if a==3:
h=min(h,y)
if w<=w_d or h<=h_d:
print(0)
else:
print((w-w_d)*(h-h_d))
|
s704668565
|
Accepted
| 18
| 3,064
| 269
|
w,h,n=map(int,input().split())
w_d=0
h_d=0
for i in range(n):
x,y,a=map(int,input().split())
if a==1:
w_d=max(w_d,x)
if a==2:
w=min(w,x)
if a==3:
h_d=max(h_d,y)
if a==4:
h=min(h,y)
if w<=w_d or h<=h_d:
print(0)
else:
print((w-w_d)*(h-h_d))
|
s661850355
|
p02612
|
u629201777
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,164
| 41
|
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.
|
x=int(input())
print((1000-x//1000)%1000)
|
s152745025
|
Accepted
| 27
| 9,132
| 41
|
x=int(input())
print((1000-x%1000)%1000)
|
s439277343
|
p03592
|
u844789719
| 2,000
| 262,144
|
Wrong Answer
| 330
| 3,060
| 189
|
We have a grid with N rows and M columns of squares. Initially, all the squares are white. There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted. Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.
|
import itertools
N, M, K = [int(_) for _ in input().split()]
for i, j in itertools.product(range(N), range(M)):
if (N-i)*j+i*(M-j) == K:
print('YES')
exit()
print('NO')
|
s243646765
|
Accepted
| 296
| 3,188
| 193
|
import itertools
N, M, K = [int(_) for _ in input().split()]
for i, j in itertools.product(range(N+1), range(M+1)):
if (N-i)*j+i*(M-j) == K:
print('Yes')
exit()
print('No')
|
s123250573
|
p03049
|
u930149040
| 2,000
| 1,048,576
|
Wrong Answer
| 37
| 3,064
| 442
|
Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string.
|
N = int(input())
ans = 0
b_x = 0
b_a = 0
x_a = 0
y = 0
for _ in range(N):
s = input()
ans += s.find('AB')
if s[0] == 'B' and s[-1] == 'A':
b_a += 1
elif s[0] == 'B':
b_x += 1
elif s[-1] == 'A':
x_a += 1
else:
y += 1
if x_a >= 1 and b_x >= 1:
ans += b_a
elif x_a >= 1 or b_x >= 1:
ans += b_a - 1
else:
ans += b_a - 2
if x_a - 1 <= 0 or b_x -1 <= 0:
ans += 0
else:
ans += min(x_a - 1, b_x - 1)
print(ans)
|
s931172457
|
Accepted
| 38
| 3,064
| 460
|
N = int(input())
ans = 0
b_x = 0
b_a = 0
x_a = 0
y = 0
for _ in range(N):
s = input()
ans += s.count('AB')
if s[0] == 'B' and s[-1] == 'A':
b_a += 1
elif s[0] == 'B':
b_x += 1
elif s[-1] == 'A':
x_a += 1
else:
y += 1
if x_a >= 1 and b_x >= 1:
ans += b_a + 1
elif x_a >= 1 or b_x >= 1:
ans += b_a
else:
if b_a > 0:
ans += b_a - 1
if x_a - 1 <= 0 or b_x -1 <= 0:
ans += 0
else:
ans += min(x_a - 1, b_x - 1)
print(ans)
|
s278259094
|
p03673
|
u210827208
| 2,000
| 262,144
|
Wrong Answer
| 2,108
| 26,020
| 171
|
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n=int(input())
A=list(map(int,input().split()))
B=[]
for i in range(1,n+1):
C=[]
B.append(A[i-1])
for i in B[::-1]:
C.append(i)
B=C
print(B)
|
s045701939
|
Accepted
| 226
| 25,412
| 357
|
from collections import deque
n=int(input())
A=list(map(int,input().split()))
B=deque([])
if n%2==0:
for i in range(n):
if i%2==0:
B.append(A[i])
else:
B.appendleft(A[i])
else:
for i in range(n):
if i%2==0:
B.appendleft(A[i])
else:
B.append(A[i])
print(*B)
|
s973924721
|
p03759
|
u509674552
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 150
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
# coding: utf-8
if __name__ == '__main__':
a,b,c = map(int,input().split())
if b-a == c-b:
print("Yes")
else:
print("No")
|
s461940871
|
Accepted
| 17
| 2,940
| 151
|
# coding: utf-8
if __name__ == '__main__':
a,b,c = map(int,input().split())
if b-a == c-b:
print("YES")
else:
print("NO")
|
s623160752
|
p03997
|
u445509700
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
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.
|
a = int(input())
b = int(input())
c = int(input())
d = 0
d = (a + b) * c / 2
print(d)
|
s835258588
|
Accepted
| 17
| 2,940
| 96
|
a = int(input())
b = int(input())
c = int(input())
d = 0
e = 0
d = a + b
e = d * c // 2
print(e)
|
s188741706
|
p03129
|
u649427151
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 197
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
n, k = map(int, input().split())
if n % 2 == 0:
if n >= k * 2:
print("Yes")
else:
print("No")
else:
if n + 1 >= k * 2:
print("Yes")
else:
print("No")
|
s177025131
|
Accepted
| 17
| 2,940
| 197
|
n, k = map(int, input().split())
if n % 2 == 0:
if n >= k * 2:
print("YES")
else:
print("NO")
else:
if n + 1 >= k * 2:
print("YES")
else:
print("NO")
|
s969528003
|
p03672
|
u623687794
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 236
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
s=input()
ans=len(s)
def check(m):
a=len(m)
if a%2==1:
return False
else:
if s[0:m//2]==s[m//2:m]:
return True
else:
return False
for i in range(1,len(s),-1):
ans-=1
if check(s[0:i]):
print(ans)
|
s133466480
|
Accepted
| 17
| 2,940
| 254
|
s=input()
ans=len(s)
def check(m):
a=len(m)
if a%2==1:
return False
else:
if s[0:a//2]==s[a//2:a]:
return True
else:
return False
for i in range(len(s)-1,0,-1):
ans-=1
if check(s[0:i])==True:
print(ans)
break
|
s508868909
|
p03160
|
u498202416
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 92,568
| 313
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
h = list(map(int,input().split()))
dp = [float('inf') for i in range(N)]
dp[0] = 0
for i in range(1,N):
print(dp)
if dp[i] > dp[i-1]+abs(h[i]-h[i-1]):
dp[i] = dp[i-1]+abs(h[i]-h[i-1])
if i > 1 and dp[i] > dp[i-2]+abs(h[i]-h[i-2]):
dp[i] = dp[i-2]+abs(h[i]-h[i-2])
print(dp[N-1])
|
s668547000
|
Accepted
| 165
| 14,104
| 270
|
import math
N = int(input())
h = list(map(int,input().split())) + [0] * 2
cost = [float("inf")] * (N+2)
cost[0] = 0
for i in range(N):
cost[i+1] = min(cost[i+1], cost[i] + abs(h[i]-h[i+1]))
cost[i+2] = min(cost[i+2], cost[i] + abs(h[i]-h[i+2]))
print(cost[N-1])
|
s875354529
|
p03795
|
u856169020
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 94
|
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.
