message stringlengths 2 45.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 254 108k | cluster float64 3 3 | __index_level_0__ int64 508 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
Input
The first line contains two space-separated integers n and m <image>.
Each of the next m lines contains two space-separated integers xi and yi (1 β€ xi < yi β€ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to n in some order.
Output
Print a single integer β the maximum possible danger.
Examples
Input
1 0
Output
1
Input
2 1
1 2
Output
2
Input
3 2
1 2
2 3
Output
4
Note
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
Submitted Solution:
```
from sys import stdout
from sys import stdin
def get():
return stdin.readline().strip()
def getf():
return [int(i) for i in get().split()]
def put(a, end = "\n"):
stdout.write(str(a) + end)
def putf(a, sep = " ", end = "\n"):
stdout.write(sep.join([str(i) for i in a]) + end)
def bfs(g, v, i):
v[i] = False
q = [i]
usd = set()
while q:
t = q.pop()
usd.add(t)
for j in g[t]:
v[j] = False
if j not in usd:
q.append(j)
return usd
from collections import defaultdict as dd
def main():
n, k = getf()
g = dd(list)
for i in range(k):
x, y = getf()
g[x - 1] += [y - 1]
g[y - 1] += [x - 1]
res = 0
v = [True] * n
for u in range(n):
if(v[u] == True):
bfs(g, v, u)
res += 1
put(1 << (n - res))
main()
``` | instruction | 0 | 53,430 | 3 | 106,860 |
Yes | output | 1 | 53,430 | 3 | 106,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
Input
The first line contains two space-separated integers n and m <image>.
Each of the next m lines contains two space-separated integers xi and yi (1 β€ xi < yi β€ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to n in some order.
Output
Print a single integer β the maximum possible danger.
Examples
Input
1 0
Output
1
Input
2 1
1 2
Output
2
Input
3 2
1 2
2 3
Output
4
Note
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
Submitted Solution:
```
n, m = map(int, input().split())
p = [[] for _ in range(n)]
for _ in range(m):
u, v = map(int, input().split())
p[u-1].append(v-1)
p[v-1].append(u-1)
visited = [1]*n
def dfs(a):
cnt = 1
for x in p[a]:
if visited[x]:
visited[x] = 0
cnt += dfs(x)
return cnt
seq = []
for i in range(n):
if visited[i]:
visited[i] = 0
seq.append(dfs(i))
ans = 1
for x in seq:
ans *= 2**(x-1)
print(ans)
``` | instruction | 0 | 53,431 | 3 | 106,862 |
Yes | output | 1 | 53,431 | 3 | 106,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
Input
The first line contains two space-separated integers n and m <image>.
Each of the next m lines contains two space-separated integers xi and yi (1 β€ xi < yi β€ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to n in some order.
Output
Print a single integer β the maximum possible danger.
Examples
Input
1 0
Output
1
Input
2 1
1 2
Output
2
Input
3 2
1 2
2 3
Output
4
Note
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
Submitted Solution:
```
from sys import stdin ,stdout
from os import path
if(path.exists('input.txt')):
stdin = open("input.txt","r")
from collections import defaultdict
danger = 1
def dfs(visited,node,amr):
visited[node] = 1
for item in amr[node]:
if visited[item] == 0 :
global danger
danger*= 2
dfs(visited,item,amr)
x,y = map(int,stdin.readline().strip().split())
amr = defaultdict(list)
for _ in range(y):
a,b = map(int,stdin.readline().strip().split())
amr[a].append(b)
amr[b].append(a)
visited = [0]*(x+1)
for i in range(1,x+1):
if visited[i] == 0:
dfs(visited,i,amr)
print(danger)
``` | instruction | 0 | 53,432 | 3 | 106,864 |
Yes | output | 1 | 53,432 | 3 | 106,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
Input
The first line contains two space-separated integers n and m <image>.
Each of the next m lines contains two space-separated integers xi and yi (1 β€ xi < yi β€ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to n in some order.
Output
Print a single integer β the maximum possible danger.
Examples
Input
1 0
Output
1
Input
2 1
1 2
Output
2
Input
3 2
1 2
2 3
Output
4
Note
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
Submitted Solution:
```
n, m = map(int, input().split())
p = [i for i in range(n+1)]
r = [0 for i in range(n+1)]
k = 0
def get(x):
while p[x] != x:
x = p[x]
return x
def union(x, y):
x = get(x)
y = get(y)
if x == y:
return 0
if r[x] > r[y]:
x, y = y, x
p[x] = y
if r[x] == r[y]:
r[y] += 1
return 1
for i in range(m):
x, y = map(int, input().split())
k += union(x, y)
print(2**k)
``` | instruction | 0 | 53,433 | 3 | 106,866 |
Yes | output | 1 | 53,433 | 3 | 106,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
Input
The first line contains two space-separated integers n and m <image>.
Each of the next m lines contains two space-separated integers xi and yi (1 β€ xi < yi β€ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to n in some order.
Output
Print a single integer β the maximum possible danger.
Examples
Input
1 0
Output
1
Input
2 1
1 2
Output
2
Input
3 2
1 2
2 3
Output
4
Note
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
Submitted Solution:
```
a,b=map(int,input().split())
s=[]
for n in range(b):
s.append(list(map(int,input().split())))
print(2*b)
``` | instruction | 0 | 53,434 | 3 | 106,868 |
No | output | 1 | 53,434 | 3 | 106,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
Input
The first line contains two space-separated integers n and m <image>.
Each of the next m lines contains two space-separated integers xi and yi (1 β€ xi < yi β€ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to n in some order.
Output
Print a single integer β the maximum possible danger.
Examples
Input
1 0
Output
1
Input
2 1
1 2
Output
2
Input
3 2
1 2
2 3
Output
4
Note
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
Submitted Solution:
```
data = [int(x) for x in input().split(' ')]
n = data[0]
m = data[1]
if n==1:
print("1")
else:
dados = []
for case in range(m):
dados.append([int(x) for x in input().split(' ')])
print(pow(2,n-1))
``` | instruction | 0 | 53,435 | 3 | 106,870 |
No | output | 1 | 53,435 | 3 | 106,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
Input
The first line contains two space-separated integers n and m <image>.
Each of the next m lines contains two space-separated integers xi and yi (1 β€ xi < yi β€ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to n in some order.
Output
Print a single integer β the maximum possible danger.
Examples
Input
1 0
Output
1
Input
2 1
1 2
Output
2
Input
3 2
1 2
2 3
Output
4
Note
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
Submitted Solution:
```
import bisect
import collections
import copy
import functools
import heapq
import itertools
import math
import random
import re
import sys
import time
import string
from typing import List
# sys.setrecursionlimit(999999)
n,m = map(int,input().split())
tree = list(range(n+1))
def find(st):
if tree[st]!=st:
tree[st] = find(tree[st])
return tree[st]
for _ in range(m):
u,v = map(int,input().split())
tree[find(u)] = find(tree[v])
cnt = collections.defaultdict(int)
for i in range(1,n+1):
cnt[find(i)]+=1
ans = 0
for k,v in cnt.items():
ans+= (1<<(v-1))
print(ans)
``` | instruction | 0 | 53,436 | 3 | 106,872 |
No | output | 1 | 53,436 | 3 | 106,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
Input
The first line contains two space-separated integers n and m <image>.
Each of the next m lines contains two space-separated integers xi and yi (1 β€ xi < yi β€ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to n in some order.
Output
Print a single integer β the maximum possible danger.
