text stringlengths 216 39.6k | conversation_id int64 219 108k | embedding list | cluster int64 11 11 |
|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob got a job as a system administrator in X corporation. His first task was to connect n servers with the help of m two-way direct connection so that it becomes possible to transmit data from one server to any other server via these connections. Each direct connection has to link two different servers, each pair of servers should have at most one direct connection. Y corporation, a business rival of X corporation, made Bob an offer that he couldn't refuse: Bob was asked to connect the servers in such a way, that when server with index v fails, the transmission of data between some other two servers becomes impossible, i.e. the system stops being connected. Help Bob connect the servers.
Input
The first input line contains 3 space-separated integer numbers n, m, v (3 β€ n β€ 105, 0 β€ m β€ 105, 1 β€ v β€ n), n β amount of servers, m β amount of direct connections, v β index of the server that fails and leads to the failure of the whole system.
Output
If it is impossible to connect the servers in the required way, output -1. Otherwise output m lines with 2 numbers each β description of all the direct connections in the system. Each direct connection is described by two numbers β indexes of two servers, linked by this direct connection. The servers are numbered from 1. If the answer is not unique, output any.
Examples
Input
5 6 3
Output
1 2
2 3
3 4
4 5
1 3
3 5
Input
6 100 1
Output
-1
Submitted Solution:
```
#! /usr/bin/python3
# SUBMISSION 1 (wrong answer on test 1): big dumb forgot to print answer
# SUBMISSION 2 (runtime error on test 2): big dumb again forget to handle if m out of bounds
# SUBMISSION 3 (TLE on test 10): need to optimize, maybe change handling of calculating factorial? SOLUTION: tried eliminating factorial calculation
import sys
import math
import copy
big = set()
small = []
def main():
input_list = get_input()
line = [int(x) for x in input_list[0].split(" ")]
solution = solve(line[0], line[1], line[2])
if solution != -1:
for i in solution:
print(i)
else:
print(-1)
def get_input():
input_list = []
for line in sys.stdin:
input_list.append(line.rstrip("\n"))
return input_list
def generate_sets(n, m, v):
global big
global small
if v == 1:
small = [2, 1]
for i in range(1, n + 1):
big.add(i)
big.remove(2)
big = list(big)
else:
small = [1, v]
for i in range(1, n + 1):
big.add(i)
big.remove(1)
big = list(big)
return None
def generate_basic_network(n, m, v):
global big
global small
edges = set()
temp = copy.deepcopy(big)
temp = set(temp)
temp.remove(v)
temp = list(temp)
basic = small + temp
for i in range(len(basic) - 1):
edges.add(str(basic[i]) + " " + str(basic[i + 1]))
return edges
def solve(n, m, v):
global big
global small
big = set()
small = []
edges = set()
# m_max = 1 + (math.factorial(n - 1) / (2 * math.factorial(n - 3)))
m_min = n - 1
if m >= m_min:
generate_sets(n, m, v)
edges = generate_basic_network(n, m, v)
for j in range(len(big)):
if len(edges) == m:
break
for k in range(len(big[j + 1:])):
if len(edges) == m:
break
edge = str(big[j]) + " " + str(big[j + 1:][k])
reverse_edge = str(big[j + 1:][k]) + " " + str(big[j])
if not edge in edges and not reverse_edge in edges:
edges.add(edge)
if len(edges) != m:
return -1
return edges
else:
return -1
if __name__ == "__main__":
main()
```
Yes
| 11,923 | [
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-0.79... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 β€ n β€ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 β€ wi β€ 106 for each 1 β€ i β€ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
Submitted Solution:
```
def main():
n = int(input())
w = list(map(int, input().split()))
bits = [0] * (10 ** 6 + 100)
for e in w:
bits[e] += 1
cur, res = 0, 0
for e in bits:
cur += e
if cur % 2:
res += 1
cur //= 2
print(res)
if __name__ == '__main__':
main()
```
Yes
| 12,065 | [
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... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 β€ n β€ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 β€ wi β€ 106 for each 1 β€ i β€ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
Submitted Solution:
```
n = int(input())
l = list(map(int,input().split()))
bit = [0]*(10**6+101)
for i in l:
bit[i] = bit[i]+1
#print(bit[0],bit[1],bit[2],bit[3])
ans = 0
s = 0
for i in bit:
s = s+i
ans = ans+(s%2)
s = s//2
print(ans)
```
Yes
| 12,067 | [
0.2822265625,
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0.052154541015625,
-1.078125,
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0.3486328125,
-0.55615234375,
-0.96923828125... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 β€ n β€ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 β€ wi β€ 106 for each 1 β€ i β€ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from collections import Counter
n = int(input())
weights = sorted(list(map(int,input().split())))
counter = Counter(weights)
minus_counter = Counter()
ans = 0
for weight, count in sorted(counter.items()):
count -= max(0, minus_counter[2**weight])
if count > 0 and count // 2 > 0:
minus_counter[2**weight*(count//2)*2] += 1
count -= (count//2)*2
ans += 1
ans += count
print(ans)
```
No
| 12,068 | [
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0.334228515625,
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-0.97509765... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 β€ n β€ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 β€ wi β€ 106 for each 1 β€ i β€ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from collections import Counter
n = int(input())
weights = sorted(list(map(int,input().split())))
counter = [0] * 10** 7
for weight, count in Counter(weights).items():
counter[weight] = count
ans = 0
for i in range(10**7):
weight, count = (i, counter[i])
if count > 0 and count // 2 > 0:
counter[weight+(count//2)] += 1
count -= (count//2)*2
ans += count
print(ans)
```
No
| 12,071 | [
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0.024688720703125,
-1.0498046875,
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0.343505859375,
-0.6103515625,
-0.96533... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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 β€ a, b, c, d, e, f β€ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
import math
import sys
while True:
try:
a,b,c,d,e,f = map(float, input().split())
if a*e == b*d:
continue
x = (c*e - b*f)/(a*e - b*d)
y = (c*d - a*f)/(b*d - a*e)
if x == 0:
x = 0
if y == 0:
y = 0
print('{0:.3f} {1:.3f}'.format(x,y))
except EOFError:
break
```
Yes
| 12,398 | [
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0.266845703125,
0.32666015625,
0.755859375,
0.2484130859375,
0.3408203125,
0.1571044921875,
-0.71533203125,
-0.59765625,
0.01224517822265625,
-0.56494140625,
-1... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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 β€ a, b, c, d, e, f β€ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
while(True):
try:
a,b,c,d,e,f = map(float, input().split())
y = (a*f-c*d)/(a*e-b*d)
x = (c-b*y)/a
print("%.3f %.3f"%(x,y))
except:
break
```
Yes
| 12,399 | [
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0.0313720703125,
-0.51513671875,
-0.95703125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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 β€ a, b, c, d, e, f β€ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
# coding: utf-8
import sys
for line in sys.stdin:
a, b, c, d, e, f = map(float, line.strip().split())
y = (c*d-a*f)/(b*d-a*e)
x = (c - b*y) / a
print("%.3lf %.3lf" % (x, y))
```
Yes
| 12,401 | [
0.291259765625,
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0.07086181640625,
-0.5400390625,
-0.9609... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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 β€ a, b, c, d, e, f β€ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
import sys
[print("{0[0]:.3f} {0[1]:.3f}".format([round(y, 3) for y in [(x[4] * x[2] - x[1] * x[5]) / (x[0] * x[4] - x[1] * x[3]), (x[0] * x[5] - x[2] * x[3]) / (x[0] * x[4] - x[1] * x[3])]])) for x in [[float(y) for y in x.split()] for x in sys.stdin]]
```
No
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0.12255859375,
-0.62646484375,
-0.995... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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 β€ a, b, c, d, e, f β€ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
print('ok')
```
No
| 12,403 | [
0.3369140625,
0.29296875,
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0.1412353515625,
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-0.08038330078125,
0.28125,
0.2470703125,
0.7548828125,
0.2381591796875,
0.38037109375,
0.028045654296875,
-0.74951171875,
-0.69140625,
0.060150146484375,
-0.54345703125,
-0.99609375,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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 β€ a, b, c, d, e, f β€ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
def jisuan(lis):
ns = lis.split(" ")
for i in range(6):
ns[i] = float(ns[i])
a, b, c, d, e, f = ns[0], ns[1], ns[2], ns[3], ns[4], ns[5]
x = (c * e - b * f) / (a * e - b * d)
y = (c * d - a * f) / (b * d - a * e)
return [x, y]
while 1:
try:
inpu = str(input())
except:
break
nums = inpu.split("\n")
for l in range(len(nums)):
k0 = jisuan("1 2 3 4 5 6")[0]
k1 = jisuan("1 2 3 4 5 6")[1]
print("%.3f" % k0, "%.3f" % k1)
```
No
| 12,404 | [
0.187744140625,
0.26904296875,
-0.08575439453125,
0.021453857421875,
-0.70703125,
-0.322998046875,
-0.0087432861328125,
0.36474609375,
0.282470703125,
0.779296875,
0.3525390625,
0.31005859375,
0.10931396484375,
-0.80712890625,
-0.62744140625,
0.127685546875,
-0.56103515625,
-0.9628... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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 β€ a, b, c, d, e, f β€ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
# coding: utf-8
# Your code here!
import numpy as np
while True:
try:
s=input().split()
l=[int(s[0]),int(s[1]),int(s[3]),int(s[4])]
lb=[int(s[2]),int(s[5])]
l=np.dot(np.linalg.inv(np.array(l).reshape(2,2)),np.array(lb))
print('%.3f %.3f' % (l[0],l[1]))
except:
break
```
No
| 12,405 | [
0.302978515625,
0.241455078125,
-0.11871337890625,
0.083984375,
-0.7822265625,
-0.30078125,
-0.072998046875,
0.2493896484375,
0.212646484375,
0.8076171875,
0.283203125,
0.298828125,
0.1146240234375,
-0.685546875,
-0.67822265625,
0.1405029296875,
-0.61669921875,
-1.0361328125,
-0.... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
3
NNN
NNN
NNN
Output
Taro
"Correct Solution:
```
from collections import deque
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N = int(readline())
G = [[] for i in range(N)]
for i in range(N):
*vs, = readline()
for j in range(N):
if vs[j] == "Y":
G[i].append(j)
d = 0
C = [0, 0]
used = [0]*N
for i in range(N):
d += len(G[i])
if used[i]:
continue
used[i] = 1
que = deque([i])
c = 0
while que:
v = que.popleft()
c += 1
for w in G[v]:
if used[w]:
continue
used[w] = 1
que.append(w)
C[c % 2] += 1
r0 = (N*(N-1)//2 - d//2) & 1
memo = {}
def dfs(i, p, q):
key = (p, q)
if key in memo:
return memo[key]
if p+q == 2:
r = (q == 2) ^ r0
memo[key] = e = r ^ (i & 1)
return e
r = 0
if p > 1 or (p and q):
if dfs(i+1, p-1, q) == 0:
r = 1
if q > 1:
if dfs(i+1, p+1, q-2) == 0:
r = 1
memo[key] = r
return r
if dfs(0, C[0], C[1]):
write("Taro\n")
else:
write("Hanako\n")
solve()
```
| 12,442 | [
0.338134765625,
0.37158203125,
-0.0986328125,
0.031707763671875,
-0.65234375,
-0.354736328125,
-0.08062744140625,
0.324951171875,
0.6630859375,
0.7802734375,
0.45361328125,
0.1719970703125,
0.1165771484375,
-0.63720703125,
-0.814453125,
0.0380859375,
-0.296875,
-0.580078125,
-0.2... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
c,l=0,[]
for i in range(n):
l.append(int(input()))
for i in l:
if i%2!=0:
c+=1
cnt,p=int(c/2),0
for i in l:
if i%2==0:
print(int(i/2))
elif i%2!=0 and p<cnt and i<0:
print(int(i/2))
p+=1
elif i%2!=0 and p<cnt and i>0:
print(int(i/2)+1)
p+=1
elif i%2!=0 and p>=cnt and i<0:
print(int(i/2)-1)
p+=1
else:
print(int(i/2))
```
| 12,569 | [
0.2283935546875,
-0.1934814453125,
-0.328125,
0.27880859375,
-0.52685546875,
-0.6533203125,
0.0406494140625,
0.0279388427734375,
-0.0692138671875,
0.70947265625,
0.6572265625,
0.1624755859375,
0.20703125,
-1.064453125,
-0.4716796875,
0.0318603515625,
-0.8583984375,
-0.7568359375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
st=1
for i in range(n):
a=int(input())
if a%2==0:
print(a//2)
continue
if a==0:
print(0)
if a<0:
if st==1:
print(a//2)
st=0
else:
print(a//2+1)
st=1
if a>0:
if st==1:
print(a//2)
st=0
else:
print(a//2+1)