|
N = int(input())
power = 1
for i in range(N):
power = (power*(i+1)) % (10**9+7)
print(power)
|
s322806045
|
Accepted
| 17
| 2,940
| 57
|
N = int(input())
back = (N//15) * 200
print(800*N - back)
|
s053343404
|
p03852
|
u095094246
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
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`.
|
ps='aeiuo'
c=input()
for i in range(5):
if c==ps[i]:
print('vowel')
else:
print('consonant')
|
s996262759
|
Accepted
| 17
| 2,940
| 118
|
import sys
ps='aeiuo'
c=input()
for i in range(5):
if c==ps[i]:
print('vowel')
sys.exit()
print('consonant')
|
s187061975
|
p03681
|
u479638406
| 2,000
| 262,144
|
Wrong Answer
| 24
| 3,316
| 110
|
Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys. How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.
|
n, m = map(int, input().split())
if n-m != 0 or abs(n-m) != 1:
print(0)
else:
print((2*n*m)%(int(1e9)+7))
|
s891990589
|
Accepted
| 704
| 5,312
| 217
|
import math
n, m = map(int, input().split())
o = math.factorial(n)
p = math.factorial(m)
if abs(n-m) > 1:
print(0)
elif abs(n-m) == 1:
print((o*p)%(int(1e9)+7))
elif n-m == 0:
print((2*o*p)%(int(1e9)+7))
|
s110220541
|
p03469
|
u823044869
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 73
|
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.
|
y,m,d = map(int,input().split('/'))
print("2018"+'/'+str(m)+'/'+str(d))
|
s954677566
|
Accepted
| 19
| 2,940
| 54
|
y,m,d = input().split('/')
print("2018"+'/'+m+'/'+d)
|
s300280700
|
p03493
|
u316386814
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 43
|
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.
|
a = int(input())
print(bin(a).count('1'))
|
s387663977
|
Accepted
| 18
| 2,940
| 46
|
a = int(input(), 2)
print(bin(a).count('1'))
|
s251992855
|
p03162
|
u055764432
| 2,000
| 1,048,576
|
Wrong Answer
| 1,050
| 45,416
| 310
|
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
n=int(input())
l=list()
for i in range(n):
li=[int(i) for i in input().split()]
l.append(li)
dp=[[0,0,0] for i in range(n)]
dp[0][0]=l[0][0]
dp[0][1]=l[0][1]
dp[0][2]=l[0][2]
for i in range(1,n):
for j in range(3):
for k in range(3):
dp[i][j]=max(dp[i-1][k]+l[i][k],dp[i][j])
print(max(dp[-1]))
|
s534047363
|
Accepted
| 970
| 44,128
| 334
|
n=int(input())
l=list()
for i in range(n):
li=[int(i) for i in input().split()]
l.append(li)
dp=[[0,0,0] for i in range(n)]
dp[0][0]=l[0][0]
dp[0][1]=l[0][1]
dp[0][2]=l[0][2]
for i in range(1,n):
for j in range(3):
for k in range(3):
if j!=k:
dp[i][j]=max(dp[i-1][k]+l[i][j],dp[i][j])
print(max(dp[-1]))
|
s418470222
|
p03563
|
u202619899
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.
|
R = float(input())
G = float(input())
print(2*G-R)
|
s509068161
|
Accepted
| 17
| 2,940
| 47
|
R = int(input())
G = int(input())
print(2*G-R)
|
s822833624
|
p02606
|
u273339216
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,108
| 239
|
How many multiples of d are there among the integers between L and R (inclusive)?
|
l, r, d = map(int, input().split())
res = 0
iter = 0
p = 0
if d <= l:
while res<=l:
res = d*(p+1)
p+=1
if res>=l:
iter+=1
#print(iter)
while res<=r:
res = res+ d*(iter)
if res <= r:
iter+=1
#print(res)
print(iter)
|
s551097891
|
Accepted
| 32
| 9,156
| 110
|
l, r, d = map(int, input().split())
count = 0
for i in range(l, r+1):
if i%d == 0:
count+=1
print(count)
|
s546132007
|
p02419
|
u358919705
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,296
| 184
|
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
|
W = input()
num = 0
for line in input():
if line == 'END_OF_TEXT':
break
for word in line.split():
if word.lower() == W.lower():
num += 1
print(num)
|
s019098507
|
Accepted
| 20
| 7,456
| 194
|
W = input()
num = 0
while True:
line = input()
if line == 'END_OF_TEXT':
break
for word in line.split():
if word.lower() == W.lower():
num += 1
print(num)
|
s099324361
|
p04029
|
u051237313
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,140
| 40
|
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)
|
s859868152
|
Accepted
| 25
| 8,964
| 48
|
n = int(input())
print(int((n * (n + 1)) / 2))
|
s136571514
|
p03644
|
u379692329
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
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())
result = 1
while int(result/2) < N:
result *= 2
print(result)
|
s703148071
|
Accepted
| 17
| 2,940
| 80
|
N = int(input())
result = 1
while result*2 <= N:
result *= 2
print(result)
|
s435830321
|
p02297
|
u089116225
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,724
| 975
|
For a given polygon g, computes the area of the polygon. g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n-1) are sides of g. The line segment connecting pn and p1 is also a side of the polygon. Note that the polygon is not necessarily convex.