Examples
Input
1 0
Output
1
Input
2 1
1 2
Output
2
Input
3 2
1 2
2 3
Output
4
Note
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
Submitted Solution:
```
class DisjointSet: #only support query, NO construction supported
def __init__(self, n):
self.num_sets = n #--@union
self.parents = list(range(n))
self.ranks = [0] * n
self.sizes = [1] * n #used@union: size of set@root! 0 if NOT repsentive
def find(self,j): #no compress
while self.parents[j]!=j:
j = self.parents[j]
return j
def union(self,i,j):
i = self.find(i)
j = self.find(j)
if i == j:
return False
rd = self.ranks[i] - self.ranks[j]
if rd == 0: # Increment repr0's rank if both nodes have same rank
self.ranks[i] += 1
elif rd < 0: # Swap to ensure that repr0's rank >= repr1's rank
i, j = j, i
self.parents[j] = i #merge
self.sizes[i] += self.sizes[j]
self.sizes[j] = 0
self.num_sets -= 1
return True
def main():
n,m = list(map(int,input().split())) #100,100
cl = [list(map(int,input().split())) for _ in range(m)]
ds = DisjointSet(n)
for a,b in cl:
ds.union(a-1,b-1)
print(2**(max(ds.sizes)-1))
main()
``` | instruction | 0 | 53,437 | 3 | 106,874 |
No | output | 1 | 53,437 | 3 | 106,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2Β·n people in the group (including Vadim), and they have exactly n - 1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. i-th person's weight is wi, and weight is an important matter in kayaking β if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
Input
The first line contains one number n (2 β€ n β€ 50).
The second line contains 2Β·n integer numbers w1, w2, ..., w2n, where wi is weight of person i (1 β€ wi β€ 1000).
Output
Print minimum possible total instability.
Examples
Input
2
1 2 3 4
Output
1
Input
4
1 3 4 6 3 4 100 200
Output
5
Submitted Solution:
```
n = int(input())
A = list(map(int, input().split()))
D = []
for i in range(2*n-1):
for j in range(i+1, 2*n):
d = abs(A[i]-A[j])
D.append((d, i, j))
D.sort()
cnt = 0
ans = 0
for i in range(len(D)):
d, j, k = D[i]
if A[j] != 0 and A[k] != 0:
ans += d
cnt += 1
A[j] = 0
A[k] = 0
if cnt == n-1:
break
print(ans)
``` | instruction | 0 | 53,601 | 3 | 107,202 |
No | output | 1 | 53,601 | 3 | 107,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare x^y with y^x. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare x^y with y^x for Vasya, maybe then other androids will respect him.
Input
On the only line of input there are two integers x and y (1 β€ x, y β€ 10^{9}).
Output
If x^y < y^x, then print '<' (without quotes). If x^y > y^x, then print '>' (without quotes). If x^y = y^x, then print '=' (without quotes).
Examples
Input
5 8
Output
>
Input
10 3
Output
<
Input
6 6
Output
=
Note
In the first example 5^8 = 5 β
5 β
5 β
5 β
5 β
5 β
5 β
5 = 390625, and 8^5 = 8 β
8 β
8 β
8 β
8 = 32768. So you should print '>'.
In the second example 10^3 = 1000 < 3^{10} = 59049.
In the third example 6^6 = 46656 = 6^6.
Submitted Solution:
```
a, b = map(int, input().split())
if a == 2 and b == 3:
print("<")
elif a == 2 and b == 4:
print("=")
elif a == 3 and b == 2:
print(">")
elif a == 4 and b == 2:
print("=")
elif a == b:
print("=")
elif a == 1:
print("<")
elif b == 1:
print(">")
elif a<b:
print(">")
else:
print("<")
``` | instruction | 0 | 53,648 | 3 | 107,296 |
Yes | output | 1 | 53,648 | 3 | 107,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare x^y with y^x. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare x^y with y^x for Vasya, maybe then other androids will respect him.
Input
On the only line of input there are two integers x and y (1 β€ x, y β€ 10^{9}).
Output
If x^y < y^x, then print '<' (without quotes). If x^y > y^x, then print '>' (without quotes). If x^y = y^x, then print '=' (without quotes).
Examples
Input
5 8
Output
>
Input
10 3
Output
<
Input
6 6
Output
=
Note
In the first example 5^8 = 5 β
5 β
5 β
5 β
5 β
5 β
5 β
5 = 390625, and 8^5 = 8 β
8 β
8 β
8 β
8 = 32768. So you should print '>'.
In the second example 10^3 = 1000 < 3^{10} = 59049.
In the third example 6^6 = 46656 = 6^6.
Submitted Solution:
```
import math
x, y=map(int, input().split())
if x==2 and y==4:
print('=')
elif y==2 and x==4:
print('=')
elif x==y:
print('=')
elif (y*math.log(x)<x*math.log(y)):
print ('<')
else:
print('>')
``` | instruction | 0 | 53,649 | 3 | 107,298 |
Yes | output | 1 | 53,649 | 3 | 107,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare x^y with y^x. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare x^y with y^x for Vasya, maybe then other androids will respect him.
Input
On the only line of input there are two integers x and y (1 β€ x, y β€ 10^{9}).
Output
If x^y < y^x, then print '<' (without quotes). If x^y > y^x, then print '>' (without quotes). If x^y = y^x, then print '=' (without quotes).
Examples
Input
5 8
Output
>
Input
10 3
Output
<
Input
6 6
Output
=
Note
In the first example 5^8 = 5 β
5 β
5 β
5 β
5 β
5 β
5 β
5 = 390625, and 8^5 = 8 β
8 β
8 β
8 β
8 = 32768. So you should print '>'.
In the second example 10^3 = 1000 < 3^{10} = 59049.
In the third example 6^6 = 46656 = 6^6.
Submitted Solution:
```
import math
x,y=map(int,input().split())
a=y*math.log(x)
b=x*math.log(y)
if(a>b):
print(">")
elif(a<b):
print("<")
else:
print("=")
``` | instruction | 0 | 53,650 | 3 | 107,300 |
Yes | output | 1 | 53,650 | 3 | 107,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare x^y with y^x. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare x^y with y^x for Vasya, maybe then other androids will respect him.
Input
On the only line of input there are two integers x and y (1 β€ x, y β€ 10^{9}).
Output
If x^y < y^x, then print '<' (without quotes). If x^y > y^x, then print '>' (without quotes). If x^y = y^x, then print '=' (without quotes).
Examples
Input
5 8
Output
>
Input
10 3
Output
<
Input
6 6
Output
=
Note
In the first example 5^8 = 5 β
5 β
5 β
5 β
5 β
5 β
5 β
5 = 390625, and 8^5 = 8 β
8 β
8 β
8 β
8 = 32768. So you should print '>'.
In the second example 10^3 = 1000 < 3^{10} = 59049.
In the third example 6^6 = 46656 = 6^6.
Submitted Solution:
```
x, y = map(int, input().split())
if x > 2 and y > 2:
if x > y:
print("<")
elif y > x:
print(">")
else:
print("=")
elif x == 2 and y == 4:
print("=")
elif x == 4 and y == 2:
print("=")
elif x > 2:
print(">")
elif y > 2:
print("<")
else:
if x ** y > y ** x:
print(">")
elif y ** x > x ** y:
print("<")
else:
print("=")
``` | instruction | 0 | 53,651 | 3 | 107,302 |
No | output | 1 | 53,651 | 3 | 107,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare x^y with y^x. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare x^y with y^x for Vasya, maybe then other androids will respect him.
Input
On the only line of input there are two integers x and y (1 β€ x, y β€ 10^{9}).
Output
If x^y < y^x, then print '<' (without quotes). If x^y > y^x, then print '>' (without quotes). If x^y = y^x, then print '=' (without quotes).
Examples
Input
5 8
Output
>
Input
10 3
Output
<
Input
6 6
Output
=
Note
In the first example 5^8 = 5 β
5 β
5 β
5 β
5 β
5 β
5 β
5 = 390625, and 8^5 = 8 β
8 β
8 β
8 β
8 = 32768. So you should print '>'.
In the second example 10^3 = 1000 < 3^{10} = 59049.
In the third example 6^6 = 46656 = 6^6.
Submitted Solution:
```
import math
x,y=map(int, input().split())
if (x==y):
print('=')
elif x*math.log(y)>y*math.log(x):
print('<')
else:
print('>')
``` | instruction | 0 | 53,652 | 3 | 107,304 |
No | output | 1 | 53,652 | 3 | 107,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare x^y with y^x. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare x^y with y^x for Vasya, maybe then other androids will respect him.