st=1
```
| 12,570 | [
0.2283935546875,
-0.1934814453125,
-0.328125,
0.27880859375,
-0.52685546875,
-0.6533203125,
0.0406494140625,
0.0279388427734375,
-0.0692138671875,
0.70947265625,
0.6572265625,
0.1624755859375,
0.20703125,
-1.064453125,
-0.4716796875,
0.0318603515625,
-0.8583984375,
-0.7568359375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
l=[]
no=0
po=0
for i in range(n):
x=int(input())
if(x<0 and x%2==1):
no+=1
elif(x>0 and x%2==1):
po+=1
l.append(x)
ans=[]
no//=2
po//=2
for i in l:
x=i
if(x%2==1):
if(x<0 and no>0):
# print("....",x)
x-=1
no-=1
elif(x>0 and po>0):
x+=1
po-=1
# print(x,x//2)
if(x<0):
ans.append(-(abs(x)//2))
else:
ans.append(x//2)
for i in ans:
print(i)
```
| 12,571 | [
0.2283935546875,
-0.1934814453125,
-0.328125,
0.27880859375,
-0.52685546875,
-0.6533203125,
0.0406494140625,
0.0279388427734375,
-0.0692138671875,
0.70947265625,
0.6572265625,
0.1624755859375,
0.20703125,
-1.064453125,
-0.4716796875,
0.0318603515625,
-0.8583984375,
-0.7568359375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
l1=[]
odd=0
for i in range (n):
l1.append(int(input()))
l=[]
sum1=0
for i in range (n):
if(l1[i]%2==0):
sum1+=int(l1[i]/2)
l.append(int(l1[i]/2))
elif(l1[i]%2!=0 and odd==0):
l.append(int(l1[i]/2))
sum1+=int(l1[i]/2)
if(sum1==0):
for i in range (n):
print(l[i])
elif(sum1<0):
for i in range (n):
if(l1[i]%2!=0 and l1[i]>0):
sum1+=1
l[i]+=1
if(sum1==0):
break
for i in range (n):
print(l[i])
elif(sum1>0):
for i in range (n):
if(l1[i]%2!=0 and l1[i]<0):
sum1-=1
l[i]-=1
if(sum1==0):
break
for i in range (n):
print(l[i])
```
| 12,572 | [
0.2283935546875,
-0.1934814453125,
-0.328125,
0.27880859375,
-0.52685546875,
-0.6533203125,
0.0406494140625,
0.0279388427734375,
-0.0692138671875,
0.70947265625,
0.6572265625,
0.1624755859375,
0.20703125,
-1.064453125,
-0.4716796875,
0.0318603515625,
-0.8583984375,
-0.7568359375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
mod=1
for _ in range(n):
a=int(input())
if(a%2)==0:
print(a//2)
else:
print(a//2 + mod)
mod=1-mod
```
| 12,573 | [
0.2283935546875,
-0.1934814453125,
-0.328125,
0.27880859375,
-0.52685546875,
-0.6533203125,
0.0406494140625,
0.0279388427734375,
-0.0692138671875,
0.70947265625,
0.6572265625,
0.1624755859375,
0.20703125,
-1.064453125,
-0.4716796875,
0.0318603515625,
-0.8583984375,
-0.7568359375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
l=[]
for i in range(int(input())):
l.append(int(input()))
ans=[]
f=1
for i in l:
if i%2==0:
ans.append(i//2)
else:
ans.append((i+f)//2)
f*=-1
for i in ans:
print(i)
```
| 12,574 | [
0.2283935546875,
-0.1934814453125,
-0.328125,
0.27880859375,
-0.52685546875,
-0.6533203125,
0.0406494140625,
0.0279388427734375,
-0.0692138671875,
0.70947265625,
0.6572265625,
0.1624755859375,
0.20703125,
-1.064453125,
-0.4716796875,
0.0318603515625,
-0.8583984375,
-0.7568359375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
from collections import Counter
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
#for _ in range(int(input())):
#import math
import sys
# from collections import deque
#from collections import Counter
# ls=list(map(int,input().split()))
# for i in range(m):
# for i in range(int(input())):
#n,k= map(int, input().split())
#arr=list(map(int,input().split()))
#n=sys.stdin.readline()
#n=int(n)
#n,k= map(int, input().split())
#arr=list(map(int,input().split()))
#n=int(inaput())
#for _ in range(int(input())):
import math
n = int(input())
f=0
for i in range(n):
u= int(input())
if u%2==0:
print(u//2)
else:
if f==0:
print(math.floor(u/2))
f=1
else:
print(math.ceil(u/2))
f=0
#arr=list(map(int,input().split()))
#for _ in range(int(input())):
#n, k = map(int, input().split())
#arr=list(map(int,input().split()))
```
| 12,575 | [
0.2283935546875,
-0.1934814453125,
-0.328125,
0.27880859375,
-0.52685546875,
-0.6533203125,
0.0406494140625,
0.0279388427734375,
-0.0692138671875,
0.70947265625,
0.6572265625,
0.1624755859375,
0.20703125,
-1.064453125,
-0.4716796875,
0.0318603515625,
-0.8583984375,
-0.7568359375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
sum=0
for i in range(int(input())):
x=int(input())
y=x//2
if y==x/2:
print(y)
else:
if sum==0:
sum=x-y*2
print(y)
else:
y=y+sum
sum=0
print(y)
```
| 12,576 | [
0.2283935546875,
-0.1934814453125,
-0.328125,
0.27880859375,
-0.52685546875,
-0.6533203125,
0.0406494140625,
0.0279388427734375,
-0.0692138671875,
0.70947265625,
0.6572265625,
0.1624755859375,
0.20703125,
-1.064453125,
-0.4716796875,
0.0318603515625,
-0.8583984375,
-0.7568359375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
import sys
import math
def main():
#for _ in range(int(input())):
n = int(sys.stdin.readline())
#r, p, s= [int(x) for x in sys.stdin.readline().split()]
#l = list(set(int(x) for x in sys.stdin.readline().split()))
#str = sys.stdin.readline()
flag=0
for i in range(n):
x=int(sys.stdin.readline())
if x%2==0:
print(x//2)
else:
if flag==0:
print(x//2)
flag=1
else:
q=x//2
q+=1
print(q)
flag=0
if __name__ == "__main__":
main()
```
Yes
| 12,577 | [
0.357177734375,
-0.11895751953125,
-0.3291015625,
0.1282958984375,
-0.583984375,
-0.5087890625,
-0.0521240234375,
0.043731689453125,
-0.088623046875,
0.6982421875,
0.5107421875,
0.1732177734375,
0.155029296875,
-0.97998046875,
-0.453125,
-0.046783447265625,
-0.82177734375,
-0.65673... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
n=int(input())
l=[]
low=0
even=0
for i in range(0,n):
t=int(input())
if t%2==0:
even=even+t//2
l.append((t//2,0))
else:
low=low+(t//2)
l.append((t//2))
if low!=-1*even:
count=(-1*even)-low
for i in range(0,len(l)):
if count==0:
break
if type(l[i])!=tuple:
l[i]=l[i]+1
count=count-1
for i in range(0,len(l)):
if type(l[i])==tuple:
print(l[i][0])
else:
print(l[i])
```
Yes
| 12,578 | [
0.357177734375,
-0.11895751953125,
-0.3291015625,
0.1282958984375,
-0.583984375,
-0.5087890625,
-0.0521240234375,
0.043731689453125,
-0.088623046875,
0.6982421875,
0.5107421875,
0.1732177734375,
0.155029296875,
-0.97998046875,
-0.453125,
-0.046783447265625,
-0.82177734375,
-0.65673... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
from math import *
n = int(input())
k = []
p = 0
for x in range(n):
a = int(input())
if a%2 == 0:
a = int(a/2)
k.append(a)
else:
if p%2 == 0:
a = floor(a/2)
k.append(a)
else:
a = ceil(a/2)
k.append(a)
p += 1
for i in range(len(k)):
print(k[i])
```
Yes
| 12,579 | [
0.357177734375,
-0.11895751953125,
-0.3291015625,
0.1282958984375,
-0.583984375,
-0.5087890625,
-0.0521240234375,
0.043731689453125,
-0.088623046875,
0.6982421875,
0.5107421875,
0.1732177734375,
0.155029296875,
-0.97998046875,
-0.453125,
-0.046783447265625,
-0.82177734375,
-0.65673... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
import math
n = int(input())
check1 = 0
while n:
n -= 1
a = int(input())
if a % 2 == 0: print(a//2)
elif a % 2 and check1 % 2 == 0:
print(int(math.floor(a / 2)))
check1 += 1
elif a % 2 and check1 % 2:
print(int(math.ceil(a / 2)))
check1 += 1
else: print(a)
```
Yes
| 12,580 | [
0.357177734375,
-0.11895751953125,
-0.3291015625,
0.1282958984375,
-0.583984375,
-0.5087890625,
-0.0521240234375,
0.043731689453125,
-0.088623046875,
0.6982421875,
0.5107421875,
0.1732177734375,
0.155029296875,
-0.97998046875,
-0.453125,
-0.046783447265625,
-0.82177734375,
-0.65673... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
n=int(input())
o=0
for i in range(n):
l=int(input())
if(l>=0):
if(l%2==0):
print(l//2)
else:
if(o==0):
o=o+1
print(l//2)
else:
print(l//2+1)
o+=1
else:
l=abs(l)
if(l%2==0):
print(-(l//2))
else:
if(o==0):
o=o+1
print(-(l//2))
else:
print(-(l//2+1))
o=o+1
```
No
| 12,581 | [
0.357177734375,
-0.11895751953125,
-0.3291015625,
0.1282958984375,
-0.583984375,
-0.5087890625,
-0.0521240234375,
0.043731689453125,
-0.088623046875,
0.6982421875,
0.5107421875,
0.1732177734375,
0.155029296875,
-0.97998046875,
-0.453125,
-0.046783447265625,
-0.82177734375,
-0.65673... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
from math import ceil,floor
n=int(input())
cou=0
while n:
a=int(input())
if(cou%2==0 and a%2==1):
print(ceil(a/2))
else:
print(floor(a/2))
cou+=1
n-=1
```
No
| 12,582 | [
0.357177734375,
-0.11895751953125,
-0.3291015625,
0.1282958984375,
-0.583984375,
-0.5087890625,
-0.0521240234375,
0.043731689453125,
-0.088623046875,
0.6982421875,
0.5107421875,
0.1732177734375,
0.155029296875,
-0.97998046875,
-0.453125,
-0.046783447265625,
-0.82177734375,
-0.65673... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
from math import *
n = int(input())
sum = 0
found = 1
for i in range(n):
x = int(input())
if x&1:
if found == 1:
print(floor(x / 2))
found = 0
else:
print(ceil(x / 2))
found = 1
else:
print(x / 2)
```
No
| 12,583 | [
0.357177734375,
-0.11895751953125,
-0.3291015625,
0.1282958984375,
-0.583984375,
-0.5087890625,
-0.0521240234375,
0.043731689453125,
-0.088623046875,
0.6982421875,
0.5107421875,
0.1732177734375,
0.155029296875,
-0.97998046875,
-0.453125,
-0.046783447265625,
-0.82177734375,
-0.65673... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced β their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. In particular, if a_i is even, b_i = (a_i)/(2). Here β x β denotes rounding down to the largest integer not greater than x, and β x β denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced β their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 β€ n β€ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 β€ a_i β€ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = β (a_i)/(2) β or b_i = β (a_i)/(2) β. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
n=int(input())
a=[]
b=[]
for i in range(n):
x=int(input())
a.append(x)
if x>=0:
b.append(x//2)
else:
b.append(0-(abs(x)//2))
if sum(b)==(-1):
for i in range(n):
if a[i]>0 and a[i]%2==1:
b[i]+=1
break
elif sum(b)==1:
for i in range(n):
if a[i]<0 and a[i]%2==1:
b[i]-=1
break
for i in b:
print(i)
```
No
| 12,584 | [
0.357177734375,
-0.11895751953125,
-0.3291015625,
0.1282958984375,
-0.583984375,
-0.5087890625,
-0.0521240234375,
0.043731689453125,
-0.088623046875,
0.6982421875,
0.5107421875,
0.1732177734375,
0.155029296875,
-0.97998046875,
-0.453125,
-0.046783447265625,
-0.82177734375,
-0.65673... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys
from collections import deque, Counter
input = sys.stdin.buffer.readline
MOD = 1000000007
T = int(input())
for _ in range(T):
n, p = map(int, input().split())
ls = list(map(int, input().split()))
ls.sort(reverse=True)
if p == 1: print(1 if n%2 else 0); continue
cp, cc = 0, 0
ok, pos = 1, 0
for i, u in enumerate(ls):
if cc == 0:
cp, cc = u, 1
else:
tp, tc = cp, cc
while tp > u and tc <= n:
tp -= 1; tc *= p
if tp == u:
cp, cc = tp, tc-1
else:
ok, pos = 0, i; break
if ok:
print((pow(p, cp, MOD)*cc)%MOD)
else:
a = (pow(p, cp, MOD)*cc)%MOD
b = 0
for u in ls[pos:]:
b += pow(p, u, MOD)
b %= MOD
print((a-b)%MOD)
```
Yes
| 12,610 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
from sys import stdin, stdout
import math
from collections import defaultdict
def main():
MOD7 = 1000000007
t = int(stdin.readline())
pw = [0] * 21
for w in range(20,-1,-1):
pw[w] = int(math.pow(2,w))
for ks in range(t):
n,p = list(map(int, stdin.readline().split()))
arr = list(map(int, stdin.readline().split()))
if p == 1:
if n % 2 ==0:
stdout.write("0\n")
else:
stdout.write("1\n")
continue
arr.sort(reverse=True)
left = -1
i = 0
val = [0] * 21
tmp = p
val[0] = p
slot = defaultdict(int)
for x in range(1,21):
tmp = (tmp * tmp) % MOD7
val[x] = tmp
while i < n:
x = arr[i]
if left == -1:
left = x
else:
slot[x] += 1
tmp = x
if x == left:
left = -1
slot.pop(x)
else:
while slot[tmp] % p == 0:
slot[tmp+1] += 1
slot.pop(tmp)
tmp += 1
if tmp == left:
left = -1
slot.pop(tmp)
i+=1
if left == -1:
stdout.write("0\n")
continue
res = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= left:
left -= pww
res = (res * val[w]) % MOD7
if left == 0:
break
for x,c in slot.items():
tp = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= x:
x -= pww
tp = (tp * val[w]) % MOD7
if x == 0:
break
res = (res - tp * c) % MOD7
stdout.write(str(res)+"\n")
main()
```
Yes
| 12,611 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
from math import log2
def main():
t=int(input())
allAns=[]
MOD=10**9+7
for _ in range(t):
n,p=readIntArr()
a=readIntArr()
if p==1: #put half in each group
if n%2==1:
ans=1
else:
ans=0
else:
a.sort(reverse=True)
totalMOD=0
totalExact=0 #store exact total//p**a[i] for current i. total must be a multiple of p**a[i]
prevPow=a[0]
i=0
while i<n:
x=a[i]
#if totalExact//p**x>n, then just subtract all the remaining elements since totalExact can't reach 0
if totalExact!=0 and log2(totalExact)+(prevPow-x)*log2(p)>log2(n): #using log or else number may get too big
while i<n:
x=a[i]
totalMOD-=pow(p,x,MOD)
totalMOD=(totalMOD+MOD)%MOD
i+=1