|
import math
def vec(s,e):
return (e[0]-s[0], e[1]-s[1])
def dot(p,q):
return p[0]*p[1] + q[0]*q[1]
def is_obtuse(p,q,r):
b = vec(q,r)
a = vec(q,p)
c = (-b[1],b[0])
bc = dot(a,b)
pc = dot(c,b)
return math.atan2(pc,bc) < 0
def dist(p,q):
return ((p[0]-q[0])**2 + (p[1]-q[1])**2)**0.5
def area(p,q,r):
a = dist(p,q)
b = dist(q,r)
c = dist(r,p)
z = (a+b+c)/2
return (z*(z-a)*(z-b)*(z-c))**0.5
n = int(input())
v = []
for _ in range(n):
v.append([int(x) for x in input().split()])
sub_area = 0
v_ignore = [0 for _ in range(n)]
for i in range(n-2):
if is_obtuse(v[i],v[i+1],v[i+2]):
sub_list += area(v[i],v[i+1],v[i+2])
v_ignore[i+1] = 1
else:
pass
v_cover = []
for i,x in enumerate(v_ignore):
if x == 0:
v_cover.append(v[i])
cover_area = 0
for i in range(1,len(v_cover)-1):
cover_area += area(v_cover[0],v_cover[i],v_cover[i+1])
print(cover_area - sub_area)
|
s791162365
|
Accepted
| 20
| 5,624
| 323
|
def area(p,q,r):
a = (q[0]-p[0], q[1]-p[1])
b = (r[0]-p[0], r[1]-p[1])
return (a[0]*b[1] - b[0]*a[1]) / 2
n = int(input())
v = []
for _ in range(n):
v.append([float(x) for x in input().split()])
v = list(reversed(v))
s = 0
for i in range(1,n-1):
s += area(v[0],v[i],v[i+1])
print(-1 * round(s,1))
|
s018329606
|
p03814
|
u117629640
| 2,000
| 262,144
|
Wrong Answer
| 34
| 4,840
| 208
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
# -*- coding: utf-8 -*-
def main():
a = list(input())
i = a.index('A')
while(a.index('Z') < i):
a.pop(a.index('Z'))
print(a.index('Z') - i + 2)
if __name__ == '__main__':
main()
|
s319111024
|
Accepted
| 24
| 3,512
| 141
|
# -*- coding: utf-8 -*-
def main():
a = str(input())
print(a.rindex('Z') - a.index('A') + 1)
if __name__ == '__main__':
main()
|
s777594970
|
p02742
|
u368270116
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,164
| 87
|
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:
|
h,w=map(int,input().split())
if (h*w)%2==0:
print((h*w)/2)
else:
print(((h*w)/2)+1)
|
s476389631
|
Accepted
| 25
| 9,104
| 142
|
import math
h,w=map(int,input().split())
if h==1 or w==1:
print(1)
elif (h*w)%2==0:
print(int((h*w)/2))
else:
print(math.ceil((h*w)/2))
|
s039020618
|
p02393
|
u585035894
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,588
| 48
|
Write a program which reads three integers, and prints them in ascending order.
|
print(sorted([int(i) for i in input().split()]))
|
s638490403
|
Accepted
| 20
| 5,592
| 49
|
print(*sorted([int(i) for i in input().split()]))
|
s588963282
|
p03693
|
u672898046
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 96
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
a, b, c = map(int, input().split())
e = b*10+c
if e//4 == 0:
print("YES")
else:
print("NO")
|
s722683593
|
Accepted
| 17
| 2,940
| 101
|
a, b, c = map(int, input().split())
e = a*100+b*10+c
if e%4 == 0:
print("YES")
else:
print("NO")
|
s325792733
|
p02393
|
u698693989
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,476
| 84
|
Write a program which reads three integers, and prints them in ascending order.
|
x=input().split()
y=list(map(int,x))
a=y[0]
b=y[1]
c=y[2]
d=sorted([a,b,c])
print(d)
|
s434472518
|
Accepted
| 20
| 5,548
| 70
|
list=input().split()
list=sorted(list)
print(list[0],list[1],list[2])
|
s495529256
|
p03433
|
u662449766
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 218
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
import sys
input = sys.stdin.readline
def main():
n = int(input())
a = int(input())
n = n // 500
if n <= a:
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
|
s888552608
|
Accepted
| 17
| 2,940
| 216
|
import sys
input = sys.stdin.readline
def main():
n = int(input())
a = int(input())
n = n % 500
if n <= a:
print("Yes")
else:
print("No")
if __name__ == "__main__":
main()
|
s768433511
|
p03493
|
u240055120
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 79
|
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.
|
a = list(input())
count = 0
for i in a:
if i == 1:
count+=1
print(count)
|
s669278114
|
Accepted
| 17
| 2,940
| 92
|
a = input()
count = 0
for i in range(len(a)):
if int(a[i]) == 1:
count+=1
print(count)
|
s136886694
|
p02417
|
u789974879
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,552
| 277
|
Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.
|
text = input()
text_new = text.lower()
alphabet = ["a","b","c","d","e","f","g","h","i","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
for i in alphabet:
tmp = 0
for j in text_new:
if j == i:
tmp += 1
print(i + ":" + str(tmp))
|
s853697521
|
Accepted
| 20
| 5,560
| 369
|
text = ""
while 1:
try:
x = input()
text += x
except EOFError:
break
text_new = text.lower()
alphabet = ["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"]
for i in alphabet:
tmp = 0
for j in text_new:
if j == i:
tmp += 1
print(i + " : " + str(tmp))
|
s480410157
|
p03636
|
u851704997
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
s = input()
x = len(s)
print(s[0] + str(x) + s[-1])
|
s184552668
|
Accepted
| 17
| 2,940
| 56
|
s = input()
x = len(s) - 2
print(s[0] + str(x) + s[-1])
|
s623384573
|
p03194
|
u623819879
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 115
|
There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P. Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.