Input
On the only line of input there are two integers x and y (1 β€ x, y β€ 10^{9}).
Output
If x^y < y^x, then print '<' (without quotes). If x^y > y^x, then print '>' (without quotes). If x^y = y^x, then print '=' (without quotes).
Examples
Input
5 8
Output
>
Input
10 3
Output
<
Input
6 6
Output
=
Note
In the first example 5^8 = 5 β
5 β
5 β
5 β
5 β
5 β
5 β
5 = 390625, and 8^5 = 8 β
8 β
8 β
8 β
8 = 32768. So you should print '>'.
In the second example 10^3 = 1000 < 3^{10} = 59049.
In the third example 6^6 = 46656 = 6^6.
Submitted Solution:
```
# maa chudaaye duniya
from math import log
x, y = map(int, input().split())
if x == y:
print('=')
exit()
if y * log(x, 10) > x * log(y, 10):
print('>')
else:
print('<')
``` | instruction | 0 | 53,653 | 3 | 107,306 |
No | output | 1 | 53,653 | 3 | 107,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare x^y with y^x. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare x^y with y^x for Vasya, maybe then other androids will respect him.
Input
On the only line of input there are two integers x and y (1 β€ x, y β€ 10^{9}).
Output
If x^y < y^x, then print '<' (without quotes). If x^y > y^x, then print '>' (without quotes). If x^y = y^x, then print '=' (without quotes).
Examples
Input
5 8
Output
>
Input
10 3
Output
<
Input
6 6
Output
=
Note
In the first example 5^8 = 5 β
5 β
5 β
5 β
5 β
5 β
5 β
5 = 390625, and 8^5 = 8 β
8 β
8 β
8 β
8 = 32768. So you should print '>'.
In the second example 10^3 = 1000 < 3^{10} = 59049.
In the third example 6^6 = 46656 = 6^6.
Submitted Solution:
```
def power_(x, n):
if n == 0:
return 1
elif n < 0:
return 1 / power_(x, -n)
else:
answer = power_(x, n // 2)
answer *= answer
if n % 2:
answer *= x
return answer
x, y = map(int, input().split())
a = power_(x, y)
b = power_(y, x)
if (a < b):
print(">")
elif (a == b):
print("=")
else:
print("<")
``` | instruction | 0 | 53,654 | 3 | 107,308 |
No | output | 1 | 53,654 | 3 | 107,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 β
a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 β€ n β€ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 β€ a_i, b_i β€ 5) β the parameters of the i-th light.
Output
Print a single integer β the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4.
Submitted Solution:
```
numBulbs = int(input())
initalState = list(map(int, list(input())))
maxInterval = 100000
delta = [0] * maxInterval
currentOn = 0
ON = 1
OFF = 0
for bulbIndex in range(numBulbs):
if(initalState[bulbIndex] == ON):
delta[0] += 1
for bulbIndex in range(numBulbs):
interval, offset = map(int, input().split(' '))
currentState = initalState[bulbIndex]
for intervalIndex in range(15000):
currentValue = interval * intervalIndex + offset
newState = 0 if currentState else 1
delta[currentValue] += newState - currentState
currentState = newState
maxOn = 0
for index in range(maxInterval):
currentOn += delta[index]
if(currentOn > maxOn):
maxOn = currentOn
print(maxOn)
``` | instruction | 0 | 54,032 | 3 | 108,064 |
No | output | 1 | 54,032 | 3 | 108,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[MisoilePunchβͺ - 彩](https://www.youtube.com/watch?v=5VleIdkEeak)
This is an interactive problem!
On a normal day at the hidden office in A.R.C. Markland-N, Rin received an artifact, given to her by the exploration captain Sagar.
After much analysis, she now realizes that this artifact contains data about a strange flower, which has existed way before the New Age. However, the information about its chemical structure has been encrypted heavily.
The chemical structure of this flower can be represented as a string p. From the unencrypted papers included, Rin already knows the length n of that string, and she can also conclude that the string contains at most three distinct letters: "C" (as in Carbon), "H" (as in Hydrogen), and "O" (as in Oxygen).
At each moment, Rin can input a string s of an arbitrary length into the artifact's terminal, and it will return every starting position of s as a substring of p.
However, the artifact has limited energy and cannot be recharged in any way, since the technology is way too ancient and is incompatible with any current A.R.C.'s devices. To be specific:
* The artifact only contains 7/5 units of energy.
* For each time Rin inputs a string s of length t, the artifact consumes (1)/(t^2) units of energy.
* If the amount of energy reaches below zero, the task will be considered failed immediately, as the artifact will go black forever.
Since the artifact is so precious yet fragile, Rin is very nervous to attempt to crack the final data. Can you give her a helping hand?
Interaction
The interaction starts with a single integer t (1 β€ t β€ 500), the number of test cases. The interaction for each testcase is described below:
First, read an integer n (4 β€ n β€ 50), the length of the string p.
Then you can make queries of type "? s" (1 β€ |s| β€ n) to find the occurrences of s as a substring of p.
After the query, you need to read its result as a series of integers in a line:
* The first integer k denotes the number of occurrences of s as a substring of p (-1 β€ k β€ n). If k = -1, it means you have exceeded the energy limit or printed an invalid query, and you need to terminate immediately, to guarantee a "Wrong answer" verdict, otherwise you might get an arbitrary verdict because your solution will continue to read from a closed stream.
* The following k integers a_1, a_2, β¦, a_k (1 β€ a_1 < a_2 < β¦ < a_k β€ n) denote the starting positions of the substrings that match the string s.
When you find out the string p, print "! p" to finish a test case. This query doesn't consume any energy. The interactor will return an integer 1 or 0. If the interactor returns 1, you can proceed to the next test case, or terminate the program if it was the last testcase.
If the interactor returns 0, it means that your guess is incorrect, and you should to terminate to guarantee a "Wrong answer" verdict.
Note that in every test case the string p is fixed beforehand and will not change during the queries, i.e. the interactor is not adaptive.
After printing any query do not forget to print end of line and flush the output. Otherwise, you might get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see the documentation for other languages.
Hacks
For hack, use the following format. Note that you can only hack with one test case:
The first line should contain a single integer t (t = 1).
The second line should contain an integer n (4 β€ n β€ 50) β the string's size.
The third line should contain a string of size n, consisting of characters "C", "H" and "O" only. This is the string contestants will have to find out.
Examples
Input
1
4
2 1 2
1 2
0
1
Output
? C
? CH
? CCHO
! CCHH
Input
2
5
0
2 2 3
1
8
1 5
1 5
1 3
2 1 2
1
Output
? O
? HHH
! CHHHH
? COO
? COOH
? HCCOO
? HH
! HHHCCOOH
Note
Note that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
ans = [''] * (n+1)
print('? CO')
co = list(map(int, input().split()))[1:]
print('? CH')
ch = list(map(int, input().split()))[1:]
print('? CC')
cc = list(map(int, input().split()))[1:]
for d in co:
ans[d] = 'C'
ans[d + 1] = 'O'
for d in ch:
ans[d] = 'C'
ans[d + 1] = 'H'
for d in cc:
ans[d] = 'C'
ans[d + 1] = 'C'
print('? OO')
oo = list(map(int, input().split()))[1:]
print('? OH')
oh = list(map(int, input().split()))[1:]
for d in oo:
ans[d] = 'O'
ans[d + 1] = 'O'
for d in oh:
ans[d] = 'O'
ans[d + 1] = 'H'
for i in range(1, n):
if ans[i] == 'O' and ans[i + 1] == '':
ans[i + 1] = 'C'
for i in range(1, n):
if ans[i] == '' and ans[i + 1] == 'O':
ans[i] = 'H'
for i in range(1, n):
if ans[i] == '' and ans[i + 1] == 'H':
ans[i] = 'H'
print('? HHH')
hhh = list(map(int, input().split()))[1:]
for d in hhh:
ans[d] = 'H'
ans[d + 1] = 'H'
ans[d + 2] = 'H'
for d in range(1, n):
if ans[d] == '' and ans[d + 1] == '':
ans[d+1] = 'O'
kek = ""
for d in range(n, 0, -1):
if d == n:
if ans[d] == '':
ans[d] = 'H'
elif ans[d] != '':
kek += ans[d]
else:
print('?', 'H' + kek[::-1])
ddi = input()[1:]
if str(d) in ddi:
ans[d] = 'H'
else:
ans[d] = 'O'
print('!', ''.join(ans[1:]))
kek = input()
# print(5 / 4 + 1 / 9)
``` | instruction | 0 | 54,066 | 3 | 108,132 |
No | output | 1 | 54,066 | 3 | 108,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[MisoilePunchβͺ - 彩](https://www.youtube.com/watch?v=5VleIdkEeak)
This is an interactive problem!