else:
if totalExact>0:
totalExact*=(p**(prevPow-x))
prevPow=x
if totalExact==0: #add
totalMOD+=pow(p,x,MOD)
# totalMOD+=mod_pow(p,x)
totalMOD%=MOD
totalExact+=1
else:
totalMOD-=pow(p,x,MOD)
# totalMOD-=mod_pow(p,x)
totalMOD=(totalMOD+MOD)%MOD
totalExact-=1
# print('x:{} total:{}'.format(x,total))
i+=1
ans=totalMOD
allAns.append(ans)
multiLineArrayPrint(allAns)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
main()
```
Yes
| 12,612 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from math import log
def main():
mod = 10**9+7
for _ in range(int(input())):
n,p = map(int,input().split())
k = sorted(map(int,input().split()),reverse=1)
a,b = 0,0
# power,num
sign = [-1]*n
for ind,i in enumerate(k):
if not b:
a,b = i,1
sign[ind] = 1
else:
if a-i > log(n,p)-log(b,p):
break
b,a = b*pow(p,a-i)-1,i
ans = 0
for a,b in zip(sign,k):
ans = (ans+a*pow(p,b,mod))%mod
print(ans)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
Yes
| 12,613 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
from sys import stdin, stdout
import math
def main():
MOD7 = 1000000007
t = int(stdin.readline())
pw = [0] * 21
for w in range(20,-1,-1):
pw[w] = int(math.pow(2,w))
for ks in range(t):
n,p = list(map(int, stdin.readline().split()))
arr = list(map(int, stdin.readline().split()))
if p == 1:
if n % 2 ==0:
stdout.write("0\n")
else:
stdout.write("1\n")
continue
arr.sort(reverse=True)
res = 0
i = 0
val = [0] * 21
tmp = p
val[0] = p
for x in range(1,21):
tmp = (tmp * tmp) % MOD7
val[x] = tmp
while i < n:
if res == 0 and i + 1 < n and arr[i] == arr[i+1]:
i +=2
continue
tp = 1
x = arr[i]
for w in range(20,-1,-1):
pww = pw[w]
if pww <= x:
x -= pww
tp = (tp * val[w]) % MOD7
if x == 0:
break
if res == 0:
res = tp
else:
res = (res - tp) % MOD7
if res == 0 and t == 9999 and ks == 9998:
print(i,res,tp)
i+=1
if t != 9999:
stdout.write(str(res)+"\n")
elif ks == 9998:
print(n,p)
main()
```
No
| 12,614 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
from math import log
import copy
#sys.setrecursionlimit(10**6)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
def li():
return [int(x) for x in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def bo(i):
return ord(i)-ord('a')
from copy import *
from bisect import *
t=fi()
while t>0:
t-=1
n,p=mi()
a=li()
mod=10**9+7
a.sort()
d=[0 for i in range(10**6+1)]
for i in range(n):
d[a[i]]+=1
if p==1:
print(0 if n%2==0 else 1)
continue
#print(len(d))
pp=[]
for i in range(len(d)):
l=p
#print(d[i],p)
if d[i]==0:
continue
z=d[i]
for j in range(int(log(z,p))+2,-1,-1):
if d[i]>p**j:
d[j+i]+=d[i]//(p**j)
d[i]-=(d[i]//(p**j))*(p**j)
if d[i]>0:
pp.append([i,d[i]])
c=pow(p,pp[-1][0],mod)*pp[-1][1]
#print(pp)
for i in range(len(pp)-1):
c-= pow(p,pp[i][0],mod)*pp[i][1]
c=(c+mod)%mod
print(c)
```
No
| 12,615 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys;input=sys.stdin.readline
mod = 10**9+7
T, = map(int, input().split())
for _ in range(T):
N, M = map(int, input().split())
X = list(map(int, input().split()))
sX = sorted(list(set(X)))[::-1]
CC = dict()
for x in X:
if x not in CC:
CC[x] = 0
CC[x] += 1
rmn = 0
flag = False
bi=sX[0]
for i in sX:
cc = CC[i]
if rmn:
tmp = rmn
for _ in range(bi-i):
tmp*=M
if tmp>N:
flag = True
break
if flag:
rmn *= pow(M, bi-i, mod)
else:
rmn *= pow(M, bi-i)
if flag:
rmn = (rmn-cc)%mod
else:
rmn-=cc
if rmn <= cc:
rmn %= 2
bi = i
rmn = rmn*pow(M, bi, mod)%mod
print(rmn)
```
No
| 12,616 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 β€ t β€ 10^5) β the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 β€ n, p β€ 10^6). The second line contains n integers k_i (0 β€ k_i β€ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer β the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
MOD = 10 ** 9 + 7
def compress(string):
string.append(-1)
n = len(string)
begin, end, cnt = 0, 1, 1
while end < n:
if string[begin] == string[end]:
end, cnt = end + 1, cnt + 1
else:
yield string[begin], cnt
begin, end, cnt = end, end + 1, 1
t = int(input())
LIMIT = 10 ** 6
for _ in range(t):
n, p = map(int, input().split())
k = list(map(int, input().split()))
k.sort(reverse=True)
k = compress(k)
if p == 1:
print(n % 2)
continue
over_cnt = 0
over_ans = 1
while over_ans <= LIMIT:
over_ans *= p
over_cnt += 1
ans = -1
ans_cnt = -1
flag = False
res = 0
for val, cnt in k:
if flag:
res -= cnt * pow(p, val, MOD)
res %= MOD
if ans == -1 and cnt % 2 == 0:
continue
if ans == -1:
ans_cnt = 1
ans = val
continue
if ans - val >= over_cnt or ans_cnt >= LIMIT:
flag = True
res = ans_cnt * pow(p, ans, MOD) % MOD
res -= cnt * pow(p, val, MOD)
res %= MOD
continue
nokori = ans_cnt * p ** (ans - val)
nokori -= cnt
if nokori == 0:
ans = -1
ans_cnt = -1
elif nokori < 0:
ans = val
ans_cnt = -nokori % 2
else:
ans = val
ans_cnt = nokori
if res != 0:
print(res)
elif ans == -1:
print(0)
else:
print(ans_cnt * pow(p, ans, MOD) % MOD)
```
No
| 12,617 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java β System.out.flush(), in Pascal β flush(output) and in Python β sys.stdout.flush().
Mr. Chanek wants to buy a flamingo to accompany his chickens on his farm. Before going to the pet shop, Mr. Chanek stops at an animal festival to have fun. It turns out there is a carnival game with a flamingo as the prize.
There are N mysterious cages, which are numbered from 1 to N. Cage i has A_i (0 β€ A_i β€ 10^3) flamingoes inside (1 β€ i β€ N). However, the game master keeps the number of flamingoes inside a secret. To win the flamingo, Mr. Chanek must guess the number of flamingoes in each cage.
Coincidentally, Mr. Chanek has N coins. Each coin can be used to ask once, what is the total number of flamingoes inside cages numbered L to R inclusive? With L < R.
Input
Use standard input to read the responses of your questions.
Initially, the judge will give an integer N (3 β€ N β€ 10^3), the number of cages, and the number of coins Mr. Chanek has.
For each of your questions, the jury will give an integer that denotes the number of flamingoes from cage L to R inclusive.
If your program does not guess the flamingoes or ask other questions, you will get "Wrong Answer". Of course, if your program asks more questions than the allowed number, your program will get "Wrong Answer".
Output
To ask questions, your program must use standard output.
Then, you can ask at most N questions. Questions are asked in the format "? L R", (1 β€ L < R β€ N).
To guess the flamingoes, print a line that starts with "!" followed by N integers where the i-th integer denotes the number of flamingo in cage i. After answering, your program must terminate or will receive the "idle limit exceeded" verdict. You can only guess the flamingoes once.
Examples
Input
6
Β
5
Β
15
Β
10
Β
Output
Β
? 1 2
Β
? 5 6
Β
? 3 4
Β
! 1 4 4 6 7 8
Note
In the sample input, the correct flamingoes amount is [1, 4, 4, 6, 7, 8].
Submitted Solution:
```
# DEFINING SOME GOOD STUFF
import heapq
import sys
from math import *
import threading
from heapq import *
from itertools import count
from pprint import pprint
from collections import defaultdict
from heapq import heapify, heappop, heappush
# threading.stack_size(10**8)
# sys.setrecursionlimit(300000)
'''
-> if you are increasing recursionlimit then remember submitting using python3 rather pypy3
-> sometimes increasing stack size don't work locally but it will work on CF
'''
mod = 10 ** 9+7
inf = 10 ** 15
decision = ['NO', 'YES']
yes = 'YES'
no = 'NO'
# ------------------------------FASTIO----------------------------
import os
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n")+(not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ________________________FAST FACTORIAL______________________________#
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was "+str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n+1-len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n+1):
prev = nextArr[i-initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was "+str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n+1-len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n+1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was "+str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n+1-len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n+1):
prev = nextArr[i-initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n-k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n-k, k)) * f.invFactorial(min(k, n-k)) % self.MOD
def npr(self, n, k):
if k < 0 or n < k:
return 0
f = self.factorial
return (f.calc(n) * f.invFactorial(n-k)) % self.MOD
#_______________SEGMENT TREE ( logn range modifications )_____________#
class SegmentTree:
def __init__(self, data, default = 0, func = lambda a, b: max(a, b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len-1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size+self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i+i], self.data[i+i+1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx+self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx+1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# ____________________MY FAVOURITE FUNCTIONS_______________________#
def lower_bound(li, num):
answer = len(li)
start = 0
end = len(li)-1
while (start <= end):
middle = (end+start) // 2
if li[middle] >= num:
answer = middle
end = middle-1
else:
start = middle+1
return answer # min index where x is not less than num
def upper_bound(li, num):
answer = len(li)
start = 0
end = len(li)-1
while (start <= end):
middle = (end+start) // 2
if li[middle] <= num:
start = middle+1
else:
answer = middle
end = middle-1
return answer # max index where x is greater than num
def abs(x):
return x if x >= 0 else -x
def binary_search(li, val):
# print(lb, ub, li)
ans = -1
lb = 0
ub = len(li)-1
while (lb <= ub):
mid = (lb+ub) // 2
# print('mid is',mid, li[mid])
if li[mid] > val:
ub = mid-1
elif val > li[mid]:
lb = mid+1
else:
ans = mid # return index
break
return ans
def kadane(x): # maximum sum contiguous subarray
sum_so_far = 0
current_sum = 0
for i in x:
current_sum += i
if current_sum < 0:
current_sum = 0
else:
sum_so_far = max(sum_so_far, current_sum)
return sum_so_far
def pref(li):
pref_sum = [0]
for i in li:
pref_sum.append(pref_sum[-1]+i)
return pref_sum
def SieveOfEratosthenes(n):
prime = [{1, i} for i in range(n+1)]
p = 2
while (p <= n):
for i in range(p * 2, n+1, p):
prime[i].add(p)
p += 1
return prime
def primefactors(n):
factors = []
while (n % 2 == 0):
factors.append(2)
n //= 2
for i in range(3, int(sqrt(n))+1, 2): # only odd factors left
while n % i == 0:
factors.append(i)
n //= i
if n > 2: # incase of prime
factors.append(n)
return factors
def prod(li):
ans = 1
for i in li:
ans *= i
return ans
def sumk(a, b):
print('called for', a, b)
ans = a * (a+1) // 2
ans -= b * (b+1) // 2
return ans
def sumi(n):
ans = 0
if len(n) > 1:
for x in n:
ans += int(x)
return ans
else:
return int(n)
def checkwin(x, a):
if a[0][0] == a[1][1] == a[2][2] == x:
return 1
if a[0][2] == a[1][1] == a[2][0] == x:
return 1
if (len(set(a[0])) == 1 and a[0][0] == x) or (len(set(a[1])) == 1 and a[1][0] == x) or (len(set(a[2])) == 1 and a[2][0] == x):
return 1
if (len(set(a[0][:])) == 1 and a[0][0] == x) or (len(set(a[1][:])) == 1 and a[0][1] == x) or (len(set(a[2][:])) == 1 and a[0][0] == x):
return 1
return 0
# _______________________________________________________________#
inf = 10**9 + 7
def main():
# karmanya = int(input())
karmanya = 1
# divisors = SieveOfEratosthenes(200010)
# print(divisors)
while karmanya != 0:
karmanya -= 1
n = int(input())
# a,b,c,d = map(int, input().split())
# s = [int(x) for x in list(input())]
# s = list(input())
# a = list(map(int, input().split()))
# b = list(map(int, input().split()))
# c = list(map(int, input().split()))
# d = defaultdict(list)
ans = [0]*n
# print(ans)
l,r = 1, n
print('?',l,r)
sys.stdout.flush()
s = int(input())
for i in range(n-2):
r -= 1
print('?',l,r)
sys.stdout.flush()
x = int(input())
ans[n-1-i] = s - x
s = x
ans[1] = s-1
ans[0] = 1
print('!', *ans)
sys.stdout.flush()
main()
# t = threading.Thread(target=main)
# t.start()
# t.join()
```
No
| 12,650 | [
0.50537109375,
0.216796875,
0.00708770751953125,
0.10888671875,
-0.475830078125,
-0.327880859375,
-0.212646484375,
0.31640625,
0.007625579833984375,
0.8525390625,
0.439208984375,
-0.1854248046875,
0.1630859375,
-0.5458984375,
-0.72265625,
-0.27197265625,
-0.47412109375,
-1.15332031... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java β System.out.flush(), in Pascal β flush(output) and in Python β sys.stdout.flush().
Mr. Chanek wants to buy a flamingo to accompany his chickens on his farm. Before going to the pet shop, Mr. Chanek stops at an animal festival to have fun. It turns out there is a carnival game with a flamingo as the prize.