|
n,p=map(int, input().split())
c=1
for i in range(int(pow(n,1/p))):
if n % (i+1)**p == 0:
c=i+1
print(c)
|
s644964689
|
Accepted
| 49
| 3,064
| 441
|
n,p=map(int, input().split())
ans=1
m=int(pow(p,1/n))+1
if n==1:
print(p)
else:
m=int(pow(p,1/n))+1
i=2
c=0
while(i<=p and i<=m):
#print('i,p=',i,p)
if p % i == 0:
p=p/i
c=c+1
if p % i != 0:
#print('i**(c//n)=',i**(c//n))
ans = ans * ( i**(c//n) )
i=i+1
c=0
else:
i=i+1
print(ans)
|
s819606087
|
p03729
|
u075155299
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 93
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a,b,c = input().split()
if a[:-1]==b[0] and b[:-1]==c[0]:
print("YES")
else:
print("NO")
|
s477023258
|
Accepted
| 17
| 2,940
| 93
|
a,b,c = input().split()
if a[-1:]==b[0] and b[-1:]==c[0]:
print("YES")
else:
print("NO")
|
s613684556
|
p02398
|
u035064179
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,580
| 99
|
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
|
a, b, c = map(int, input().split())
for i in range(a, b + 1):
if i % c == 0:
print(i)
|
s462174626
|
Accepted
| 20
| 5,596
| 136
|
a, b, c = map(int, input().split())
count = int(0)
for i in range(a, b + 1):
if c % i == 0:
count = count + 1
print(count)
|
s848363536
|
p02613
|
u854931881
| 2,000
| 1,048,576
|
Wrong Answer
| 146
| 9,168
| 234
|
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())
c1=c2=c3=c4=0
for i in range(n):
x=input()
if x=="AC":
c1+=1
elif x=="WA":
c2+=1
elif x=="TLE":
c3+=1
elif x=="RE":
c4+=1
print("AC*"+str(c1))
print("WA*"+str(c2))
print("TLE*"+str(c3))
print("RE*"+str(c4))
|
s180887965
|
Accepted
| 150
| 8,940
| 242
|
n=int(input())
c1=c2=c3=c4=0
for i in range(n):
x=input()
if x=="AC":
c1+=1
elif x=="WA":
c2+=1
elif x=="TLE":
c3+=1
elif x=="RE":
c4+=1
print("AC x "+str(c1))
print("WA x "+str(c2))
print("TLE x "+str(c3))
print("RE x "+str(c4))
|
s293529203
|
p03813
|
u234154017
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 248
|
Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise.
|
s = list(input())
cnt = 0
n = 1
for i in range(len(s)):
if s[i] == "A":
while s[i+n] != "Z":
n += 1
try:
while s[i+n] == "Z":
n += 1
except:
pass
break
print(n)
|
s253498450
|
Accepted
| 20
| 3,316
| 71
|
x = int(input())
if x >= 1200:
print("ARC")
else:
print("ABC")
|
s493838967
|
p03167
|
u227082700
| 2,000
| 1,048,576
|
Wrong Answer
| 684
| 3,064
| 283
|
There is a grid with H horizontal rows and W vertical columns. Let (i, j) denote the square at the i-th row from the top and the j-th column from the left. For each i and j (1 \leq i \leq H, 1 \leq j \leq W), Square (i, j) is described by a character a_{i, j}. If a_{i, j} is `.`, Square (i, j) is an empty square; if a_{i, j} is `#`, Square (i, j) is a wall square. It is guaranteed that Squares (1, 1) and (H, W) are empty squares. Taro will start from Square (1, 1) and reach (H, W) by repeatedly moving right or down to an adjacent empty square. Find the number of Taro's paths from Square (1, 1) to (H, W). As the answer can be extremely large, find the count modulo 10^9 + 7.
|
h,w=map(int,input().split())
dp=[[0]*w]*h
for i in range(h):
s=input()
for j in range(w):
if s[j]==".":
if i==j==0:dp[i][j]=1
elif j==0:dp[i][j]=dp[i-1][j]
elif i==0:dp[i][j]=dp[i][j-1]
else:dp[i][j]=(dp[i-1][j]+dp[i][j-1])%(10**9+7)
print(dp[-1][-1])
|
s779216606
|
Accepted
| 772
| 43,476
| 314
|
h,w=map(int,input().split())
dp=[[0for i in range(w)] for i in range(h)]
for i in range(h):
s=input()
for j in range(w):
if s[j]==".":
if i==j==0:dp[i][j]=1
elif j==0:dp[i][j]=dp[i-1][j]
elif i==0:dp[i][j]=dp[i][j-1]
else:dp[i][j]=(dp[i-1][j]+dp[i][j-1])%(10**9+7)
print(dp[-1][-1])
|
s069706010
|
p03997
|
u856169020
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
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.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s341298461
|
Accepted
| 17
| 2,940
| 68
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s877729298
|
p03477
|
u482078166
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 274
|
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.
|
s=input()
z="0"
while z in s:
if z in s:
z+="0"
else:
break
o="1"
while o in s:
if o in s:
o+="1"
else:
break
if len(z) == 2 and len(o) == 2:
print(2)
elif len(z) >= len(o):
print(len(z)-1)
else:
print(len(o))
|
s698596262
|
Accepted
| 17
| 3,060
| 144
|
w=list(map(int,input().split(" ")))
l=w[0]+w[1]
r=w[2]+w[3]
if l>r:
print("Left")
elif l==r:
print("Balanced")
else:
print("Right")
|
s340438534
|
p03338
|
u272557899
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 3,064
| 468
|
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
n = int(input())
s = input()
t = list(s)
s1 = []
s2 = []
for i in range(n):
s1.append([])
s2.append([])
for i in range(n):
for j in range(n):
if j < i:
s1[i].append(s[i])
else:
s2[i].append(s[i])
count = 0
com = []
slv = []
for i in range(n):
for j in range(i):
if s1[i][j] in s2[i] and s1[i][j] not in com:
count += 1
com.append(s1[i][j])
slv.append(count)
count = 0
com = []
print(max(slv))
|
s447286782
|
Accepted
| 25
| 3,064
| 469
|
n = int(input())
s = input()
t = list(s)
s1 = []
s2 = []
for i in range(n):
s1.append([])
s2.append([])
for i in range(n):
for j in range(n):
if j < i:
s1[i].append(t[j])
else:
s2[i].append(t[j])
count = 0
com = []
slv = []
for i in range(n):
for j in range(i):
if s1[i][j] in s2[i] and s1[i][j] not in com:
count += 1
com.append(s1[i][j])
slv.append(count)
count = 0
com = []
print(max(slv))
|
s603305568
|
p03369
|
u634384370
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 120
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s = str(input())
y = 700
if s[0] == "○":
y += 100
if s[1] == "○":
y += 100
if s[2] == "○":
y += 100
print(y)
|
s531275495
|
Accepted
| 17