On a normal day at the hidden office in A.R.C. Markland-N, Rin received an artifact, given to her by the exploration captain Sagar.
After much analysis, she now realizes that this artifact contains data about a strange flower, which has existed way before the New Age. However, the information about its chemical structure has been encrypted heavily.
The chemical structure of this flower can be represented as a string p. From the unencrypted papers included, Rin already knows the length n of that string, and she can also conclude that the string contains at most three distinct letters: "C" (as in Carbon), "H" (as in Hydrogen), and "O" (as in Oxygen).
At each moment, Rin can input a string s of an arbitrary length into the artifact's terminal, and it will return every starting position of s as a substring of p.
However, the artifact has limited energy and cannot be recharged in any way, since the technology is way too ancient and is incompatible with any current A.R.C.'s devices. To be specific:
* The artifact only contains 7/5 units of energy.
* For each time Rin inputs a string s of length t, the artifact consumes (1)/(t^2) units of energy.
* If the amount of energy reaches below zero, the task will be considered failed immediately, as the artifact will go black forever.
Since the artifact is so precious yet fragile, Rin is very nervous to attempt to crack the final data. Can you give her a helping hand?
Interaction
The interaction starts with a single integer t (1 β€ t β€ 500), the number of test cases. The interaction for each testcase is described below:
First, read an integer n (4 β€ n β€ 50), the length of the string p.
Then you can make queries of type "? s" (1 β€ |s| β€ n) to find the occurrences of s as a substring of p.
After the query, you need to read its result as a series of integers in a line:
* The first integer k denotes the number of occurrences of s as a substring of p (-1 β€ k β€ n). If k = -1, it means you have exceeded the energy limit or printed an invalid query, and you need to terminate immediately, to guarantee a "Wrong answer" verdict, otherwise you might get an arbitrary verdict because your solution will continue to read from a closed stream.
* The following k integers a_1, a_2, β¦, a_k (1 β€ a_1 < a_2 < β¦ < a_k β€ n) denote the starting positions of the substrings that match the string s.
When you find out the string p, print "! p" to finish a test case. This query doesn't consume any energy. The interactor will return an integer 1 or 0. If the interactor returns 1, you can proceed to the next test case, or terminate the program if it was the last testcase.
If the interactor returns 0, it means that your guess is incorrect, and you should to terminate to guarantee a "Wrong answer" verdict.
Note that in every test case the string p is fixed beforehand and will not change during the queries, i.e. the interactor is not adaptive.
After printing any query do not forget to print end of line and flush the output. Otherwise, you might get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see the documentation for other languages.
Hacks
For hack, use the following format. Note that you can only hack with one test case:
The first line should contain a single integer t (t = 1).
The second line should contain an integer n (4 β€ n β€ 50) β the string's size.
The third line should contain a string of size n, consisting of characters "C", "H" and "O" only. This is the string contestants will have to find out.
Examples
Input
1
4
2 1 2
1 2
0
1
Output
? C
? CH
? CCHO
! CCHH
Input
2
5
0
2 2 3
1
8
1 5
1 5
1 3
2 1 2
1
Output
? O
? HHH
! CHHHH
? COO
? COOH
? HCCOO
? HH
! HHHCCOOH
Note
Note that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
if n >= 5:
ans = [''] * (n+1)
print('? CO')
co = list(map(int, input().split()))[1:]
print('? CH')
ch = list(map(int, input().split()))[1:]
print('? CC')
cc = list(map(int, input().split()))[1:]
for d in co:
ans[d] = 'C'
ans[d + 1] = 'O'
for d in ch:
ans[d] = 'C'
ans[d + 1] = 'H'
for d in cc:
ans[d] = 'C'
ans[d + 1] = 'C'
print('? OO')
oo = list(map(int, input().split()))[1:]
print('? OH')
oh = list(map(int, input().split()))[1:]
for d in oo:
ans[d] = 'O'
ans[d + 1] = 'O'
for d in oh:
ans[d] = 'O'
ans[d + 1] = 'H'
for i in range(1, n):
if ans[i] == 'O' and ans[i + 1] == '':
ans[i + 1] = 'C'
for i in range(1, n):
if ans[i] == '' and ans[i + 1] == 'O':
ans[i] = 'H'
for i in range(1, n):
if ans[i] == '' and ans[i + 1] == 'H':
ans[i] = 'H'
for i in range(1, n):
if ans[i] == '' and ans[i + 1] == '':
ans[i] = 'H'
print('? HOC')
hoc = list(map(int, input().split()))[1:]
for d in hoc:
ans[d] = 'H'
ans[d + 1] = 'O'
ans[d + 2] = 'C'
for d in range(2, n):
if ans[d] == '':
ans[d] = 'H'
if ans[1] == '':
print('?', 'H' + ''.join(ans[2:-1]))
if input()[0] != '0':
ans[1] = 'H'
else:
ans[1] = 'O'
if ans[-1] == '':
print('?', ''.join(ans[1:-1]) + 'H')
if input()[0] != '0':
ans[-1] = 'H'
else:
print('?', ''.join(ans[1:-1]) + 'O')
if input()[0] != '0':
ans[-1] = 'O'
else:
ans[-1] = 'C'
print('!', ''.join(ans[1:]))
else:
print('! CCHH')
kek = input()
``` | instruction | 0 | 54,067 | 3 | 108,134 |
No | output | 1 | 54,067 | 3 | 108,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[MisoilePunchβͺ - 彩](https://www.youtube.com/watch?v=5VleIdkEeak)
This is an interactive problem!
On a normal day at the hidden office in A.R.C. Markland-N, Rin received an artifact, given to her by the exploration captain Sagar.
After much analysis, she now realizes that this artifact contains data about a strange flower, which has existed way before the New Age. However, the information about its chemical structure has been encrypted heavily.
The chemical structure of this flower can be represented as a string p. From the unencrypted papers included, Rin already knows the length n of that string, and she can also conclude that the string contains at most three distinct letters: "C" (as in Carbon), "H" (as in Hydrogen), and "O" (as in Oxygen).
At each moment, Rin can input a string s of an arbitrary length into the artifact's terminal, and it will return every starting position of s as a substring of p.
However, the artifact has limited energy and cannot be recharged in any way, since the technology is way too ancient and is incompatible with any current A.R.C.'s devices. To be specific:
* The artifact only contains 7/5 units of energy.
* For each time Rin inputs a string s of length t, the artifact consumes (1)/(t^2) units of energy.
* If the amount of energy reaches below zero, the task will be considered failed immediately, as the artifact will go black forever.
Since the artifact is so precious yet fragile, Rin is very nervous to attempt to crack the final data. Can you give her a helping hand?
Interaction
The interaction starts with a single integer t (1 β€ t β€ 500), the number of test cases. The interaction for each testcase is described below:
First, read an integer n (4 β€ n β€ 50), the length of the string p.
Then you can make queries of type "? s" (1 β€ |s| β€ n) to find the occurrences of s as a substring of p.
After the query, you need to read its result as a series of integers in a line:
* The first integer k denotes the number of occurrences of s as a substring of p (-1 β€ k β€ n). If k = -1, it means you have exceeded the energy limit or printed an invalid query, and you need to terminate immediately, to guarantee a "Wrong answer" verdict, otherwise you might get an arbitrary verdict because your solution will continue to read from a closed stream.