There are N mysterious cages, which are numbered from 1 to N. Cage i has A_i (0 β€ A_i β€ 10^3) flamingoes inside (1 β€ i β€ N). However, the game master keeps the number of flamingoes inside a secret. To win the flamingo, Mr. Chanek must guess the number of flamingoes in each cage.
Coincidentally, Mr. Chanek has N coins. Each coin can be used to ask once, what is the total number of flamingoes inside cages numbered L to R inclusive? With L < R.
Input
Use standard input to read the responses of your questions.
Initially, the judge will give an integer N (3 β€ N β€ 10^3), the number of cages, and the number of coins Mr. Chanek has.
For each of your questions, the jury will give an integer that denotes the number of flamingoes from cage L to R inclusive.
If your program does not guess the flamingoes or ask other questions, you will get "Wrong Answer". Of course, if your program asks more questions than the allowed number, your program will get "Wrong Answer".
Output
To ask questions, your program must use standard output.
Then, you can ask at most N questions. Questions are asked in the format "? L R", (1 β€ L < R β€ N).
To guess the flamingoes, print a line that starts with "!" followed by N integers where the i-th integer denotes the number of flamingo in cage i. After answering, your program must terminate or will receive the "idle limit exceeded" verdict. You can only guess the flamingoes once.
Examples
Input
6
Β
5
Β
15
Β
10
Β
Output
Β
? 1 2
Β
? 5 6
Β
? 3 4
Β
! 1 4 4 6 7 8
Note
In the sample input, the correct flamingoes amount is [1, 4, 4, 6, 7, 8].
Submitted Solution:
```
import sys
import time
t=int(input())
print("? 1 2",flush=True)
a=int(input())
print("? 1 3",flush=True)
b=int(input())
print("? 2 3",flush=True)
c=int(input())
l=[(a+b-c)//2,(a+c-b)//2,(b+c-a)//2]
d=(b+c-a)//2
for i in range(2,t-1):
print(f"? {i+1} {i+2} ",flush=True)
e=int(input())
l.append(e-d)
d=e-d
print("!",end= " ")
for i in range(t):
print(l[i],end=" ")
print()
```
No
| 12,651 | [
0.50537109375,
0.216796875,
0.00708770751953125,
0.10888671875,
-0.475830078125,
-0.327880859375,
-0.212646484375,
0.31640625,
0.007625579833984375,
0.8525390625,
0.439208984375,
-0.1854248046875,
0.1630859375,
-0.5458984375,
-0.72265625,
-0.27197265625,
-0.47412109375,
-1.15332031... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java β System.out.flush(), in Pascal β flush(output) and in Python β sys.stdout.flush().
Mr. Chanek wants to buy a flamingo to accompany his chickens on his farm. Before going to the pet shop, Mr. Chanek stops at an animal festival to have fun. It turns out there is a carnival game with a flamingo as the prize.
There are N mysterious cages, which are numbered from 1 to N. Cage i has A_i (0 β€ A_i β€ 10^3) flamingoes inside (1 β€ i β€ N). However, the game master keeps the number of flamingoes inside a secret. To win the flamingo, Mr. Chanek must guess the number of flamingoes in each cage.
Coincidentally, Mr. Chanek has N coins. Each coin can be used to ask once, what is the total number of flamingoes inside cages numbered L to R inclusive? With L < R.
Input
Use standard input to read the responses of your questions.
Initially, the judge will give an integer N (3 β€ N β€ 10^3), the number of cages, and the number of coins Mr. Chanek has.
For each of your questions, the jury will give an integer that denotes the number of flamingoes from cage L to R inclusive.
If your program does not guess the flamingoes or ask other questions, you will get "Wrong Answer". Of course, if your program asks more questions than the allowed number, your program will get "Wrong Answer".
Output
To ask questions, your program must use standard output.
Then, you can ask at most N questions. Questions are asked in the format "? L R", (1 β€ L < R β€ N).
To guess the flamingoes, print a line that starts with "!" followed by N integers where the i-th integer denotes the number of flamingo in cage i. After answering, your program must terminate or will receive the "idle limit exceeded" verdict. You can only guess the flamingoes once.
Examples
Input
6
Β
5
Β
15
Β
10
Β
Output
Β
? 1 2
Β
? 5 6
Β
? 3 4
Β
! 1 4 4 6 7 8
Note
In the sample input, the correct flamingoes amount is [1, 4, 4, 6, 7, 8].
Submitted Solution:
```
''' ===============================
-- @uthor : Kaleab Asfaw
-- Handle : kaleabasfaw2010
-- Bio : High-School Student
==============================='''
import sys
def inOut(x):
print("?", x)
sys.stdout.flush()
val = int(input())
return val
n = int(input())
if n%2:
lst = []
for i in range(1, n-2, 2):
val = inOut(str(i) + " " + str(i+1))
lst.append(1)
lst.append(val-1)
val = inOut(str(i+2) + " " + str(i+3))
val1 = inOut(str(i+3) + " " + str(i+4))
val2 = inOut(str(i+2) + " " + str(i+4))
y = val+val1-val2
x = val-y
z = val1-y
lst += [x, y, z]
else:
lst = []
for i in range(1, n+1, 2):
val = inOut(str(i) + " " + str(i+1))
lst.append(1)
lst.append(val-1)
ans = "! " + " ".join(list(map(str, lst)))
print(ans)
sys.stdout.flush()
# 1 4 4 6 7 8
```
No
| 12,652 | [
0.50537109375,
0.216796875,
0.00708770751953125,
0.10888671875,
-0.475830078125,
-0.327880859375,
-0.212646484375,
0.31640625,
0.007625579833984375,
0.8525390625,
0.439208984375,
-0.1854248046875,
0.1630859375,
-0.5458984375,
-0.72265625,
-0.27197265625,
-0.47412109375,
-1.15332031... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java β System.out.flush(), in Pascal β flush(output) and in Python β sys.stdout.flush().
Mr. Chanek wants to buy a flamingo to accompany his chickens on his farm. Before going to the pet shop, Mr. Chanek stops at an animal festival to have fun. It turns out there is a carnival game with a flamingo as the prize.
There are N mysterious cages, which are numbered from 1 to N. Cage i has A_i (0 β€ A_i β€ 10^3) flamingoes inside (1 β€ i β€ N). However, the game master keeps the number of flamingoes inside a secret. To win the flamingo, Mr. Chanek must guess the number of flamingoes in each cage.
Coincidentally, Mr. Chanek has N coins. Each coin can be used to ask once, what is the total number of flamingoes inside cages numbered L to R inclusive? With L < R.
Input
Use standard input to read the responses of your questions.
Initially, the judge will give an integer N (3 β€ N β€ 10^3), the number of cages, and the number of coins Mr. Chanek has.
For each of your questions, the jury will give an integer that denotes the number of flamingoes from cage L to R inclusive.
If your program does not guess the flamingoes or ask other questions, you will get "Wrong Answer". Of course, if your program asks more questions than the allowed number, your program will get "Wrong Answer".
Output
To ask questions, your program must use standard output.
Then, you can ask at most N questions. Questions are asked in the format "? L R", (1 β€ L < R β€ N).
To guess the flamingoes, print a line that starts with "!" followed by N integers where the i-th integer denotes the number of flamingo in cage i. After answering, your program must terminate or will receive the "idle limit exceeded" verdict. You can only guess the flamingoes once.
Examples
Input
6
Β
5
Β
15
Β
10
Β
Output
Β
? 1 2
Β
? 5 6
Β
? 3 4
Β
! 1 4 4 6 7 8
Note
In the sample input, the correct flamingoes amount is [1, 4, 4, 6, 7, 8].
Submitted Solution:
```
import sys
n=int(input())
a=[0]*n
print("1",n)
sys.stdout.flush()
x=int(input())
c=x
for j in range(1,n-1):
print("1",n-j)
sys.stdout.flush()
y=int(input())
a[n-j]=x-y
x=y
print("2",n)
sys.stdout.flush()
x=int(input())
a[0]=c-x
a[1]=y-a[0]
print("!",' '.join(map(str,a)))
```
No
| 12,653 | [
0.50537109375,
0.216796875,
0.00708770751953125,
0.10888671875,
-0.475830078125,
-0.327880859375,
-0.212646484375,
0.31640625,
0.007625579833984375,
0.8525390625,
0.439208984375,
-0.1854248046875,
0.1630859375,
-0.5458984375,
-0.72265625,
-0.27197265625,
-0.47412109375,
-1.15332031... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ n) β array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
import sys
input=sys.stdin.buffer.readline
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
d={}
for i in range(0,n):
if arr[i]-(i+1) not in d:
d[arr[i]-(i+1)]=1
else:
d[arr[i]-(i+1)]+=1
ans=0
for key in d.keys():
if d[key]>1:
k=d[key]
ans+=(k*(k-1))//2
print(ans)
```
Yes
| 12,706 | [
0.35302734375,
0.2119140625,
-0.1580810546875,
0.057586669921875,
-0.658203125,
-0.572265625,
-0.136474609375,
0.1981201171875,
0.147705078125,
1.099609375,
0.2081298828125,
-0.11322021484375,
0.099609375,
-0.85400390625,
-0.5693359375,
-0.044647216796875,
-0.51904296875,
-0.554199... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ n) β array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
def main():
import sys
t = int(sys.stdin.readline())
for _ in range(t):
n = int(sys.stdin.readline())
s = list(map(int, sys.stdin.readline().split()))
t = dict()
c = 0
for i in range(n):
r = s[i] - i - 1
if r in t:
c += t[r]
if r in t:
t[r] += 1
else:
t[r] = 1
sys.stdout.write(str(c) + '\n')
main()
```
Yes
| 12,707 | [
0.346923828125,
0.2110595703125,
-0.0989990234375,
-0.0021953582763671875,
-0.60498046875,
-0.54638671875,
-0.165771484375,
0.11834716796875,
0.155029296875,
1.07421875,
0.1607666015625,
-0.1473388671875,
0.07763671875,
-0.7841796875,
-0.52734375,
-0.074462890625,
-0.54541015625,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ n) β array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
t = int(input())
while t!=0:
t-=1
n = int(input())
arr = list(map(int , input().split()))
a = arr[0]-0
'''
i=1
while i <n:
arr[i] = arr[i] + a
i+= 1
'''
ans=0
j=0
a = arr[0]
while j<n:
print( a , arr[j])
if arr[j] == a:
ans += 1
a +=1
j+=1
if ans !=1:
ans = (ans*(ans-1))//2
#print('here' , ans)
print(ans)
```
No
| 12,710 | [
0.4130859375,
0.27001953125,
-0.167236328125,
0.0234527587890625,
-0.54638671875,
-0.6162109375,
-0.11248779296875,
0.186767578125,
0.279541015625,
1.072265625,
0.24169921875,
-0.06976318359375,
0.11419677734375,
-0.91845703125,
-0.55029296875,
-0.047332763671875,
-0.47998046875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ n) β array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
t = int(input())
for tc in range(t):
n = int(input())
arr = [int(z) for z in input().split()]
model = list(range(1, n+1))
c = []
for i in range(n):
if arr[i] == model[i]:
c.append(i)
ans = 0
for i in range(1, len(c)):
ans += len(c) - i
print(ans)
```
No
| 12,711 | [
0.451416015625,
0.1776123046875,
-0.136474609375,
-0.10711669921875,
-0.53564453125,
-0.493896484375,
-0.028717041015625,
0.2127685546875,
0.1383056640625,
1.076171875,
0.2724609375,
-0.120849609375,
-0.015869140625,
-0.8603515625,
-0.4580078125,
0.060882568359375,
-0.388671875,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 β€ t β€ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5).