| 2,940
| 114
|
s = str(input())
y = 700
if s[0] == "o":
y += 100
if s[1] == "o":
y += 100
if s[2] == "o":
y += 100
print(y)
|
s618617865
|
p03575
|
u625963200
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 1,107
|
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
class UnionFind():
def __init__(self,n):
self.n=n
self.parents=[-1]*n
def find(self,x):
if self.parents[x]<0:
return x
else:
self.parents[x]=self.find(self.parents[x])
return self.parents[x]
def union(self,x,y):
x=self.find(x)
y=self.find(y)
if x==y:
return
if self.parents[x]>self.parents[y]:
x,y=y,x
self.parents[x]+=self.parents[y]
self.parents[y]=x
def same(self,x,y):
return self.find(x)==self.find(y)
def members(self,x):
root=self.find(x)
return [i for i in range(self.n) if self.find(i)==root]
def roots(self):
return [i for i,x in enumerate(self.parents) if x<0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()}
def __str__(self):
return '\n'.join('{}: {}'.format(r,self.members(r)) for r in self.roots())
n,m=map(int,input().split())
a,b=[0]*m,[0]*m
res=0
for i in range(m):
UF=UnionFind(n)
for j in range(m):
if j!=i:
UF.union(a[j],b[j])
if UF.group_count()==2:
ret+=1
print(res)
|
s968757161
|
Accepted
| 29
| 3,064
| 440
|
n,m=map(int,input().split())
AB=[list(map(int,input().split())) for _ in range(m)]
def dfs(x):
if visit[x]==True:
return
visit[x]=True
for i in range(n):
if graph[x][i]==True:
dfs(i)
ans=0
for i in range(m):
graph=[[False]*n for _ in range(n)]
for j in range(m):
if j!=i:
a,b=AB[j]
graph[a-1][b-1]=True
graph[b-1][a-1]=True
visit=[False]*n
dfs(0)
if sum(visit)!=n:
ans+=1
print(ans)
|
s749151951
|
p02467
|
u548155360
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,748
| 1,031
|
Factorize a given integer n.
|
# coding=utf-8
import array
import math
def sieve_of_eratosthenes(end):
# noinspection PyUnusedLocal
is_prime = array.array('B', (True for i in range(end+1)))
is_prime[0] = False
is_prime[1] = False
primes = array.array("L")
for i in range(2, end+1):
if is_prime[i]:
primes.append(i)
for j in range(i * 2, end+1, i):
is_prime[j] = False
return primes
if __name__ == '__main__':
N = int(input())
n = N
upper_n = int(math.sqrt(n)) + 1
print(upper_n)
prime_list = sieve_of_eratosthenes(upper_n)
print(prime_list)
elementary_number_list = array.array("L")
counter = 0
while counter < len(prime_list):
if n % prime_list[counter] == 0:
elementary_number_list.append(prime_list[counter])
n //= prime_list[counter]
else:
counter += 1
print("{0}: {1}".format(N, ' '.join(map(str, elementary_number_list))))
|
s205189996
|
Accepted
| 20
| 5,788
| 1,092
|
# coding=utf-8
import array
import math
def sieve_of_eratosthenes(end):
# noinspection PyUnusedLocal
is_prime = array.array('B', (True for i in range(end+1)))
is_prime[0] = False
is_prime[1] = False
primes = array.array("L")
for i in range(2, end+1):
if is_prime[i]:
primes.append(i)
for j in range(i * 2, end+1, i):
is_prime[j] = False
return primes
if __name__ == '__main__':
N = int(input())
n = N
upper_n = int(math.sqrt(n)) + 1
# print(upper_n)
prime_list = sieve_of_eratosthenes(upper_n)
elementary_number_list = array.array("L")
counter = 0
while counter < len(prime_list):
if n % prime_list[counter] == 0:
elementary_number_list.append(prime_list[counter])
n //= prime_list[counter]
else:
counter += 1
if n != 1:
elementary_number_list.append(n)
print("{0}: {1}".format(N, ' '.join(map(str, elementary_number_list))))
|
s893182041
|
p02420
|
u299798926
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,740
| 326
|
Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).
|
while 1:
n=input()
if n=='-':
break
m=[i for i in range(len(n))]
count=int(input())
for i in range(count):
l=int(input())
for j in range(len(n)):
m[j]=n[(l+j)%len(n)]
n=m
print(n)
for i in range(len(n)):
print(n[i],end="")
print()
|
s484495030
|
Accepted
| 20
| 7,744
| 313
|
while 1:
n=input()
if n=='-':
break
m=[i for i in range(len(n))]
count=int(input())
for i in range(count):
l=int(input())
for j in range(len(n)):
m[j]=n[(l+j)%len(n)]
n=m.copy()
for i in range(len(n)):
print(n[i],end="")
print()
|
s976572902
|
p02612
|
u301812308
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,140
| 38
|
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.
|
price = int(input())
print(price%1000)
|
s721237184
|
Accepted
| 26
| 9,184
| 96
|
price = int(input())
change = 1000 - price % 1000
if change == 1000:
change = 0
print(change)
|
s440681802
|
p02796
|
u346308892
| 2,000
| 1,048,576
|
Wrong Answer
| 711
| 48,180
| 362
|
In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.
|
def acinput():
return list(map(int, input().split(" ")))
N=int(input())
bars=[]
for i in range(N):
tmp = acinput()
tmp.extend([tmp[0]-tmp[1],tmp[0]+tmp[1]])
bars.append(tmp)
bars=sorted(bars,key=lambda x:x[3])
print(bars)
count=0
R=-10000000000
for i in range(0,N):
if R<=bars[i][2]:
R=bars[i][3]
count+=1
print(count)
|
s915385247
|
Accepted
| 604
| 36,264
| 363
|
def acinput():
return list(map(int, input().split(" ")))
N=int(input())
bars=[]
for i in range(N):
tmp = acinput()
tmp.extend([tmp[0]-tmp[1],tmp[0]+tmp[1]])
bars.append(tmp)
bars=sorted(bars,key=lambda x:x[3])
#print(bars)
count=0
R=-10000000000
for i in range(0,N):
if R<=bars[i][2]:
R=bars[i][3]
count+=1
print(count)
|
s891111240
|
p03719
|
u584459098
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 92
|
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 C>=A and C<=B:
print('yes')
else:
print('no')
|
s908156431
|
Accepted
| 18
| 2,940
| 92
|
A, B, C = map(int, input().split())
if C>=A and C<=B:
print('Yes')
else:
print('No')
|
s587835124
|
p02854
|
u711238850
| 2,000
| 1,048,576
|
Wrong Answer
| 206
| 41,612
| 469
|
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length.