* The following k integers a_1, a_2, β¦, a_k (1 β€ a_1 < a_2 < β¦ < a_k β€ n) denote the starting positions of the substrings that match the string s.
When you find out the string p, print "! p" to finish a test case. This query doesn't consume any energy. The interactor will return an integer 1 or 0. If the interactor returns 1, you can proceed to the next test case, or terminate the program if it was the last testcase.
If the interactor returns 0, it means that your guess is incorrect, and you should to terminate to guarantee a "Wrong answer" verdict.
Note that in every test case the string p is fixed beforehand and will not change during the queries, i.e. the interactor is not adaptive.
After printing any query do not forget to print end of line and flush the output. Otherwise, you might get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see the documentation for other languages.
Hacks
For hack, use the following format. Note that you can only hack with one test case:
The first line should contain a single integer t (t = 1).
The second line should contain an integer n (4 β€ n β€ 50) β the string's size.
The third line should contain a string of size n, consisting of characters "C", "H" and "O" only. This is the string contestants will have to find out.
Examples
Input
1
4
2 1 2
1 2
0
1
Output
? C
? CH
? CCHO
! CCHH
Input
2
5
0
2 2 3
1
8
1 5
1 5
1 3
2 1 2
1
Output
? O
? HHH
! CHHHH
? COO
? COOH
? HCCOO
? HH
! HHHCCOOH
Note
Note that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
ans = [''] * (n+1)
print('? CO')
co = list(map(int, input().split()))[1:]
print('? CH')
ch = list(map(int, input().split()))[1:]
print('? CC')
cc = list(map(int, input().split()))[1:]
for d in co:
ans[d] = 'C'
ans[d + 1] = 'O'
for d in ch:
ans[d] = 'C'
ans[d + 1] = 'H'
for d in cc:
ans[d] = 'C'
ans[d + 1] = 'C'
print('? OO')
oo = list(map(int, input().split()))[1:]
print('? OH')
oh = list(map(int, input().split()))[1:]
for d in oo:
ans[d] = 'O'
ans[d + 1] = 'O'
for d in oh:
ans[d] = 'O'
ans[d + 1] = 'H'
for i in range(1, n):
if ans[i] == 'O' and ans[i + 1] == '':
ans[i + 1] = 'C'
for i in range(1, n):
if ans[i] == '' and ans[i + 1] == 'O':
ans[i] = 'H'
for i in range(1, n):
if ans[i] == '' and ans[i + 1] == 'H':
ans[i] = 'H'
for i in range(1, n):
if ans[i] == '' and ans[i + 1] == '':
ans[i] = 'H'
print('? HOC')
hoc = list(map(int, input().split()))[1:]
for d in hoc:
ans[d] = 'H'
ans[d + 1] = 'O'
ans[d + 2] = 'C'
for d in range(1, n+1):
if ans[d] == '':
ans[d] = 'H'
print('!', ''.join(ans[1:]))
kek = input()
# print(5 / 4 + 1 / 9)
``` | instruction | 0 | 54,068 | 3 | 108,136 |
No | output | 1 | 54,068 | 3 | 108,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[MisoilePunchβͺ - 彩](https://www.youtube.com/watch?v=5VleIdkEeak)
This is an interactive problem!
On a normal day at the hidden office in A.R.C. Markland-N, Rin received an artifact, given to her by the exploration captain Sagar.
After much analysis, she now realizes that this artifact contains data about a strange flower, which has existed way before the New Age. However, the information about its chemical structure has been encrypted heavily.
The chemical structure of this flower can be represented as a string p. From the unencrypted papers included, Rin already knows the length n of that string, and she can also conclude that the string contains at most three distinct letters: "C" (as in Carbon), "H" (as in Hydrogen), and "O" (as in Oxygen).
At each moment, Rin can input a string s of an arbitrary length into the artifact's terminal, and it will return every starting position of s as a substring of p.
However, the artifact has limited energy and cannot be recharged in any way, since the technology is way too ancient and is incompatible with any current A.R.C.'s devices. To be specific:
* The artifact only contains 7/5 units of energy.
* For each time Rin inputs a string s of length t, the artifact consumes (1)/(t^2) units of energy.
* If the amount of energy reaches below zero, the task will be considered failed immediately, as the artifact will go black forever.
Since the artifact is so precious yet fragile, Rin is very nervous to attempt to crack the final data. Can you give her a helping hand?
Interaction
The interaction starts with a single integer t (1 β€ t β€ 500), the number of test cases. The interaction for each testcase is described below:
First, read an integer n (4 β€ n β€ 50), the length of the string p.
Then you can make queries of type "? s" (1 β€ |s| β€ n) to find the occurrences of s as a substring of p.
After the query, you need to read its result as a series of integers in a line:
* The first integer k denotes the number of occurrences of s as a substring of p (-1 β€ k β€ n). If k = -1, it means you have exceeded the energy limit or printed an invalid query, and you need to terminate immediately, to guarantee a "Wrong answer" verdict, otherwise you might get an arbitrary verdict because your solution will continue to read from a closed stream.
* The following k integers a_1, a_2, β¦, a_k (1 β€ a_1 < a_2 < β¦ < a_k β€ n) denote the starting positions of the substrings that match the string s.
When you find out the string p, print "! p" to finish a test case. This query doesn't consume any energy. The interactor will return an integer 1 or 0. If the interactor returns 1, you can proceed to the next test case, or terminate the program if it was the last testcase.
If the interactor returns 0, it means that your guess is incorrect, and you should to terminate to guarantee a "Wrong answer" verdict.
Note that in every test case the string p is fixed beforehand and will not change during the queries, i.e. the interactor is not adaptive.
After printing any query do not forget to print end of line and flush the output. Otherwise, you might get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see the documentation for other languages.
Hacks
For hack, use the following format. Note that you can only hack with one test case:
The first line should contain a single integer t (t = 1).
The second line should contain an integer n (4 β€ n β€ 50) β the string's size.
The third line should contain a string of size n, consisting of characters "C", "H" and "O" only. This is the string contestants will have to find out.
Examples
Input
1
4
2 1 2
1 2
0
1
Output
? C
? CH
? CCHO
! CCHH
Input
2
5
0
2 2 3
1
8
1 5
1 5
1 3
2 1 2
1
Output
? O
? HHH
! CHHHH
? COO
? COOH
? HCCOO
? HH
! HHHCCOOH
Note
Note that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well.
Submitted Solution:
```
Q = int(input())
for _ in range(Q):
def chk(t):
print("?", t)
L = [int(a)-1 for a in input().split()][1:]
for l in L:
for i in range(len(t)):
X[l+i] = t[i]
return min(L) if len(L) else -1
N = int(input())
i = -1
X = ["_"] * N
for s in ["CH", "CO", "CC"]:
i = max(i, chk(s))
if i < 0:
chk("HO")
chk("HH")
for i in range(N-1):
if X[i] == "_": X[i] = "O"
if X[N-1] == "_":
print("?", "".join(X[:-1]) + "H")
L = [int(a)-1 for a in input().split()][1:]
if 0 in L:
X[-1] = "H"
else:
print("?", "".join(X[:-1]) + "O")
L = [int(a)-1 for a in input().split()][1:]
if 0 in L:
X[-1] = "O"
else:
X[-1] = "C"
print("!", "".join(X))
aa = int(input())
exit()
else:
for j in range(i-1, -1, -1):
if X[j] != "_": continue
print("?", "H" + "".join(X[j+1:i+2]))
L = [int(a)-1 for a in input().split()][1:]
if j in L:
X[j] = "H"
else:
X[j] = "O"
for j in range(i+2, N-1):
if j < N-1 and X[j] != "_": continue
print("?", "".join(X[:j]) + "H")
L = [int(a)-1 for a in input().split()][1:]
if 0 in L:
X[j] = "H"
else:
X[j] = "O"
if X[N-1] == "_":
print("?", "".join(X[:-1]) + "H")
L = [int(a)-1 for a in input().split()][1:]
if 0 in L:
X[-1] = "H"
else:
print("?", "".join(X[:-1]) + "O")
L = [int(a)-1 for a in input().split()][1:]
if 0 in L:
X[-1] = "O"
else:
X[-1] = "C"
print("!", "".join(X))
aa = int(input())
``` | instruction | 0 | 54,069 | 3 | 108,138 |
No | output | 1 | 54,069 | 3 | 108,139 |
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.