The second line of each test case contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ n) β array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 β
10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
ls = list(map(int,input().split()))
ls2 = []
for i,v in enumerate(ls):
ls2.append(v - i)
d = {}
for i in ls2:
if i in d:
d[i] += 1
else:
d[i] = 1
dd = []
for j in d:
if d[j] == 1:
dd.append(j)
for k in dd:
d.pop(k)
sov = sum(d.values())
print(((sov * (sov - 1)) // 2))
```
No
| 12,713 | [
0.32373046875,
0.24755859375,
-0.11407470703125,
0.020843505859375,
-0.75048828125,
-0.765625,
-0.0941162109375,
0.20849609375,
0.295654296875,
1.0751953125,
0.08465576171875,
-0.141845703125,
0.164306640625,
-0.7998046875,
-0.412841796875,
0.01776123046875,
-0.51416015625,
-0.5932... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
n = int(input())
map = {}
for i in range(0,n):
input_string = input()
input_string = input_string.split(" ")
if input_string[0] in map:
temp = map[input_string[0]]
map.pop(input_string[0])
map[input_string[1]] = temp
else:
map[input_string[1]] = input_string[0]
print(len(map))
for key in map.keys():
value = map[key]
print(f'{value} {key}')
```
Yes
| 12,867 | [
0.0212554931640625,
0.0648193359375,
-0.1558837890625,
0.3115234375,
-0.476806640625,
-0.65869140625,
-0.499755859375,
0.2005615234375,
0.01163482666015625,
0.8046875,
0.6689453125,
-0.1248779296875,
0.25830078125,
-0.7294921875,
-0.7001953125,
0.279296875,
-0.640625,
-0.5190429687... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
q = int(input())
old = dict()
new = dict()
for i in range(q):
name = input().split()
if new.get(name[0]) == None:
old[name[0]] = name[1]
new[name[1]] = name[0]
else:
old[new[name[0]]] = name[1]
new[name[1]] = new[name[0]]
print(len(old))
for i in old:
print(i, old[i])
```
Yes
| 12,868 | [
0.1043701171875,
-0.0049896240234375,
-0.1090087890625,
0.3505859375,
-0.48681640625,
-0.61962890625,
-0.456298828125,
0.166748046875,
0.003803253173828125,
0.77734375,
0.66552734375,
-0.215576171875,
0.1636962890625,
-0.65576171875,
-0.76904296875,
0.33837890625,
-0.66845703125,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
from collections import defaultdict
class DisjSet:
def __init__(self, n):
self.rank = defaultdict(lambda:1)
self.parent = defaultdict(lambda:None)
def sol(self,a):
if self.parent[a]==None:
self.parent[a]=a
def find(self, x):
#ans=1
if (self.parent[x] != x):
self.parent[x] = self.find(self.parent[x])
#ans+=1
return self.parent[x]
def Union(self, x, y):
xset = self.find(x)
yset = self.find(y)
if xset == yset:
return
if self.rank[xset] < self.rank[yset]:
self.parent[xset] = yset
elif self.rank[xset] > self.rank[yset]:
self.parent[yset] = xset
else:
self.parent[yset] = xset
self.rank[xset] = self.rank[xset] + 1
n=int(input())
obj=DisjSet(n+1)
temp=[]
visited=defaultdict(lambda:None)
for p in range(n):
a,b=input().split()
temp.append([a,b])
obj.sol(a)
obj.sol(b)
obj.Union(a,b)
#print(obj.find('MikeMirzayanov'))
cont=0
ans=[]
while temp:
a,b=temp.pop()
if visited[obj.find(b)]==None:
ans.append([obj.find(b),b])
cont+=1
visited[obj.find(b)]=True
print(cont)
for val in ans:
print(*val)
```
Yes
| 12,869 | [
0.0131072998046875,
-0.00757598876953125,
-0.091552734375,
0.361328125,
-0.41796875,
-0.55517578125,
-0.49853515625,
0.165283203125,
0.005313873291015625,
0.82470703125,
0.58837890625,
-0.26318359375,
0.2255859375,
-0.6689453125,
-0.72509765625,
0.274658203125,
-0.64892578125,
-0.4... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
n = int(input())
m, rev = {}, {}
for i in range(n):
a, b = input().split()
if a not in rev:
m[a] = b
rev[b] = a
else:
m[rev[a]] = b
rev[b] = rev[a]
print(len(m))
for key in m:
print(key, m[key])
```
Yes
| 12,870 | [
0.07293701171875,
-0.032257080078125,
-0.1085205078125,
0.296875,
-0.478271484375,
-0.60498046875,
-0.5107421875,
0.161865234375,
0.032562255859375,
0.81884765625,
0.61669921875,
-0.185546875,
0.138427734375,
-0.66845703125,
-0.71923828125,
0.293212890625,
-0.59814453125,
-0.537109... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
n = int(input())
p = []
for i in range(n):
x, y = map(str, input().split(' '))
for j in range(len(p)):
if(p[j][1] == x):
p[j][1] = y
else:
p.append([x,y])
if(p == []):
p.append([x,y])
for i in range(len(p)):
print(p[i][0] + ' ' + p[i][1])
```
No
| 12,871 | [
0.056884765625,
0.01151275634765625,
-0.0638427734375,
0.310791015625,
-0.4814453125,
-0.63671875,
-0.474853515625,
0.1800537109375,
0.01560211181640625,
0.79345703125,
0.62939453125,
-0.158203125,
0.168701171875,
-0.68505859375,
-0.72705078125,
0.2744140625,
-0.6552734375,
-0.5185... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
n = int(input())
a, b = [], []
for i in range(n):
x, y = input().split()
if x not in b:
a.append(x); b.append(y)
else:
b[b.index(x)] = y
for i in range(len(a)):
print(a[i], b[i])
```
No
| 12,872 | [
0.0662841796875,
-0.0235137939453125,
-0.07769775390625,
0.335693359375,
-0.47412109375,
-0.6162109375,
-0.484375,
0.19384765625,
0.046966552734375,
0.8154296875,
0.62841796875,
-0.169921875,
0.126220703125,
-0.68603515625,
-0.73388671875,
0.293701171875,
-0.6220703125,
-0.54931640... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
from math import sqrt,ceil
def mi():return map(int,input().split())
def li():return list(mi())
def ii():return int(input())
def si():return input()
rank=[]
#dsu(disjoint set unit) -> :)
def find(x,parent):
if(x==parent[x]):
return x
parent[x]=find(parent[x],parent)
return parent[x]
def union(x,y,parent):
x1=find(x,parent)
y1=find(y,parent)
if(x1==y1):
return
if(rank[x1]>=rank[y1]):
parent[y]=x1
rank[x1]+=1
else:
parent[x]=y1
rank[y1]+=1
def main():
q=ii()
old={}
a=[]
for i in range(q):
s1,s2=map(str,input().split())
old[s2]=s1
a.append(s2)
for i in range(q):
for j in range(q):
if(a[j]==old[a[j]]):
continue
if(a[i]==old[a[j]]):
old[a[j]]=old[a[i]]
a[j]=a[i]
old[a[j]]=a[i]
b=[]
for i in old.keys():
if i!=old[i]:
b.append(i)
print(len(b))
for i in b:
print(old[i],i)
if __name__=="__main__":
main()
```
No
| 12,873 | [
0.07171630859375,
0.03485107421875,
-0.09967041015625,
0.348388671875,
-0.482666015625,
-0.60400390625,
-0.4833984375,
0.202880859375,
-0.0011568069458007812,
0.81103515625,
0.62744140625,
-0.130615234375,
0.1610107421875,
-0.705078125,
-0.671875,
0.321533203125,
-0.59814453125,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
users = dict()
nombres = list()
cant = input()
handles = []
class Handle:
def __init__(self, old):
self.old = old
self.new = None
def actualizar(self, new):
self.new = new
for i in range(int(cant)):
nombre = input().split(" ")
if nombre[0] not in nombres:
nombres.append(nombre[0])
u = Handle(nombre[0])
handles.append(u)
if nombre[1] not in nombres:
u.actualizar(nombre[1])
nombres.append(nombre[1])
else:
if nombre[1] not in nombres:
for i in handles:
if i.new == nombre[0]:
i.actualizar(nombre[1])
handles.remove(i)
handles.append(i)
break
print(len(handles))
a = len(handles) - 1
while a >= 0:
print("{} {}".format(handles[a].old, handles[a].new))
a -= 1
```
No
| 12,874 | [
0.07928466796875,
-0.022369384765625,
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0.25390625,
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0.1553955078125,
0.0013980865478515625,
0.87548828125,
0.74072265625,
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0.147705078125,
-0.64013671875,
-0.6845703125,
0.365478515625,
-0.54345703125,
-0.... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Tags: implementation, sortings
Correct Solution:
```
n = int(input())
r = 1
l = []
c = []
for i in range(n):
x, y = map(int, input().split())
if x != y:
r = 0
break
l.append(x)
c.append(x)
c.sort()
if r == 0:
print("rated")
else:
if c == l[::-1]:
print("maybe")
else:
print("unrated")
```
| 12,998 | [
0.219970703125,
-0.2802734375,
-0.491455078125,
0.419189453125,
-0.52294921875,
-0.5751953125,
-0.28125,
0.124755859375,
-0.2099609375,
0.92578125,
0.99609375,
0.208251953125,
0.389892578125,
-1.263671875,
-0.30078125,
0.03436279296875,
-0.7353515625,
-0.73046875,
-0.333984375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
d=list(range(n))
u=True
e=True
for i in d:
a,b=map(int,list(input().split()))
if (a!=b):
print("rated")
e=False
break
d[i]=a
if e:
for i in range(n):
if (i>0):
if d[i]>d[i-1]:
print("unrated")
u=False
break
if u:
print("maybe")
```
| 12,999 | [
0.219970703125,
-0.2802734375,
-0.491455078125,
0.419189453125,
-0.52294921875,
-0.5751953125,
-0.28125,
0.124755859375,
-0.2099609375,
0.92578125,
0.99609375,
0.208251953125,
0.389892578125,
-1.263671875,
-0.30078125,
0.03436279296875,
-0.7353515625,
-0.73046875,
-0.333984375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
m=4126
f=False
for i in range(n):
a,b=[int(i) for i in input().split()]
if a!=b:
print("rated")
exit(0)
if a>m:
f=True
m=min(a,m)
if f:
print("unrated")
exit(0)
print("maybe")
```
| 13,000 | [
0.219970703125,
-0.2802734375,
-0.491455078125,
0.419189453125,
-0.52294921875,
-0.5751953125,
-0.28125,
0.124755859375,
-0.2099609375,
0.92578125,
0.99609375,
0.208251953125,
0.389892578125,
-1.263671875,
-0.30078125,
0.03436279296875,
-0.7353515625,
-0.73046875,
-0.333984375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Tags: implementation, sortings
Correct Solution:
```
r = [tuple(map(int, input().split())) for _ in range(int(input()))]
if any([p[0] != p[1] for p in r]):
print('rated')
elif r != list(sorted(r, reverse=True)):
print('unrated')
else:
print('maybe')
```
| 13,001 | [
0.219970703125,
-0.2802734375,
-0.491455078125,
0.419189453125,
-0.52294921875,
-0.5751953125,
-0.28125,
0.124755859375,
-0.2099609375,
0.92578125,
0.99609375,
0.208251953125,
0.389892578125,
-1.263671875,
-0.30078125,
0.03436279296875,
-0.7353515625,
-0.73046875,
-0.333984375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
first_rate=[]
second_rate=[]
for i in range(n):
k=[int(i) for i in input().split()]
first_rate.append(k[0])
second_rate.append(k[1])
if first_rate == second_rate :
first_rate_sorted=sorted(first_rate,reverse=True)
if first_rate_sorted == first_rate:
print("maybe")
else :
print("unrated")
else :
print("rated")
```
| 13,002 | [
0.219970703125,
-0.2802734375,
-0.491455078125,
0.419189453125,
-0.52294921875,
-0.5751953125,
-0.28125,
0.124755859375,
-0.2099609375,
0.92578125,
0.99609375,
0.208251953125,
0.389892578125,
-1.263671875,
-0.30078125,
0.03436279296875,
-0.7353515625,
-0.73046875,
-0.333984375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Tags: implementation, sortings
Correct Solution:
```
# ===================================
# (c) MidAndFeed aka ASilentVoice
# ===================================
# import math, fractions, collections
# ===================================
n = int(input())
q = [[int(x) for x in input().split()] for i in range(n)]
if any(x[0] != x[1] for x in q):
print("rated")
else:
unrated = 0
for i in range(n-1):
piv = q[i]
for j in range(i+1, n):
temp = q[j]
if piv < temp:
unrated = 1
break
if unrated:
break
print("unrated" if unrated else "maybe")
```
| 13,003 | [
0.219970703125,
-0.2802734375,
-0.491455078125,
0.419189453125,
-0.52294921875,
-0.5751953125,
-0.28125,
0.124755859375,
-0.2099609375,
0.92578125,
0.99609375,
0.208251953125,
0.389892578125,
-1.263671875,
-0.30078125,
0.03436279296875,
-0.7353515625,
-0.73046875,
-0.333984375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
a=[]
for i in range(n):
r=list(map(int,input().split()))
a.append(r)
b=sorted(a,reverse=True)
for i in a:
if(i[1]-i[0]!=0):
print("rated")
exit(0)
for i in range(n):
if a[i]!=b[i]:
print("unrated")
exit(0)
c=0
for i in a:
if(i[1]-i[0]==0):
c+=1
if(c==n):
print("maybe")
```
| 13,004 | [
0.219970703125,
-0.2802734375,
-0.491455078125,
0.419189453125,
-0.52294921875,
-0.5751953125,
-0.28125,
0.124755859375,
-0.2099609375,
0.92578125,
0.99609375,
0.208251953125,
0.389892578125,
-1.263671875,
-0.30078125,
0.03436279296875,
-0.7353515625,
-0.73046875,
-0.333984375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Tags: implementation, sortings
Correct Solution:
```
a=[]
for _ in range(int(input())):
a.append(list(map(int,input().split())))
a=list(zip(*a))
if a[0]!=a[1]:
print('rated')
else:
if list(a[0])==sorted(a[0],reverse=True):
print('maybe')
else:
print('unrated')
```
| 13,005 | [
0.219970703125,
-0.2802734375,
-0.491455078125,
0.419189453125,
-0.52294921875,
-0.5751953125,
-0.28125,
0.124755859375,
-0.2099609375,
0.92578125,
0.99609375,
0.208251953125,
0.389892578125,
-1.263671875,
-0.30078125,
0.03436279296875,
-0.7353515625,
-0.73046875,
-0.333984375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
R = lambda:map(int,input().split())
n, = R()
rate = []
for i in range(n):
x, y = R()
if x != y:
exit(print('rated'))
rate.append(y)
if rate == sorted(rate, reverse=True):
print("maybe")
else:
print("unrated")
```
Yes
| 13,006 | [
0.254638671875,
-0.184814453125,
-0.4150390625,
0.299072265625,
-0.61328125,
-0.484130859375,
-0.30859375,
0.177001953125,
-0.262451171875,
0.8720703125,
0.88623046875,
0.225341796875,
0.2978515625,
-1.21875,
-0.226318359375,
-0.060699462890625,
-0.68701171875,
-0.73583984375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
from math import inf
def solve():
global rates
for b, a in rates:
if b != a: return 'rated'
for i in range(1, len(rates)):
if rates[i-1] < rates[i]:
return 'unrated'
return 'maybe'
def main():
global rates
n = int(input())
rates = [list(map(int, input().split())) for _ in range(n)]
print(solve())
main()
```
Yes
| 13,007 | [
0.254638671875,
-0.184814453125,
-0.4150390625,
0.299072265625,
-0.61328125,
-0.484130859375,
-0.30859375,
0.177001953125,
-0.262451171875,
0.8720703125,
0.88623046875,
0.225341796875,
0.2978515625,
-1.21875,
-0.226318359375,
-0.060699462890625,
-0.68701171875,
-0.73583984375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
n = int(input())
check = []
for i in range(n):
x, y = map(int, input().split())
if x != y:
print('rated')
exit()
check.append(x)
comp = sorted(check, reverse=True)
if comp == check:
print('maybe')
else: print('unrated')
```
Yes
| 13,008 | [
0.254638671875,
-0.184814453125,
-0.4150390625,
0.299072265625,
-0.61328125,
-0.484130859375,
-0.30859375,
0.177001953125,
-0.262451171875,
0.8720703125,
0.88623046875,
0.225341796875,
0.2978515625,
-1.21875,
-0.226318359375,
-0.060699462890625,
-0.68701171875,
-0.73583984375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
k=[];n=0
for i in " "*int(input()):a,b=map(int,input().split());k+=[b];n+=a!=b
if n:print("rated")
elif all(i==j for i,j in zip(k,sorted(k,reverse=True))):print("maybe")
else:print("unrated")
```
Yes
| 13,009 | [
0.254638671875,
-0.184814453125,
-0.4150390625,
0.299072265625,
-0.61328125,
-0.484130859375,
-0.30859375,
0.177001953125,
-0.262451171875,
0.8720703125,
0.88623046875,
0.225341796875,
0.2978515625,
-1.21875,
-0.226318359375,
-0.060699462890625,
-0.68701171875,
-0.73583984375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
n = int(input())
count = 0
unrated = 0
may = 0
listX = []
listY = []
x , y = input().split()
listX.append(x)
listY.append(y)
for a in range(1 , n):
x , y = input().split()
listX.append(x)
listY.append(y)
if x != y:
count += 1
else:
if int(listX[a]) > int(listX[a-1]):
unrated += 1
elif int(listX[a]) <= int(listX[a-1]):
may += 1
if count > 0:
print("rated")
elif unrated > 0:
print("unrated")
elif may > 0:
print("maybe")
```
No
| 13,010 | [
0.254638671875,
-0.184814453125,
-0.4150390625,
0.299072265625,
-0.61328125,
-0.484130859375,
-0.30859375,
0.177001953125,
-0.262451171875,
0.8720703125,
0.88623046875,
0.225341796875,
0.2978515625,
-1.21875,
-0.226318359375,
-0.060699462890625,
-0.68701171875,
-0.73583984375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
#collaborated with Bhumi Patel
n = int(input())
finalarray=[]
finalarray1=[]
rating=False
temp=True
for i in range(n):
temp_var=input()
temp_var=temp_var.split()
finalarray.append(int(temp_var[0]))
finalarray1.append(int(temp_var[1]))
if finalarray[i]!=finalarray1[i]:
rating=True
for i in range(1,len(finalarray)):
if finalarray[i]>finalarray[i-1] or finalarray1[i]>finalarray1[i-1]:
temp=True
break
if rating==True:
print("rated")
elif temp==False and rating==False:
print("unrated")
else:
print("maybe")
```
No
| 13,011 | [
0.254638671875,
-0.184814453125,
-0.4150390625,
0.299072265625,
-0.61328125,
-0.484130859375,
-0.30859375,
0.177001953125,
-0.262451171875,
0.8720703125,
0.88623046875,
0.225341796875,
0.2978515625,
-1.21875,
-0.226318359375,
-0.060699462890625,
-0.68701171875,
-0.73583984375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
from bisect import bisect_right as br
import sys
from collections import *
from math import *
import re
def sieve(n):
prime=[True for i in range(n+1)]
p=2
while p*p<=n:
if prime[p]==True:
for i in range(p*p,n+1,p):
prime[i]=False
p+=1
c=0
for i in range(2,n):
if prime[i]:
#print(i)
c+=1
return c
def totient(n):
res,p=n,2
while p*p<=n:
if n%p==0:
while n%p==0:
n=n//p
res-=int(res/p)
p+=1
if n>1:res-=int(res/n)
return res
def iseven(n):return[False,True][0 if n%2 else 1]
def inp_matrix(n):return list([input().split()] for i in range(n))
def inp_arr():return list(map(int,input().split()))
def inp_integers():return map(int,input().split())
def inp_strings():return input().split()
def lcm(a,b):return (a*b)/gcd(a,b)
max_int = sys.maxsize
mod = 10**9+7
flag1=False
flag2=False
n=int(input())
a=[input().split() for i in range(n)]
for i in range(n):
if a[i][0]!=a[i][1]:flag1=True
for i in range(1,n):
if a[i][0]>a[i-1][0]:flag2=True
#print(flag1,flag2)
if flag1:print('rated')
elif not flag1 and flag2:print('unrated')
else:print('maybe')
```
No
| 13,012 | [
0.254638671875,
-0.184814453125,
-0.4150390625,
0.299072265625,
-0.61328125,
-0.484130859375,
-0.30859375,
0.177001953125,
-0.262451171875,
0.8720703125,
0.88623046875,
0.225341796875,
0.2978515625,
-1.21875,
-0.226318359375,
-0.060699462890625,
-0.68701171875,
-0.73583984375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
def main():
ans = 'unrated'
n_old = None
quan = int(input())
while quan:
num = input().split()
if num[0] != num[1]:
ans = 'rated'
if n_old != None and n_old == num[0] and ans == 'unrated':
ans = 'maybe'
n_old = num[0]
quan -= 1
return ans
print(main())
```
No
| 13,013 | [
0.254638671875,
-0.184814453125,
-0.4150390625,
0.299072265625,
-0.61328125,
-0.484130859375,
-0.30859375,
0.177001953125,
-0.262451171875,
0.8720703125,
0.88623046875,
0.225341796875,
0.2978515625,
-1.21875,
-0.226318359375,
-0.060699462890625,
-0.68701171875,
-0.73583984375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo Kingdom Congress is voting on a bill.
N members are present, and the i-th member (1 β€ i β€ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box.
After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected.
In order for the bill to pass, at least how many members must be in favor of it?
Constraints
* 1 β€ N β€ 10^5
* 1 β€ P β€ 100
* 1 β€ w_i β€ 10^9
* 1 β€ b_i β€ 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N P
w_1 b_1
w_2 b_2
:
w_N b_N
Output
Print the minimum number of members in favor of the bill required for passage.
Examples
Input
4 75
1 1
1 1
1 1
1 1
Output
3
Input
4 75
1 1
1 1
1 1
100 1
Output
1
Input
5 60
6 3
5 9
3 4
7 8
4 7
Output
3
Submitted Solution:
```
from itertools import accumulate
def solve(l, r):
if l > r:
return l
m = (l + r) // 2
w, b = whites[m], blues[m]
if w * 100 // (w + b) < p:
l = m + 1
else:
r = m - 1
return solve(l, r)
n, p = map(int, input().split())
wbs = [tuple(map(int, input().split())) for _ in range(n)]
ass = [(sum(t), i) for i, t in enumerate(wbs)]
ass.sort(reverse=True)
whites = list(accumulate(wbs[i][0] for s, i in ass))
blues = [wbs[i][1] for s, i in ass[1:]]
blues = list(reversed(list(accumulate(reversed(blues))))) + [0]
# print(list(whites))
# print(list(blues))
print(solve(0, n - 1) + 1)
```
No
| 13,184 | [
0.5537109375,
-0.156982421875,
-0.0079345703125,
-0.03326416015625,
-0.339111328125,
0.01177978515625,
0.050689697265625,
0.419921875,
0.171630859375,
0.9677734375,
0.5947265625,
0.20068359375,
0.2183837890625,
-0.483154296875,
-0.348388671875,
0.387451171875,
-0.814453125,
-0.5532... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo Kingdom Congress is voting on a bill.
N members are present, and the i-th member (1 β€ i β€ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box.
After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected.
In order for the bill to pass, at least how many members must be in favor of it?
Constraints
* 1 β€ N β€ 10^5
* 1 β€ P β€ 100
* 1 β€ w_i β€ 10^9
* 1 β€ b_i β€ 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N P
w_1 b_1
w_2 b_2
:
w_N b_N
Output
Print the minimum number of members in favor of the bill required for passage.
Examples
Input
4 75
1 1
1 1
1 1
1 1
Output
3
Input
4 75
1 1
1 1
1 1
100 1
Output
1
Input
5 60
6 3
5 9
3 4
7 8
4 7
Output
3
Submitted Solution:
```
from itertools import accumulate
def solve(l, r):
if l > r:
return l
m = (l + r) // 2
w, b = whites[m], blues[m]
if w * 100 // (w + b) < p:
l = m + 1
else:
r = m - 1
return solve(l, r)
n, p = map(int, input().split())
wbs = [tuple(map(int, input().split())) for _ in range(n)]
ass = [(sum(t), t[0], i) for i, t in enumerate(wbs)]
ass.sort(reverse=True)
whites = list(accumulate(wbs[i][0] for s, t, i in ass))
blues = [wbs[i][1] for s, t, i in ass[1:]]
blues = list(reversed(list(accumulate(reversed(blues))))) + [0]
# print(list(whites))
# print(list(blues))
print(solve(0, n - 1) + 1)
```
No
| 13,185 | [
0.55517578125,
-0.143310546875,
-0.005001068115234375,
-0.027069091796875,
-0.333740234375,
0.006011962890625,
0.053924560546875,
0.403564453125,
0.1728515625,
0.96240234375,
0.59033203125,
0.2109375,
0.2266845703125,
-0.48388671875,
-0.34765625,
0.390869140625,
-0.81494140625,
-0.... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
"Correct Solution:
```
N = int(input())
A = [int(x) for x in input().split()]
sum = A[0]
ans = A[0]
for i in range(1, N):
sum += A[i]
ans = min(ans, sum // (i + 1))
print(ans)
```
| 13,260 | [
0.7060546875,
0.333984375,
-0.32861328125,
0.1689453125,
-0.60302734375,
-0.8017578125,
-0.046600341796875,
0.26904296875,
0.443359375,
0.94482421875,
0.43017578125,
-0.048187255859375,
0.34765625,
-0.5947265625,
-0.392578125,
-0.31201171875,
-0.59814453125,
-0.79443359375,
-0.51... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
"Correct Solution:
```
n = map(int,input().split())
a = list(map(int,input().split()))
ok,ng = 1,10000
while ng-ok > 1:
mid = (ok+ng)//2
valid = True
now = 0
for i in a:
now += i
now -= mid
if now < 0:
valid = False
if valid:
ok = mid
else:
ng = mid
print(ok)
```
| 13,261 | [
0.84228515625,
0.39453125,
-0.25537109375,
0.275634765625,
-0.6025390625,
-0.84033203125,
0.040435791015625,
0.2626953125,
0.4482421875,
1.10546875,
0.436279296875,
0.052978515625,
0.50244140625,
-0.552734375,
-0.1962890625,
-0.268798828125,
-0.51123046875,
-0.70849609375,
-0.500... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
"Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
r = 100
for i in range(n):
r = min(r, sum(a[:i+1])//(i+1))
print(r)
```
| 13,263 | [
0.7021484375,
0.302001953125,
-0.345703125,
0.2137451171875,
-0.63916015625,
-0.80126953125,
-0.059539794921875,
0.2181396484375,
0.38232421875,
0.95263671875,
0.45654296875,
-0.01171112060546875,
0.4384765625,
-0.59716796875,
-0.302978515625,
-0.28662109375,
-0.62451171875,
-0.756... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
"Correct Solution:
```
N = int(input())
A = [int(x) for x in input().split()]
def check(x):
work = 0
for a in A:
work += a
if work < x:
return 0
work -= x
return 1
ans = 0
for i in range(101):
if check(i):
ans = i
print(ans)
```
| 13,265 | [
0.70263671875,
0.302001953125,
-0.260009765625,
0.217041015625,
-0.603515625,
-0.8046875,
-0.060882568359375,
0.2578125,
0.487060546875,
1.01953125,
0.499755859375,
-0.0290679931640625,
0.368408203125,
-0.65869140625,
-0.302978515625,
-0.3134765625,
-0.61474609375,
-0.7939453125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
Submitted Solution:
```
from itertools import accumulate
n=int(input())
a=list(accumulate(map(int,input().split())))
for i in range(100,0,-1):
for j in range(n):
if i*(j+1)>a[j]:break
else:
print(i)
break
```
Yes
| 13,267 | [
0.445068359375,
0.2861328125,
-0.388427734375,
0.1932373046875,
-0.65185546875,
-0.59423828125,
-0.2366943359375,
0.32177734375,
0.433349609375,
1.0654296875,
0.2210693359375,
0.0335693359375,
0.45166015625,
-0.546875,
-0.2314453125,
-0.31005859375,
-0.53759765625,
-0.9150390625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
Submitted Solution:
```
N=int(input())
A=list(map(int,input().split()))
SUM=[A[0]]
for i in range(1,N):
SUM.append(SUM[-1]+A[i])
B=[SUM[i]//(i+1) for i in range(N)]
print(min(B))
```
Yes
| 13,268 | [
0.52099609375,
0.339599609375,
-0.333984375,
0.218994140625,
-0.68408203125,
-0.62841796875,
-0.17626953125,
0.409912109375,
0.339111328125,
0.99462890625,
0.368896484375,
0.059295654296875,
0.357666015625,
-0.5517578125,
-0.352783203125,
-0.3154296875,
-0.56103515625,
-0.828125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
ans = 10 ** 7
sum = 0
for i in range(n):
sum+=a[i]
ans = min(ans,sum // (i + 1))
print(ans)
```
Yes
| 13,269 | [
0.5205078125,
0.370361328125,
-0.326904296875,
0.2413330078125,
-0.6904296875,
-0.6474609375,
-0.187255859375,
0.427978515625,
0.368896484375,
0.97802734375,
0.375244140625,
0.073974609375,
0.36328125,
-0.5380859375,
-0.38037109375,
-0.308837890625,
-0.541015625,
-0.80908203125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
ans=float("inf")
s=0
for i in range(n):
s+=a[i]
ans=min(ans,int(s/(i+1)))
print(ans)
```
Yes
| 13,270 | [
0.5322265625,
0.369140625,
-0.32763671875,
0.2425537109375,
-0.7080078125,
-0.62255859375,
-0.1805419921875,
0.428466796875,
0.364013671875,
1.0361328125,
0.400390625,
0.0677490234375,
0.401611328125,
-0.57177734375,
-0.340087890625,
-0.305419921875,
-0.52685546875,
-0.8212890625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sereja has got an array, consisting of n integers, a1, a2, ..., an. Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms:
1. Make vi-th array element equal to xi. In other words, perform the assignment avi = xi.
2. Increase each array element by yi. In other words, perform n assignments ai = ai + yi (1 β€ i β€ n).
3. Take a piece of paper and write out the qi-th array element. That is, the element aqi.
Help Sereja, complete all his operations.