|
import math
from collections import deque
def main():
n = int(input())
a = [int(t)for t in input().split()]
a_rev = a[::-1]
cusum = [a[0]]
rcusum = [a_rev[0]]
for i in range(1,n):
cusum.append(cusum[i-1]+a[i])
rcusum.append(rcusum[i-1]+a[i])
rcusum = rcusum[::-1]
result = []
for i in range(n):
result.append(abs(cusum[i]-rcusum[i]))
print(min(result))
if __name__ == "__main__":
main()
|
s008235164
|
Accepted
| 202
| 41,612
| 473
|
import math
from collections import deque
def main():
n = int(input())
a = [int(t)for t in input().split()]
a_rev = a[::-1]
cusum = [a[0]]
rcusum = [a_rev[0]]
for i in range(1,n):
cusum.append(cusum[i-1]+a[i])
rcusum.append(rcusum[i-1]+a_rev[i])
rcusum = rcusum[::-1]
result = []
for i in range(1,n):
result.append(abs(cusum[i-1]-rcusum[i]))
print(min(result))
if __name__ == "__main__":
main()
|
s595480526
|
p03719
|
u788023488
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
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())
ans='NO'
if a<=c<=b:
ans='YES'
print(ans)
|
s089399565
|
Accepted
| 17
| 2,940
| 77
|
a,b,c = map(int, input().split())
ans='No'
if a<=c<=b:
ans='Yes'
print(ans)
|
s754150522
|
p02678
|
u469936642
| 2,000
| 1,048,576
|
Wrong Answer
| 739
| 39,084
| 554
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
from collections import defaultdict as d, deque
def bfs(adj, start, n):
visited = [0] * (n + 1)
tab = [0] * n
q = deque([start])
visited[start] = 1
while q:
s = q.popleft()
for i in adj[s]:
if visited[i] == 0:
q.append(i)
visited[i] = 1
tab[i - 1] = tab[s - 1] + 1
return tab
n, m = map(int, input().split())
adj = d(list)
for i in range(m):
a, b = map(int, input().split())
adj[a].append(b)
adj[b].append(a)
t = bfs(adj, 1, n)
if 0 in t[1:]:
print('No')
else:
print('Yes')
for i in range(1, n):
print(adj[i+1][0])
|
s423799048
|
Accepted
| 689
| 39,312
| 534
|
from collections import defaultdict as d, deque
def bfs(adj, start, n):
visited = [0] * (n + 1)
tab = [0] * n
q = deque([start])
visited[start] = 1
while q:
s = q.popleft()
for i in adj[s]:
if visited[i] == 0:
q.append(i)
visited[i] = 1
tab[i - 1] = s
return tab
n, m = map(int, input().split())
adj = d(list)
for i in range(m):
a, b = map(int, input().split())
adj[a].append(b)
adj[b].append(a)
t = bfs(adj, 1, n)
if 0 in t[1:]:
print('No')
else:
print('Yes')
for i in range(1, n):
print(t[i])
|
s995932356
|
p02261
|
u554503378
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,608
| 788
|
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def bubble_sort(cards):
list_length = len(cards)
for i in range(list_length):
for j in range(list_length-1,i,-1):
if int(cards[j][1:]) < int(cards[j-1][1:]):
cards[j],cards[j-1] = cards[j-1],cards[j]
return cards
def selection_sort(l, n):
for i in range(n):
minj = i
for j in range(i+1, n):
if l[j][0] < l[minj][0]:
minj = j
l[i], l[minj] = l[minj], l[i]
return l[:]
n = int(input())
cards = input().split()
bubble_cards = [i for i in bubble_sort(cards)]
selection_cards = [i for i in selection_sort(cards,n)]
print(' '.join(bubble_cards))
print('Stable')
print(' '.join(selection_cards))
if bubble_cards == selection_cards:
print('Stable')
else:
print('Not stable')
|
s702725452
|
Accepted
| 20
| 5,612
| 872
|
def bubble_sort(num_list):
flag = 1
while flag:
flag = 0
for i in range(len(num_list)-1,0,-1):
if int(num_list[i][1:]) < int(num_list[i-1][1:]):
num_list[i],num_list[i-1] = num_list[i-1],num_list[i]
flag = 1
def selectin_sort(num_list):
for i in range(len(num_list)):
min_idx = i
for j in range(i,len(num_list)):
if int(num_list[j][1:]) < int(num_list[min_idx][1:]):
min_idx = j
num_list[i],num_list[min_idx] = num_list[min_idx],num_list[i]
n = input()
card_list_1 = [i for i in input().split()]
card_list_2 = [i for i in card_list_1]
bubble_sort (card_list_1)
selectin_sort(card_list_2)
print(' '.join(card_list_1))
print('Stable')
print(' '.join(card_list_2))
if card_list_1 == card_list_2:
print('Stable')
else:
print('Not stable')
|
s734215604
|
p03555
|
u693211869
| 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.