Kochiya Sanae is playing with magnets. Realizing that some of those magnets are demagnetized, she is curious to find them out.
There are n magnets, which can be of the following 3 types:
* N
* S
* - β these magnets are demagnetized.
Note that you don't know the types of these magnets beforehand.
You have a machine which can measure the force between the magnets, and you can use it at most n+β log_2nβ times.
You can put some magnets to the left part of the machine and some to the right part of the machine, and launch the machine. Obviously, you can put one magnet to at most one side (you don't have to put all magnets). You can put the same magnet in different queries.
Then the machine will tell the force these magnets produce. Formally, let n_1,s_1 be the number of N and S magnets correspondently on the left and n_2,s_2 β on the right. Then the force between them would be n_1n_2+s_1s_2-n_1s_2-n_2s_1. Please note that the force is a signed value.
However, when the absolute value of the force is strictly larger than n, the machine will crash into pieces.
You need to find all magnets of type - (all demagnetized ones), without breaking the machine.
Note that the interactor is not adaptive. The types of the magnets are fixed before the start of the interaction and do not change with queries.
It is guaranteed that there are at least 2 magnets whose type is not -, and at least 1 magnet of type -.
Input
The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
Interaction
For each test case you should start by reading an integer n (3 β€ n β€ 2000) β the number of the magnets.
It is guaranteed that the total sum of all n over all test cases doesn't exceed 2000.
After that you can put some magnets into the machine and make a query from the statement.
You have to print each query in three lines:
* In the first line print "? l r" (without quotes) where l and r (1 β€ l,r < n, l+r β€ n) respectively denote the number of the magnets you put to left and right.
* In the second line print l integers a_1, ..., a_l (1 β€ a_i β€ n, a_i β a_j if i β j) β the indices of the magnets you put to left.
* In the third line print r integers b_1, ..., b_r (1 β€ b_i β€ n, b_i β b_j if i β j) β the indices of the magnets you put to right.
* The same magnet can't be put to both sides in the same query. Formally, you should guarantee that a_i β b_j for any i and j. However, you may leave some magnets unused.
After printing a query do not forget to output end of line and flush the output. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
After this, you should read an integer F β the force these magnets produce.
Note that if your query is invalid(either the query limit exceeds, the machine crashes or the arguments are invalid), the interactor will terminate immediately. In this case terminate your program to receive verdict Wrong Answer instead of arbitrary verdicts.
If you are confident about your answer, use the following format to report it:
* "! k A", where k is the number of magnets you found, and A is an array consisting of k different integers from 1 to n denoting the indices of the magnets of type - that you found. You may print elements of A in arbitrary order.
* After that, if this is the last test case, you have to terminate your program; otherwise you should immediately continue to deal with the next test case.
Note that the interactor is not adaptive. The types of the magnets are fixed before the start of interaction and do not change with queries.
Hacks
To hack a solution, use the following format:
The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases in your hack.
Then follow the descriptions of the t test cases, each printed in two lines:
* The first line contains a single integer n (3 β€ n β€ 2000) β the number of magnets.
* The second line contains a string S of length n consisting of only N, S and -, denoting the magnets' types.
* Each of your test case should guarantee that there are at least 2 magnets whose type is not -, and at least 1 magnet of type -. Meanwhile, the total sum of n in all test cases should not exceed 2000.
Example
Input
1
4
0
1
0
0
Output
? 1 1
3
4
? 1 2
1
2 3
? 1 1
1
4
? 1 1
1
3
! 2 3 4
Note
The empty lines in the sample are just for you to better understand the interaction process. You're not required to print them.
In the sample, the types of the magnets are NN--.
At first, you put the third magnet on the left and the fourth one on the right. Both of them have type -, thus no force is produced.
Then you put the first magnet on the left and the second and third one on the right. The third magnet has type -, while the other two magnets are of type N, so the force produced is 1.
In the following two queries, the force is 0 since there is only a magnet with property - on the right.
Then we can determine that the magnets of type - are the third and the fourth one, so we should print ! 2 3 4 and exit.
Submitted Solution:
```
import random
import sys
def solve(n):
sqrtn = 1
for i in range(n):
if i * i <= n:
sqrtn = i
guaranteedl = []
guaranteedr = []
v = [i for i in range(1, n + 1)]
for t in range(n):
sample_v = random.sample(v, k=2*sqrtn)
l = sample_v[:sqrtn]
r = sample_v[sqrtn:]
print(f"? {sqrtn} {sqrtn}")
for i in l:
print(i)
for i in r:
print(i)
sys.stdout.flush()
res = int(input())
if res != 0:
guaranteedl = l
guaranteedr = r
break
while len(guaranteedl) > 1:
mid = len(guaranteedl) // 2
l = guaranteedl[:mid]
print(f"? {mid} {sqrtn}")
for i in l:
print(i)
for i in guaranteedr:
print(i)
sys.stdout.flush()
res = int(input())
if res == 0:
guaranteedl = guaranteedl[mid:]
else:
guaranteedl = guaranteedl[:mid]
t = guaranteedl[0]
zeros = []
for i in range(1, n + 1):
if i == t:
continue
print(f"? 1 1")##
print(t)
print(i)
sys.stdout.flush()
res = int(input())
if res == 0:
zeros.append(i)
zeros_string = ' '.join([str(i) for i in zeros])
print(f"! {str(len(zeros))} {zeros_string}")
sys.stdout.flush()
testcases = int(input())
for _ in range(testcases):
n = int(input())
solve(n)
``` | instruction | 0 | 54,180 | 3 | 108,360 |
No | output | 1 | 54,180 | 3 | 108,361 |
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.
Kochiya Sanae is playing with magnets. Realizing that some of those magnets are demagnetized, she is curious to find them out.
There are n magnets, which can be of the following 3 types:
* N
* S
* - β these magnets are demagnetized.
Note that you don't know the types of these magnets beforehand.
You have a machine which can measure the force between the magnets, and you can use it at most n+β log_2nβ times.
You can put some magnets to the left part of the machine and some to the right part of the machine, and launch the machine. Obviously, you can put one magnet to at most one side (you don't have to put all magnets). You can put the same magnet in different queries.
Then the machine will tell the force these magnets produce. Formally, let n_1,s_1 be the number of N and S magnets correspondently on the left and n_2,s_2 β on the right. Then the force between them would be n_1n_2+s_1s_2-n_1s_2-n_2s_1. Please note that the force is a signed value.
However, when the absolute value of the force is strictly larger than n, the machine will crash into pieces.
You need to find all magnets of type - (all demagnetized ones), without breaking the machine.
Note that the interactor is not adaptive. The types of the magnets are fixed before the start of the interaction and do not change with queries.
It is guaranteed that there are at least 2 magnets whose type is not -, and at least 1 magnet of type -.
Input
The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
Interaction
For each test case you should start by reading an integer n (3 β€ n β€ 2000) β the number of the magnets.
It is guaranteed that the total sum of all n over all test cases doesn't exceed 2000.
After that you can put some magnets into the machine and make a query from the statement.
You have to print each query in three lines:
* In the first line print "? l r" (without quotes) where l and r (1 β€ l,r < n, l+r β€ n) respectively denote the number of the magnets you put to left and right.
* In the second line print l integers a_1, ..., a_l (1 β€ a_i β€ n, a_i β a_j if i β j) β the indices of the magnets you put to left.
* In the third line print r integers b_1, ..., b_r (1 β€ b_i β€ n, b_i β b_j if i β j) β the indices of the magnets you put to right.