Input
The first line contains integers n, m (1 β€ n, m β€ 105). The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the original array.
Next m lines describe operations, the i-th line describes the i-th operation. The first number in the i-th line is integer ti (1 β€ ti β€ 3) that represents the operation type. If ti = 1, then it is followed by two integers vi and xi, (1 β€ vi β€ n, 1 β€ xi β€ 109). If ti = 2, then it is followed by integer yi (1 β€ yi β€ 104). And if ti = 3, then it is followed by integer qi (1 β€ qi β€ n).
Output
For each third type operation print value aqi. Print the values in the order, in which the corresponding queries follow in the input.
Examples
Input
10 11
1 2 3 4 5 6 7 8 9 10
3 2
3 9
2 10
3 1
3 10
1 1 10
2 10
2 10
3 1
3 10
3 9
Output
2
9
11
20
30
40
39
Submitted Solution:
```
#python3
import sys, threading, os.path
import collections, heapq, math,bisect
import string
from platform import python_version
import itertools
sys.setrecursionlimit(10**6)
threading.stack_size(2**27)
def main():
if os.path.exists('input.txt'):
input = open('input.txt', 'r')
else:
input = sys.stdin
#--------------------------------INPUT---------------------------------
n,m = list(map(int, input.readline().split()))
lis = list(map(int, input.readline().split()))
sumall = 0
sol=[]
for i in range(m):
temlis = list(map(int, input.readline().split()))
#print(temlis)
if temlis[0]==1:
lis[temlis[1]-1]=temlis[2]-sumall
elif temlis[0]==2:
sumall+=temlis[1]
elif temlis[0]==3:
sol.append(lis[temlis[1]-1]+sumall)
#print(lis,sumall)
output = '\n'.join(map(str, sol))
#-------------------------------OUTPUT----------------------------------
if os.path.exists('output.txt'):
open('output.txt', 'w').writelines(str(output))
else:
sys.stdout.write(str(output))
if __name__ == '__main__':
main()
#threading.Thread(target=main).start()
```
Yes
| 13,608 | [
-0.097412109375,
0.25537109375,
-0.2174072265625,
0.204833984375,
-0.7958984375,
-0.06988525390625,
-0.369384765625,
0.1287841796875,
0.2335205078125,
0.9013671875,
0.482421875,
0.307861328125,
-0.2330322265625,
-0.732421875,
-0.4765625,
0.00955963134765625,
-0.5654296875,
-0.875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nathan O. Davis has been running an electronic bulletin board system named JAG-channel. He is now having hard time to add a new feature there --- threaded view.
Like many other bulletin board systems, JAG-channel is thread-based. Here a thread (also called a topic) refers to a single conversation with a collection of posts. Each post can be an opening post, which initiates a new thread, or a reply to a previous post in an existing thread.
Threaded view is a tree-like view that reflects the logical reply structure among the posts: each post forms a node of the tree and contains its replies as its subnodes in the chronological order (i.e. older replies precede newer ones). Note that a post along with its direct and indirect replies forms a subtree as a whole.
Let us take an example. Suppose: a user made an opening post with a message `hoge`; another user replied to it with `fuga`; yet another user also replied to the opening post with `piyo`; someone else replied to the second post (i.e. `fuga`β) with `foobar`; and the fifth user replied to the same post with `jagjag`. The tree of this thread would look like:
hoge
ββfuga
βγββfoobar
βγββjagjag
ββpiyo
For easier implementation, Nathan is thinking of a simpler format: the depth of each post from the opening post is represented by dots. Each reply gets one more dot than its parent post. The tree of the above thread would then look like:
hoge
.fuga
..foobar
..jagjag
.piyo
Your task in this problem is to help Nathan by writing a program that prints a tree in the Nathan's format for the given posts in a single thread.
Input
Input contains a single dataset in the following format:
n
k_1
M_1
k_2
M_2
:
:
k_n
M_n
The first line contains an integer n (1 β€ n β€ 1,000), which is the number of posts in the thread. Then 2n lines follow. Each post is represented by two lines: the first line contains an integer k_i (k_1 = 0, 1 β€ k_i < i for 2 β€ i β€ n) and indicates the i-th post is a reply to the k_i-th post; the second line contains a string M_i and represents the message of the i-th post. k_1 is always 0, which means the first post is not replying to any other post, i.e. it is an opening post.
Each message contains 1 to 50 characters, consisting of uppercase, lowercase, and numeric letters.
Output
Print the given n messages as specified in the problem statement.
Sample Input 1
1
0
icpc
Output for the Sample Input 1
icpc
Sample Input 2
5
0
hoge
1
fuga
1
piyo
2
foobar
2
jagjag
Output for the Sample Input 2
hoge
.fuga
..foobar
..jagjag
.piyo
Sample Input 3
8
0
jagjag
1
hogehoge
1
buhihi
2
fugafuga
4
ponyoponyo
5
evaeva
4
nowawa
5
pokemon
Output for the Sample Input 3
jagjag
.hogehoge
..fugafuga
...ponyoponyo
....evaeva
....pokemon
...nowawa
.buhihi
Sample Input 4
6
0
nakachan
1
fan
2
yamemasu
3
nennryou2
4
dannyaku4
5
kouzai11
Output for the Sample Input 4
nakachan
.fan
..yamemasu
...nennryou2
....dannyaku4
.....kouzai11
Sample Input 5
34
0
LoveLive
1
honoka
2
borarara
2
sunohare
2
mogyu
1
eri
6
kasikoi
7
kawaii
8
eriichika
1
kotori
10
WR
10
haetekurukotori
10
ichigo
1
umi
14
love
15
arrow
16
shoot
1
rin
18
nyanyanya
1
maki
20
6th
20
star
22
nishikino
1
nozomi
24
spiritual
25
power
1
hanayo
27
darekatasukete
28
chottomattete
1
niko
30
natsuiro
30
nikkonikkoni
30
sekaino
33
YAZAWA
Output for the Sample Input 5
LoveLive
.honoka
..borarara
..sunohare
..mogyu
.eri
..kasikoi
...kawaii
....eriichika
.kotori
..WR
..haetekurukotori
..ichigo
.umi
..love
...arrow
....shoot
.rin
..nyanyanya
.maki
..6th
..star
...nishikino
.nozomi
..spiritual
...power
.hanayo
..darekatasukete
...chottomattete
.niko
..natsuiro
..nikkonikkoni
..sekaino
...YAZAWA
Sample Input 6
6
0
2ch
1
1ostu
1
2get
1
1otsu
1
1ostu
3
pgr
Output for the Sample Input 6
2ch
.1ostu
.2get
..pgr
.1otsu
.1ostu
Example
Input
1
0
icpc
Output
icpc
Submitted Solution:
```
N = int(input())
src = []
for i in range(N):
k = int(input())
s = input()
src.append((s,[]))
if i == 0: continue
src[k-1][1].append(i)
def dfs(i,depth):
s,ch = src[i]
print('.'*depth + s)
for c in ch:
dfs(c,depth+1)
dfs(0,0)
```
No
| 14,101 | [
0.6162109375,
-0.1417236328125,
-0.06439208984375,
0.21875,
-0.36669921875,
-0.138671875,
-0.294189453125,
0.130615234375,
0.127197265625,
0.67041015625,
0.400390625,
0.036468505859375,
-0.004817962646484375,
-0.60986328125,
-0.292724609375,
0.0001691579818725586,
-0.5859375,
-0.68... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
n=int(input())
Z=[]
for i in range(n):
S,R,H,C=map(int,input().split())
Z.append((S,R,H,C))
Y=[]
for i in range(n):
for j in range(n):
if(Z[i][0]<Z[j][0] and Z[i][1]<Z[j][1] and Z[i][2]<Z[j][2]):
Y.append(Z[i])
Y=set(Y)
minn=100000
ans=0
for i in range(n):
item=Z[i]
if(item not in Y and item[3]<minn):
minn=item[3]
ans=i+1
print(ans)
```
Yes
| 14,162 | [
0.380126953125,
0.07666015625,
-0.11273193359375,
0.1722412109375,
-0.5380859375,
-0.1456298828125,
0.29931640625,
0.10601806640625,
-0.0210723876953125,
0.77783203125,
0.48583984375,
-0.09625244140625,
0.1697998046875,
-0.61962890625,
-0.45263671875,
-0.0640869140625,
-0.66357421875... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
n=int(input())
A=[]
for i in range(n):
A+=[list(map(int,input().split()))]
cost=10**18
I=0
for i in range(n):
ans=True
for j in range(n):
if(A[i][0]<A[j][0] and A[i][1]<A[j][1] and A[i][2]<A[j][2]):
ans=False
if(ans):
cost=min(cost,A[i][3])
if(cost==A[i][3]):
I=i+1
print(I)
```
Yes
| 14,163 | [
0.3876953125,
0.0731201171875,
-0.1092529296875,
0.1717529296875,
-0.52294921875,
-0.151611328125,
0.2978515625,
0.111572265625,
-0.01165771484375,
0.78515625,
0.4775390625,
-0.08740234375,
0.1558837890625,
-0.6064453125,
-0.43505859375,
-0.046661376953125,
-0.66015625,
-0.92529296... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
def chck(m,l):
p,q,r,s=l
for i in m:
a,b,e,d=i
if a>p and b>q and e>r:return 0
else:return 1
n=int(input());l,c=[],[];m=10**4
for i in range(n):
t=list(map(int,input().split()))
l.append(t);c.append(t[-1])
for i in range(n):
if chck(l,l[i]):m=min(l[i][-1],m)
print(c.index(m)+1)
```
Yes
| 14,164 | [
0.39794921875,
0.048126220703125,
-0.11236572265625,
0.1961669921875,
-0.55029296875,
-0.133544921875,
0.2685546875,
0.07269287109375,
-0.0234222412109375,
0.7646484375,
0.445556640625,
-0.10626220703125,
0.1610107421875,
-0.55615234375,
-0.4560546875,
-0.051025390625,
-0.67578125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
def readln(): return tuple(map(int, input().split()))
n, = readln()
ans = 0
best = (0, 0, 0, 0)
lst = [readln() + (_ + 1,) for _ in range(n)]
lst = [v for v in lst if not [1 for u in lst if v[0] < u[0] and v[1] < u[1] and v[2] < u[2]]]
lst.sort(key=lambda x: x[3])
print(lst[0][4])
```
Yes
| 14,165 | [
0.4208984375,
0.0517578125,
-0.10125732421875,
0.169677734375,
-0.5712890625,
-0.1759033203125,
0.32373046875,
0.06756591796875,
-0.001491546630859375,
0.77001953125,
0.43359375,
-0.09417724609375,
0.21240234375,
-0.564453125,
-0.451171875,
-0.058380126953125,
-0.728515625,
-0.9560... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
n = int(input())
l = []
mspec = [0 for i in range(4)]
index = 0
price = 1000
for i in range(n):
spec = list(map(int, input().split()))
l.append(spec)
if spec[0] >= mspec[0] and spec[1] >= mspec[1] and spec[2] >= mspec[2]:
mspec = spec
for i in range(n):
if (l[i][0] >= mspec[0] or l[i][1] >= mspec[1] or l[i][2] >= mspec[2]) and l[i][
3
] <= price:
price = l[i][3]
index = i + 1
print(index)
```
No
| 14,166 | [
0.399658203125,
0.046173095703125,
-0.0701904296875,
0.1485595703125,
-0.5048828125,
-0.1444091796875,
0.2166748046875,
0.103271484375,
-0.06353759765625,
0.77099609375,
0.435302734375,
-0.12841796875,
0.1475830078125,
-0.6923828125,
-0.403564453125,
-0.060302734375,
-0.564453125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
# ========= /\ /| |====/|
# | / \ | | / |
# | /____\ | | / |
# | / \ | | / |
# ========= / \ ===== |/====|
# code
if __name__ == "__main__":
p = []
c = []
r = []
n = int(input())
x = range(n)
i = 0
for i in x:
a,b,e,d = map(int,input().split())
c.append(d)
p.append((a,b,e))
for i in x:
for j in x:
if p[j] < p[i]:
r.append(j)
for i in r:
c[i] = 1000000
print(c.index(min(c)) + 1)
```
No
| 14,167 | [
0.37255859375,
0.050262451171875,
-0.1624755859375,
0.302978515625,
-0.62353515625,
-0.1409912109375,
0.2337646484375,
0.1937255859375,
-0.041839599609375,
0.81201171875,
0.50927734375,
-0.07598876953125,
0.1806640625,
-0.59521484375,
-0.48291015625,
-0.053924560546875,
-0.6069335937... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
from sys import*
input= stdin.readline
t=int(input())
speed=[]
ram=[]
hdd=[]
cost=[]
res=[]
for i in range(t):
l=[0]*3
res.append(l)
a,b,c,d=map(int,input().split())
speed.append([a,i])
ram.append([b,i])
hdd.append([c,i])
cost.append(d)
speed.sort()
ram.sort()
hdd.sort()
for i in range(t):
res[speed[i][1]][0]=i
res[ram[i][1]][1]=i
res[hdd[i][1]][2]=i
for i in range(t):
if(t-1 not in res[i]):
cost[i]=10000
print(cost.index(min(cost))+1)
```
No
| 14,168 | [
0.376708984375,
0.08819580078125,
-0.101318359375,
0.1549072265625,
-0.5517578125,
-0.1285400390625,
0.270263671875,
0.07781982421875,
-0.034759521484375,
0.75146484375,
0.4384765625,
-0.09429931640625,
0.1539306640625,
-0.59814453125,
-0.467529296875,
-0.0606689453125,
-0.6577148437... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
n = int(input())
lap = [list(map(int,input().split())) for _ in range(n) ]
outdated = []
for i in range(n):
for j in range(n):
if lap[i][0] > lap[j][0] and lap[i][1] > lap[j][1] and lap[i][2] > lap[j][2]:
outdated.append(j)
q = len(outdated)
out = 0
min_c = float("inf")
min_i = 0
for i in range(n):
if out < q and outdated[out] == i:
out += 1
else:
if lap[i][3] < min_c:
min_c = lap[i][3]
min_i = i
print(i)
```
No
| 14,169 | [
0.3759765625,
0.07763671875,
-0.09844970703125,
0.2080078125,
-0.5126953125,
-0.14501953125,
0.297607421875,
0.0675048828125,
0.033050537109375,
0.771484375,
0.51025390625,
-0.0870361328125,
0.109130859375,
-0.59912109375,
-0.4775390625,
-0.017578125,
-0.66015625,
-0.94775390625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
def calc(A):
N = len(A)
if N == 1:
return 0
X = [0] * N
for a in A:
X[a] += 1
if max(X) > (N + 1) // 2:
return -1
Y = [0] * N
Y[A[0]] += 1
Y[A[-1]] += 1
for a, b in zip(A, A[1:]):
if a == b:
Y[a] += 2
su, ma = sum(Y), max(Y)
cc = su - ma
return su // 2 - 1 + max(ma - cc - 2, 0) // 2
T = int(input())
for _ in range(T):
N = int(input())
A = [int(a) - 1 for a in input().split()]
print(calc(A))
```
Yes
| 14,377 | [
0.1776123046875,
-0.056549072265625,
-0.1234130859375,
0.29296875,
-0.47607421875,
-0.65625,
-0.1158447265625,
0.217041015625,
0.0814208984375,
0.80517578125,
0.77978515625,
-0.276123046875,
0.144775390625,
-0.71240234375,
-0.65673828125,
-0.1729736328125,
-0.49462890625,
-0.858886... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
def calc(A):
N = len(A)
if N == 1:
return 0
X = [0] * N
for a in A:
X[a] += 1
if max(X) > (N + 1) // 2:
return -1
Y = [0] * N
Y[A[0]] += 1
Y[A[-1]] += 1
for a, b in zip(A, A[1:]):
if a == b:
Y[a] += 2
su, ma = sum(Y), max(Y)
cc = su - ma
return su // 2 - 1 + max(ma - cc - 2, 0) // 2
T = int(input())
for _ in range(T):
N = int(input())
A = [int(a) - 1 for a in input().split()]
print(calc(A))
```
Yes
| 14,378 | [
0.1776123046875,
-0.056549072265625,
-0.1234130859375,
0.29296875,
-0.47607421875,
-0.65625,
-0.1158447265625,
0.217041015625,
0.0814208984375,
0.80517578125,
0.77978515625,
-0.276123046875,
0.144775390625,
-0.71240234375,
-0.65673828125,
-0.1729736328125,
-0.49462890625,
-0.858886... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
c = [0]*(n+1)
for e in a:
c[e]+=1
if max(c) >= (n+1)//2 +1:
print(-1)
continue
count = 1
s=0
c = [0]*(n+1)
for i in range(n-1):
if a[i]==a[i+1]:
count +=1
c[a[s]]+=1
c[a[i]]+=1
s = i+1
c[a[s]]+=1
c[a[n-1]]+=1
mx = max(c)
ss = sum(c)
if mx-2 <= ss-mx :
print(count-1)
else:
count += (mx-2-(ss-mx))//2
print(count-1)
```
Yes
| 14,379 | [
0.1776123046875,
-0.056549072265625,
-0.1234130859375,
0.29296875,
-0.47607421875,
-0.65625,
-0.1158447265625,
0.217041015625,
0.0814208984375,
0.80517578125,
0.77978515625,
-0.276123046875,
0.144775390625,
-0.71240234375,
-0.65673828125,
-0.1729736328125,
-0.49462890625,
-0.858886... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys,io,os;Z=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
o=[]
for _ in range(int(Z())):
n=int(Z());a=[*map(int,Z().split())]
cn=p=a[0];pn=d=0;e=[0]*n;b=[0]*n;c=[0]*n
for i in range(n):
if a[i]==p:
if pn:
if pn!=p:b[pn-1]+=1
else:e[p-1]+=1
b[p-1]+=1;d+=1
pn=p
p=a[i];c[p-1]+=1
if pn!=p:b[pn-1]+=1
else:e[p-1]+=1
b[p-1]+=1
if 2*max(c)-1>n:o.append('-1');continue
m=0;s=sum(b)
for i in range(n):
dl=max(0,e[i]-2-d+b[i]);m=max(dl,m)
o.append(str(d+m))
print('\n'.join(o))
```
Yes
| 14,380 | [
0.1776123046875,
-0.056549072265625,
-0.1234130859375,
0.29296875,
-0.47607421875,
-0.65625,
-0.1158447265625,
0.217041015625,
0.0814208984375,
0.80517578125,
0.77978515625,
-0.276123046875,
0.144775390625,
-0.71240234375,
-0.65673828125,
-0.1729736328125,
-0.49462890625,
-0.858886... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
c = [0]*(n+1)
for e in a:
c[e]+=1
if max(c) >= (n+1)//2 +1:
print(-1)
continue
count = 1
s=0
c = [0]*(n+1)
for i in range(n-1):
if a[i]==a[i+1]:
count +=1
if a[s]!=a[i]:
c[a[s]]+=1
c[a[i]]+=1
else:
c[a[i]]+=1
s = i+1
if a[s]!=a[n-1]:
c[a[s]]+=1
c[a[n-1]]+=1
else:
c[a[s]]+=1
mx = max(c)
ss = sum(c)
if mx < (ss+1)//2 +1:
print(count-1)
else:
while mx >= (ss+1)//2 +1:
count +=1
ss += 2
print(count-1)
```
No
| 14,381 | [
0.1776123046875,
-0.056549072265625,
-0.1234130859375,
0.29296875,
-0.47607421875,
-0.65625,
-0.1158447265625,
0.217041015625,
0.0814208984375,
0.80517578125,
0.77978515625,
-0.276123046875,
0.144775390625,
-0.71240234375,
-0.65673828125,
-0.1729736328125,
-0.49462890625,
-0.858886... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
q = int(input())
for _ in range(q):
n = int(input())
l = list(map(int,input().split()))
ile = [0] * (n+1)
for i in l:
ile[i] += 1
if max(ile) >= (n+1)//2 + 1:
print(-1)
else:
if n == 1:
print(0)
else:
cyk = []
count = 0
cur = l[0]
for i in range(n):
if l[i] != cur:
cyk.append([cur,count])
count = 1
cur = l[i]
else:
count += 1
if count != 1:
cyk.append([cur, count])
if l[-1] != l[-2]:
cyk.append([cur,count])
wyn = 0
for i in cyk:
wyn += max(0, i[1]-1)
if cyk[0][1] == 1 and cyk[-1][1] == 1 and cyk[0][0] == cyk[-1][0]:
for j in range(1, len(cyk)-1):
if cyk[j][0] == cyk[0][0] and cyk[j][1] > 1:
wyn += 1
break
print(wyn)
```
No
| 14,382 | [
0.1776123046875,
-0.056549072265625,
-0.1234130859375,
0.29296875,
-0.47607421875,
-0.65625,
-0.1158447265625,
0.217041015625,
0.0814208984375,
0.80517578125,
0.77978515625,
-0.276123046875,
0.144775390625,
-0.71240234375,
-0.65673828125,
-0.1729736328125,
-0.49462890625,
-0.858886... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
q = int(input())
for _ in range(q):
n = int(input())
l = list(map(int,input().split()))
ile = [0] * (n+1)
for i in l:
ile[i] += 1
if max(ile) >= (n+1)//2 + 1:
print(-1)
else:
if n == 1:
print(0)
else:
frag = []
now = [l[0]]
for i in range(1,n):
if l[i] == l[i-1]:
frag.append(now)
now = [l[i]]
else:
now.append(l[i])
frag.append(now)
konce = []
for i in frag:
if i[0] == i[-1]:
konce.append(i[0])
k = len(frag)
d = {}
for i in konce:
d[i] = 0
for i in konce:
d[i] += 1
m = 0
for i in d:
m = max(m, d[i])
dif = max(0, 2*m-k-1)
print(k + dif-1)
```
No
| 14,383 | [
0.1776123046875,
-0.056549072265625,
-0.1234130859375,
0.29296875,
-0.47607421875,
-0.65625,
-0.1158447265625,
0.217041015625,
0.0814208984375,
0.80517578125,
0.77978515625,
-0.276123046875,
0.144775390625,
-0.71240234375,
-0.65673828125,
-0.1729736328125,
-0.49462890625,
-0.858886... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys
input = sys.stdin.readline
outL = []
t = int(input())
for _ in range(t):
n = int(input())
l = list(map(lambda x: int(x) - 1, input().split()))
intervals = []
start = 0
for i in range(1,n):
if l[i] == l[i-1]:
intervals.append((start, i-1))
start = i
intervals.append((start,n-1))
count = [0] * n
for le,ri in intervals:
count[l[le]] += 1
count[l[ri]] += 1
want = len(intervals)-1
if max(count) <= want+2:
outL.append(want)
continue
count2 = [0] * n
for v in l:
count2[v] += 1
for issue in range(n):
if count2[issue] == max(count2):
break
num_issue = max(count2)
rest = n - issue
best = max(count) - 2
if num_issue > rest + 1:
outL.append(-1)
else:
outL.append(best)
print('\n'.join(map(str,outL)))
```
No
| 14,384 | [
0.1776123046875,
-0.056549072265625,
-0.1234130859375,
0.29296875,
-0.47607421875,
-0.65625,
-0.1158447265625,
0.217041015625,
0.0814208984375,
0.80517578125,
0.77978515625,
-0.276123046875,
0.144775390625,
-0.71240234375,
-0.65673828125,
-0.1729736328125,
-0.49462890625,
-0.858886... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
People do many crazy things to stand out in a crowd. Some of them dance, some learn by heart rules of Russian language, some try to become an outstanding competitive programmers, while others collect funny math objects.
Alis is among these collectors. Right now she wants to get one of k-special tables. In case you forget, the table n Γ n is called k-special if the following three conditions are satisfied:
* every integer from 1 to n2 appears in the table exactly once;
* in each row numbers are situated in increasing order;
* the sum of numbers in the k-th column is maximum possible.
Your goal is to help Alice and find at least one k-special table of size n Γ n. Both rows and columns are numbered from 1 to n, with rows numbered from top to bottom and columns numbered from left to right.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 500, 1 β€ k β€ n) β the size of the table Alice is looking for and the column that should have maximum possible sum.
Output
First print the sum of the integers in the k-th column of the required table.
Next n lines should contain the description of the table itself: first line should contains n elements of the first row, second line should contain n elements of the second row and so on.
If there are multiple suitable table, you are allowed to print any.
Examples
Input
4 1
Output
28
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Input
5 3
Output
85
5 6 17 18 19
9 10 23 24 25
7 8 20 21 22
3 4 14 15 16
1 2 11 12 13
Submitted Solution:
```
def main():
(n, k) = (int(x) for x in input().split())
(total, table) = solver(n, k)
print(total)
for row in range(n):
for col in range(n):
print(table[row][col], end = " ")
print()
def solver(n, k):
col = k - 1
smallest = col * n + 1
table = [list(range(col * row + 1, col * (row + 1) + 1)) +
list(range((n - col) * row + smallest, (n - col) * (row + 1) + smallest))
for row in range(n)]
total = n * smallest + n * (n - 1) * (n - col) // 2
return (total, table)
#print(solver(5, 2))
main()
```
Yes
| 14,648 | [
0.6025390625,
-0.034088134765625,
-0.2420654296875,
0.14208984375,
-0.6337890625,
-0.463623046875,
-0.132568359375,
0.60400390625,
0.01001739501953125,
0.499267578125,
0.81689453125,
-0.0121612548828125,
0.00644683837890625,
-0.537109375,
-0.64306640625,
0.1729736328125,
-0.656738281... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
People do many crazy things to stand out in a crowd. Some of them dance, some learn by heart rules of Russian language, some try to become an outstanding competitive programmers, while others collect funny math objects.
Alis is among these collectors. Right now she wants to get one of k-special tables. In case you forget, the table n Γ n is called k-special if the following three conditions are satisfied:
* every integer from 1 to n2 appears in the table exactly once;
* in each row numbers are situated in increasing order;
* the sum of numbers in the k-th column is maximum possible.
Your goal is to help Alice and find at least one k-special table of size n Γ n. Both rows and columns are numbered from 1 to n, with rows numbered from top to bottom and columns numbered from left to right.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 500, 1 β€ k β€ n) β the size of the table Alice is looking for and the column that should have maximum possible sum.
Output
First print the sum of the integers in the k-th column of the required table.
Next n lines should contain the description of the table itself: first line should contains n elements of the first row, second line should contain n elements of the second row and so on.
If there are multiple suitable table, you are allowed to print any.
Examples
Input
4 1
Output
28
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Input
5 3
Output
85
5 6 17 18 19
9 10 23 24 25
7 8 20 21 22
3 4 14 15 16
1 2 11 12 13
Submitted Solution:
```
def main():
n, k = [int(t) for t in input().split()]
k -= 1
lst = [[0 for t in range(n)] for t in range(n)]
for col in range(n-1, k-1, -1):
for row in range(n):
lst[row][col] = n*n - (n-1-col) - row*(n-k)
for col in range(k-1, -1, -1):
for row in range(n):
lst[row][col] = n*k - (k-1-col) - row*k
print(sum([lst[i][k] for i in range(n)]))
out = "\n".join(" ".join(str(v) for v in lst[i]) for i in range(n))
print(out)
main()
# PYTHON3!!
```
Yes
| 14,649 | [
0.587890625,
0.008056640625,
-0.17919921875,
0.1590576171875,
-0.68994140625,
-0.5078125,
-0.061981201171875,
0.63427734375,
-0.06744384765625,
0.54296875,
0.73486328125,
0.0035381317138671875,
-0.007312774658203125,
-0.56396484375,
-0.62353515625,
0.15673828125,
-0.6865234375,
-1.... | 11 |
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