|
s = input()
t = input()
if s[0] == t[2] and s[1] == t[1] and s[2] == t[0]:
print('Yes')
else:
print('No')
|
s570445923
|
Accepted
| 18
| 2,940
| 114
|
s = input()
t = input()
if s[0] == t[2] and s[1] == t[1] and s[2] == t[0]:
print('YES')
else:
print('NO')
|
s552825916
|
p02645
|
u576335153
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,020
| 25
|
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
s = input()
print(s[0:2])
|
s981521206
|
Accepted
| 25
| 9,084
| 24
|
s= input()
print(s[0:3])
|
s172597537
|
p03079
|
u059436995
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 83
|
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
a, b, c = map(int,input().split())
if a == b and b == c:
print('Yes')
print('No')
|
s092997165
|
Accepted
| 17
| 2,940
| 88
|
if len(set(list(map(int,input().split())))) == 1:
print('Yes')
else:
print('No')
|
s660138356
|
p02578
|
u693694535
| 2,000
| 1,048,576
|
Wrong Answer
| 120
| 32,256
| 126
|
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
N=int(input())
A=list(map(int,input().split()))
F=0
for i in range(1,N):
if A[i]<A[i-1]:
F+=A[i-1]-A[i]
print(F)
|
s798324768
|
Accepted
| 168
| 32,232
| 153
|
N=int(input())
A=list(map(int,input().split()))
F=0
for i in range(1,N):
if A[i]<A[i-1]:
F+=A[i-1]-A[i]
A[i]+=A[i-1]-A[i]
print(F)
|
s002354975
|
p03129
|
u672475305
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 72
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
n,k = map(int,input().split())
print('Yes' if (n-1)//2+1 >= k else 'No')
|
s055275963
|
Accepted
| 19
| 3,060
| 72
|
n,k = map(int,input().split())
print('YES' if (n-1)//2+1 >= k else 'NO')
|
s630343322
|
p04044
|
u588081069
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 135
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
N, L = list(map(int, input().split()))
S = []
for i in range(N):
S.extend(list(map(str, input().split())))
S = sorted(S)
print(S)
|
s257062133
|
Accepted
| 17
| 3,060
| 144
|
N, L = list(map(int, input().split()))
S = []
for i in range(N):
S.extend(list(map(str, input().split())))
S = ''.join(sorted(S))
print(S)
|
s691622921
|
p02659
|
u166201488
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 8,972
| 136
|
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
A, B = 999990000000010, 9.90
B_ = int(B*100)
C = str(A*B_)
D = C[:-2]
E = C[-2:]
D,E
if E == '00':
print(D)
else:
print(D+'.'+E)
|
s106852381
|
Accepted
| 26
| 9,060
| 164
|
#A,B = 664706138336385, 9.79
A, B = input().split()
A = int(A)
B = float(B)
B_ = round(B*100)
C = A*B_
C_ = str(C)
if C < 100:
print(0)
else:
print(C_[:-2])
|
s358066012
|
p03409
|
u332906195
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,064
| 381
|
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.
|
# -*- coding: utf-8 -*-
N = int(input())
red = [tuple(map(int, input().split())) for _ in range(N)]
blu = [tuple(map(int, input().split())) for _ in range(N)]
red.sort(key=lambda x:x[1], reverse=True)
blu.sort(key=lambda x:x[0])
pairs = []
for b in blu:
for r in red:
if r[0] < b[0] and r[1] < b[1] and not r in pairs:
pairs.append(r)
print(len(pairs))
|
s598148710
|
Accepted
| 20
| 3,064
| 399
|
# -*- coding: utf-8 -*-
N = int(input())
red = [tuple(map(int, input().split())) for _ in range(N)]
blu = [tuple(map(int, input().split())) for _ in range(N)]
red.sort(key=lambda x:x[1], reverse=True)
blu.sort(key=lambda x:x[0])
pairs = []
for b in blu:
for r in red:
if r[0] < b[0] and r[1] < b[1] and not r in pairs:
pairs.append(r)
break
print(len(pairs))
|
s269404327
|
p03719
|
u539517139
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 70
|
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('Yes' if a<=b and b<=c else 'No')
|
s006137963
|
Accepted
| 17
| 2,940
| 70
|
a,b,c=map(int,input().split())
print('Yes' if a<=c and c<=b else 'No')
|
s680536325
|
p03352
|
u825528847
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 3,316
| 138
|
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
import math
X = int(input())
for i in range(X, 0, -1):
tmp = int(math.sqrt(i))
if tmp*tmp == i:
print(tmp)
break
|
s939920492
|
Accepted
| 18
| 2,940
| 172
|
X = int(input())
ans = 1
for i in range(1, X):
for j in range(2, X):
tmp = i ** j
if tmp > X:
break
ans = max(tmp, ans)
print(ans)
|
s948094172
|
p03719
|
u562669479
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 165
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
input_str = input()
input = input_str.split(' ')
a = int(input[0])
b = int(input[1])
c = int(input[2])
if a <= c and c <= b:
print('YES')
else:
print('NO')
|
s026250815
|
Accepted
| 17
| 3,064
| 165
|
input_str = input()
input = input_str.split(' ')
a = int(input[0])
b = int(input[1])
c = int(input[2])
if a <= c and c <= b:
print('Yes')
else:
print('No')
|
s237692244
|
p03047
|
u021916304
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 48
|
Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers?
|
n,k = map(int,input().split())
print(n*(n+1)//2)
|
s197802490
|
Accepted
| 17
| 2,940
| 43
|
n,k = map(int,input().split())
print(n-k+1)
|
s088728643
|
p03575
|
u856232850
| 2,000
| 262,144
|
Wrong Answer
| 2,107
| 3,064
| 585
|
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
import itertools
n,m = list(map(int,input().split()))
a = []
for i in range(m):
b = (list(map(int,input().split())))
b = [b[0]-1,b[1]-1]
a.append(b)
b = [[0 for i in range(n)] for j in range(n)]
for i in a:
b[i[0]][i[1]] = 1
b[i[1]][i[0]] = 1
c = [i+1 for i in range(n-1)]
ans = 0
for i in itertools.permutations(c,n-1):
j = [0]
j += i
count = 0
for i in range(n-1):
if b[j[i]][j[i+1]] == 1:
count += 1
if count == n-1:
ans += 1
count = 0
else:
break
print(ans)
|
s988891310
|
Accepted
| 19
| 3,188
| 411
|
n,m = list(map(int,input().split()))
a = []
for i in range(m):
a.append(list(map(int,input().split())))
b = [[0 for i in range(n)] for j in range(n)]
for i in a:
b[i[1]-1][i[0]-1] = 1
b[i[0]-1][i[1]-1] = 1
c = [sum(b[i]) for i in range(n)]
ans = 0
while 1 in c:
d = c.index(1)
e = b[d].index(1)
b[e][d] = 0
b[d][e] = 0
c = [sum(b[i]) for i in range(n)]
ans += 1
print(ans)
|
s510907694
|
p02601
|
u334242570
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 8,996
| 350
|
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
r,g,b = map(int,input().split())
k = int(input())
t=0
t+=k
for i in range(k):
if(b<=r):
b*=2
t-=1
elif(b>r and b<=g):
b*=2
t-=1
elif(b>r and b>g and g<=r):
g*=2
t-=1
if(t==0):
break
if(g>r and b>g):print("YES")
else:print("NO")
|
s516833922
|
Accepted
| 28
| 8,844
| 350
|
r,g,b = map(int,input().split())
k = int(input())
t=0
t+=k
for i in range(k):
if(b<=r):
b*=2
t-=1
elif(b>r and b<=g):
b*=2
t-=1
elif(b>r and b>g and g<=r):
g*=2
t-=1
if(t==0):
break
if(g>r and b>g):print("Yes")
else:print("No")
|
s119572085
|
p03377
|
u887087974
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 109
|
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.