* The same magnet can't be put to both sides in the same query. Formally, you should guarantee that a_i β b_j for any i and j. However, you may leave some magnets unused.
After printing a query do not forget to output end of line and flush the output. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
After this, you should read an integer F β the force these magnets produce.
Note that if your query is invalid(either the query limit exceeds, the machine crashes or the arguments are invalid), the interactor will terminate immediately. In this case terminate your program to receive verdict Wrong Answer instead of arbitrary verdicts.
If you are confident about your answer, use the following format to report it:
* "! k A", where k is the number of magnets you found, and A is an array consisting of k different integers from 1 to n denoting the indices of the magnets of type - that you found. You may print elements of A in arbitrary order.
* After that, if this is the last test case, you have to terminate your program; otherwise you should immediately continue to deal with the next test case.
Note that the interactor is not adaptive. The types of the magnets are fixed before the start of interaction and do not change with queries.
Hacks
To hack a solution, use the following format:
The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases in your hack.
Then follow the descriptions of the t test cases, each printed in two lines:
* The first line contains a single integer n (3 β€ n β€ 2000) β the number of magnets.
* The second line contains a string S of length n consisting of only N, S and -, denoting the magnets' types.
* Each of your test case should guarantee that there are at least 2 magnets whose type is not -, and at least 1 magnet of type -. Meanwhile, the total sum of n in all test cases should not exceed 2000.
Example
Input
1
4
0
1
0
0
Output
? 1 1
3
4
? 1 2
1
2 3
? 1 1
1
4
? 1 1
1
3
! 2 3 4
Note
The empty lines in the sample are just for you to better understand the interaction process. You're not required to print them.
In the sample, the types of the magnets are NN--.
At first, you put the third magnet on the left and the fourth one on the right. Both of them have type -, thus no force is produced.
Then you put the first magnet on the left and the second and third one on the right. The third magnet has type -, while the other two magnets are of type N, so the force produced is 1.
In the following two queries, the force is 0 since there is only a magnet with property - on the right.
Then we can determine that the magnets of type - are the third and the fourth one, so we should print ! 2 3 4 and exit.
Submitted Solution:
```
import random
import sys
def solve(n):
if n == 3:
print("? 1 1")
print(1)
print(2)
sys.stdout.flush()
res = int(input())
if res != 0:
print("! 1 3")
sys.stdout.flush()
return
print("? 1 1")
print(1)
print(3)
sys.stdout.flush()
res = int(input())
if res != 0:
print("! 1 2")
sys.stdout.flush()
return
print("! 1 1")
sys.stdout.flush()
return
sqrtn = 1
for i in range(n):
if i * i <= n:
sqrtn = i
guaranteedl = []
guaranteedr = []
v = [i for i in range(1, n + 1)]
for t in range(n):
sample_v = random.sample(v, k=2*sqrtn)
l = sample_v[:sqrtn]
r = sample_v[sqrtn:]
print(f"? {sqrtn} {sqrtn}")
for i in l:
print(i)
for i in r:
print(i)
sys.stdout.flush()
res = int(input())
if res != 0:
guaranteedl = l
guaranteedr = r
break
while len(guaranteedl) > 1:
mid = len(guaranteedl) // 2
l = guaranteedl[:mid]
print(f"? {mid} {sqrtn}")
for i in l:
print(i)
for i in guaranteedr:
print(i)
sys.stdout.flush()
res = int(input())
if res == 0:
guaranteedl = guaranteedl[mid:]
else:
guaranteedl = guaranteedl[:mid]
t = guaranteedl[0]
zeros = []
for i in range(1, n + 1):
if i == t:
continue
print(f"? 1 1")
print(t)
print(i)
sys.stdout.flush()
res = int(input())
if res == 0:
zeros.append(i)
zeros_string = ' '.join([str(i) for i in zeros])
print(f"! {str(len(zeros))} {zeros_string}")
sys.stdout.flush()
testcases = int(input())
for _ in range(testcases):
n = int(input())
solve(n)
``` | instruction | 0 | 54,181 | 3 | 108,362 |
No | output | 1 | 54,181 | 3 | 108,363 |
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.
Kochiya Sanae is playing with magnets. Realizing that some of those magnets are demagnetized, she is curious to find them out.
There are n magnets, which can be of the following 3 types:
* N
* S
* - β these magnets are demagnetized.
Note that you don't know the types of these magnets beforehand.
You have a machine which can measure the force between the magnets, and you can use it at most n+β log_2nβ times.
You can put some magnets to the left part of the machine and some to the right part of the machine, and launch the machine. Obviously, you can put one magnet to at most one side (you don't have to put all magnets). You can put the same magnet in different queries.
Then the machine will tell the force these magnets produce. Formally, let n_1,s_1 be the number of N and S magnets correspondently on the left and n_2,s_2 β on the right. Then the force between them would be n_1n_2+s_1s_2-n_1s_2-n_2s_1. Please note that the force is a signed value.
However, when the absolute value of the force is strictly larger than n, the machine will crash into pieces.
You need to find all magnets of type - (all demagnetized ones), without breaking the machine.
Note that the interactor is not adaptive. The types of the magnets are fixed before the start of the interaction and do not change with queries.
It is guaranteed that there are at least 2 magnets whose type is not -, and at least 1 magnet of type -.
Input
The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
Interaction
For each test case you should start by reading an integer n (3 β€ n β€ 2000) β the number of the magnets.
It is guaranteed that the total sum of all n over all test cases doesn't exceed 2000.
After that you can put some magnets into the machine and make a query from the statement.
You have to print each query in three lines:
* In the first line print "? l r" (without quotes) where l and r (1 β€ l,r < n, l+r β€ n) respectively denote the number of the magnets you put to left and right.
* In the second line print l integers a_1, ..., a_l (1 β€ a_i β€ n, a_i β a_j if i β j) β the indices of the magnets you put to left.
* In the third line print r integers b_1, ..., b_r (1 β€ b_i β€ n, b_i β b_j if i β j) β the indices of the magnets you put to right.
* The same magnet can't be put to both sides in the same query. Formally, you should guarantee that a_i β b_j for any i and j. However, you may leave some magnets unused.
After printing a query do not forget to output end of line and flush the output. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
After this, you should read an integer F β the force these magnets produce.
Note that if your query is invalid(either the query limit exceeds, the machine crashes or the arguments are invalid), the interactor will terminate immediately. In this case terminate your program to receive verdict Wrong Answer instead of arbitrary verdicts.
If you are confident about your answer, use the following format to report it:
* "! k A", where k is the number of magnets you found, and A is an array consisting of k different integers from 1 to n denoting the indices of the magnets of type - that you found. You may print elements of A in arbitrary order.
* After that, if this is the last test case, you have to terminate your program; otherwise you should immediately continue to deal with the next test case.
Note that the interactor is not adaptive. The types of the magnets are fixed before the start of interaction and do not change with queries.
Hacks
To hack a solution, use the following format:
The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases in your hack.
Then follow the descriptions of the t test cases, each printed in two lines:
* The first line contains a single integer n (3 β€ n β€ 2000) β the number of magnets.
* The second line contains a string S of length n consisting of only N, S and -, denoting the magnets' types.
* Each of your test case should guarantee that there are at least 2 magnets whose type is not -, and at least 1 magnet of type -. Meanwhile, the total sum of n in all test cases should not exceed 2000.
Example
Input
1
4
0
1
0
0
Output
? 1 1
3
4
? 1 2
1
2 3
? 1 1
1
4
? 1 1
1
3
! 2 3 4
Note
The empty lines in the sample are just for you to better understand the interaction process. You're not required to print them.
In the sample, the types of the magnets are NN--.
At first, you put the third magnet on the left and the fourth one on the right. Both of them have type -, thus no force is produced.
Then you put the first magnet on the left and the second and third one on the right. The third magnet has type -, while the other two magnets are of type N, so the force produced is 1.
In the following two queries, the force is 0 since there is only a magnet with property - on the right.