|
l = input().split()
a = l[0]
b = l[1]
x = l[2]
if a > x or a + b < x:
print('No')
else:
print('Yes')
|
s317092453
|
Accepted
| 17
| 2,940
| 129
|
l = input().split()
a = int(l[0])
b = int(l[1])
x = int(l[2])
if (a > x) or (a + b < x):
print('NO')
else:
print('YES')
|
s294967848
|
p02608
|
u894521144
| 2,000
| 1,048,576
|
Wrong Answer
| 247
| 20,060
| 548
|
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).
|
from collections import Counter
def main(N):
lst = []
for x in range(1, 142):
for y in range(1, 142):
for z in range(1, 142):
r = (x + y) ** 2 + (y + z) ** 2 + (z + x) ** 2
if r <= 2 * 10 ** 4:
lst.append(r // 2)
else:
break
c = Counter(lst)
st = set(lst)
for n in range(1, N):
if n in st:
print(c[n])
else:
print(0)
if __name__ == '__main__':
N = int(input())
main(N)
|
s339810667
|
Accepted
| 259
| 20,292
| 625
|
from collections import Counter
def main(N):
lst = []
for x in range(1, 142):
for y in range(1, 142):
for z in range(1, 142):
r = (x + y) ** 2 + (y + z) ** 2 + (z + x) ** 2
if r <= 2 * 10 ** 4:
if r % 2 == 1:
print('奇数')
lst.append(r // 2)
else:
break
c = Counter(lst)
st = set(lst)
for n in range(1, N+1):
if n in st:
print(c[n])
else:
print(0)
if __name__ == '__main__':
N = int(input())
main(N)
|
s611101333
|
p03545
|
u971124021
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 273
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
abcd = input()
def dfs(i, f, sum):
print(i,f,sum)
if i == 3:
if sum ==7:
print(f + '=7')
exit()
else:
dfs(i+1, f + '+' + abcd[i+1], sum + int(abcd[i+1]))
dfs(i+1, f + '-' + abcd[i+1], sum - int(abcd[i+1]))
f = abcd[0]
dfs(0, f, int(f))
|
s870663720
|
Accepted
| 17
| 3,060
| 272
|
s = input()
for bit in range(1<<len(s)):
f = s[0]
ans = int(s[0])
for i in range(1,len(s)):
if bit & 1<<i:
f = f + '+' + s[i]
ans += int(s[i])
else:
f = f + '-' + s[i]
ans -= int(s[i])
if ans == 7:
print(f + '=7')
exit()
|
s181896619
|
p03435
|
u943657163
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 219
|
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct.
|
c = [list(map(int, input().split())) for _ in range(3)]
b0, b1, b2 = c[0][0], c[0][1], c[0][1]
for i in range(1, 3):
if not ((c[i][0]-b0)==(c[i][1]-b1)==(c[i][2]-b2)):
print('No')
exit()
print('Yes')
|
s354191458
|
Accepted
| 17
| 3,064
| 257
|
c = [list(map(int, input().split())) for _ in range(3)]
b0 = c[0][0]
b1 = c[0][1]
b2 = c[0][2]
for i in range(1, 3):
a0 = c[i][0] - b0
a1 = c[i][1] - b1
a2 = c[i][2] - b2
if not a0 == a1 == a2:
print('No')
exit()
print('Yes')
|
s640925386
|
p02659
|
u409806558
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,076
| 73
|
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
a, b = input().split()
a = int(a)
b = float(b) * 100
print(str(a*b)[:-2])
|
s233534185
|
Accepted
| 25
| 9,120
| 78
|
a, b = input().split()
a = int(a)
b = int(b.replace(".",""))
print((a*b)//100)
|
s208964695
|
p02865
|
u606372527
| 2,000
| 1,048,576
|
Wrong Answer
| 34
| 3,696
| 1,092
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
from sys import stdin
from itertools import accumulate, dropwhile, takewhile, groupby
import bisect
import copy
import math
'''
N = int(stdin.readline().rstrip())
A = [int(x) for x in stdin.readline().rstrip().split()]
count = 0
while len([x for x in A if x % 2 == 0]) == N:
A = list(map(lambda y: y / 2, A))
count += 1
print(count)
'''
N = int(stdin.readline().rstrip())
N = (N-1) / 2
print(N)
|
s846032563
|
Accepted
| 29
| 3,568
| 189
|
from sys import stdin
from itertools import accumulate, dropwhile, takewhile, groupby
import bisect
import copy
import math
N = int(stdin.readline().rstrip())
N = int((N-1) / 2)
print(N)
|
s974529605
|
p03943
|
u320763652
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 112
|
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, b, c = map(int, input().split())
if a+b == c and a+c == b and b+c==a:
print("Yes")
else:
print("No")
|
s133244140
|
Accepted
| 17
| 2,940
| 110
|
a, b, c = map(int, input().split())
if a+b == c or a+c == b or b+c==a:
print("Yes")
else:
print("No")
|
s842119822
|
p03485
|
u769411997
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
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.
|
a, b = map(int, input().split())
if a+b%2 == 0:
print((a+b)//2)
else:
print((a+b)//2 + 1)
|
s509478943
|
Accepted
| 17
| 2,940
| 100
|
a, b = map(int, input().split())
if (a+b)%2 == 0:
print((a+b)//2)
else:
print((a+b)//2 + 1)
|
s241620964
|
p03478
|
u634046173
| 2,000
| 262,144
|
Wrong Answer
| 42
| 9,080
| 162
|
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())
su =0
for i in range(1,N+1):
si = str(i)
c = 0
for s in si:
c += int(s)
if A <= c <= B:
su += c
print(su)
|
s190182216
|
Accepted
| 37
| 9,088
| 156
|
N, A, B = map(int,input().split())
su =0
for i in range(1,N+1):
si = str(i)
c = 0
for s in si:
c += int(s)
if A <= c <= B:
su += i
print(su)
|
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