Then we can determine that the magnets of type - are the third and the fourth one, so we should print ! 2 3 4 and exit.
Submitted Solution:
```
import sys
input = sys.stdin.readline
t=int(input())
for tests in range(t):
n=int(input())
BROKEN=set()
ALL=set(range(1,n+1))
for i in range(1,n+1):
print("?",1,n-1-len(BROKEN),flush=True)
print(i,flush=True)
print(*list(ALL-BROKEN-{i}),flush=True)
x=int(input())
if x==0:
BROKEN.add(i)
L=sorted(BROKEN)
if len(L)==1:
print("!",*L,flush=True)
continue
while True:
LEN=len(L)
X=L[:LEN//2]
Y=L[LEN//2:]
print("?",len(X),len(Y),flush=True)
print(*X,flush=True)
print(*Y,flush=True)
x=int(input())
if x==0:
print("!",*L,flush=True)
break
else:
if x>0:
L=X
else:
L=Y
if len(L)==1:
ANS=BROKEN-{L[0]}
print("!",*sorted(ANS),flush=True)
break
``` | instruction | 0 | 54,182 | 3 | 108,364 |
No | output | 1 | 54,182 | 3 | 108,365 |
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.
Kochiya Sanae is playing with magnets. Realizing that some of those magnets are demagnetized, she is curious to find them out.
There are n magnets, which can be of the following 3 types:
* N
* S
* - β these magnets are demagnetized.
Note that you don't know the types of these magnets beforehand.
You have a machine which can measure the force between the magnets, and you can use it at most n+β log_2nβ times.
You can put some magnets to the left part of the machine and some to the right part of the machine, and launch the machine. Obviously, you can put one magnet to at most one side (you don't have to put all magnets). You can put the same magnet in different queries.
Then the machine will tell the force these magnets produce. Formally, let n_1,s_1 be the number of N and S magnets correspondently on the left and n_2,s_2 β on the right. Then the force between them would be n_1n_2+s_1s_2-n_1s_2-n_2s_1. Please note that the force is a signed value.
However, when the absolute value of the force is strictly larger than n, the machine will crash into pieces.
You need to find all magnets of type - (all demagnetized ones), without breaking the machine.
Note that the interactor is not adaptive. The types of the magnets are fixed before the start of the interaction and do not change with queries.
It is guaranteed that there are at least 2 magnets whose type is not -, and at least 1 magnet of type -.
Input
The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
Interaction
For each test case you should start by reading an integer n (3 β€ n β€ 2000) β the number of the magnets.
It is guaranteed that the total sum of all n over all test cases doesn't exceed 2000.
After that you can put some magnets into the machine and make a query from the statement.
You have to print each query in three lines:
* In the first line print "? l r" (without quotes) where l and r (1 β€ l,r < n, l+r β€ n) respectively denote the number of the magnets you put to left and right.
* In the second line print l integers a_1, ..., a_l (1 β€ a_i β€ n, a_i β a_j if i β j) β the indices of the magnets you put to left.
* In the third line print r integers b_1, ..., b_r (1 β€ b_i β€ n, b_i β b_j if i β j) β the indices of the magnets you put to right.
* The same magnet can't be put to both sides in the same query. Formally, you should guarantee that a_i β b_j for any i and j. However, you may leave some magnets unused.
After printing a query do not forget to output end of line and flush the output. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
After this, you should read an integer F β the force these magnets produce.
Note that if your query is invalid(either the query limit exceeds, the machine crashes or the arguments are invalid), the interactor will terminate immediately. In this case terminate your program to receive verdict Wrong Answer instead of arbitrary verdicts.
If you are confident about your answer, use the following format to report it:
* "! k A", where k is the number of magnets you found, and A is an array consisting of k different integers from 1 to n denoting the indices of the magnets of type - that you found. You may print elements of A in arbitrary order.
* After that, if this is the last test case, you have to terminate your program; otherwise you should immediately continue to deal with the next test case.
Note that the interactor is not adaptive. The types of the magnets are fixed before the start of interaction and do not change with queries.
Hacks
To hack a solution, use the following format:
The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases in your hack.
Then follow the descriptions of the t test cases, each printed in two lines:
* The first line contains a single integer n (3 β€ n β€ 2000) β the number of magnets.
* The second line contains a string S of length n consisting of only N, S and -, denoting the magnets' types.
* Each of your test case should guarantee that there are at least 2 magnets whose type is not -, and at least 1 magnet of type -. Meanwhile, the total sum of n in all test cases should not exceed 2000.
Example
Input
1
4
0
1
0
0
Output
? 1 1
3
4
? 1 2
1
2 3
? 1 1
1
4
? 1 1
1
3
! 2 3 4
Note
The empty lines in the sample are just for you to better understand the interaction process. You're not required to print them.
In the sample, the types of the magnets are NN--.
At first, you put the third magnet on the left and the fourth one on the right. Both of them have type -, thus no force is produced.
Then you put the first magnet on the left and the second and third one on the right. The third magnet has type -, while the other two magnets are of type N, so the force produced is 1.
In the following two queries, the force is 0 since there is only a magnet with property - on the right.
Then we can determine that the magnets of type - are the third and the fourth one, so we should print ! 2 3 4 and exit.
Submitted Solution:
```
def sqrt(n):
if n == 0: return 0
x = 1 << (n.bit_length() + 1) // 2
y = (x + n // x) // 2
while y < x:
x = y
y = (x + n // x) // 2
return x
def chk(A, B):
if type(A) == int:
A = [A]
if type(B) == int:
B = [B]
print("?", len(A), len(B))
print(*[a + 1 for a in A])
print(*[a + 1 for a in B])
return int(input())
def _chk(A, B):
global cnt
cnt += 1
if type(A) == int:
A = [A]
if type(B) == int:
B = [B]
ret = 0
for a in A:
for b in B:
ret += L[a] * L[b]
print(cnt, "chk", A, B, "=", ret)
return ret
def ans(A):
if type(A) == int:
A = [A]
print("!", len(A), *[a + 1 for a in A])
# print("END!!")
exit()
from random import randrange, shuffle
T = int(input())
for _ in range(T):
N = int(input())
ma = N + N.bit_length() - 1
cnt = 0
if N == 3:
L = [1, -1, 0]
if chk(0, 1):
ans(2)
if chk(0, 2):
ans(1)
if chk(1, 2):
ans(0)
if N == 4:
L = [1, 0, 0, 1]
X = [0] * 4
for i in range(4):
for j in range(i):
if chk(i, j):
X[i] = 1
X[j] = 1
ans([i for i in range(4) if X[i] == 0])
L = [randrange(3) - 1 for _ in range(N)]
while 1:
L = [randrange(3) - 1 if randrange(3) == 0 else 0 for _ in range(N)]
if sum(abs(a) for a in L) > 1: break
r = sqrt(N * 2)
ZZ = []
while 1:
X = [i for i in range(N)]
shuffle(X)
A = X[:r]
B = X[r:r*3//2]
f = chk(A, B)
if f:
break
ZZ = B
r = min(r * 2, N * 2 // 3)
while len(B) > 1:
if chk(A, B[:len(B)//2]):
B = B[:len(B)//2]
else:
B = B[len(B)//2:]
i = B[0]
Y = [0] * N
ZZZ = set(ZZ)
Y[i] = 1
for j in range(N):
if cnt >= ma: break
if i != j and j not in ZZ and chk(i, j):
Y[j] = 1
for j in range(N):
if cnt >= ma: break
if i != j and j in ZZ and chk(i, j):
Y[j] = 1
ans([j for j in range(N) if Y[j] == 0])
``` | instruction | 0 | 54,183 | 3 | 108,366 |
No | output | 1 | 54,183 | 3 | 108,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, β¦, d_n (0 β€ d_i β€ 10^9) β the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
#######puzzleVerma#######
import sys
import math
mod = 10**9+7
LI=lambda:[int(k) for k in input().split()]
input = lambda: sys.stdin.readline().rstrip()
IN=lambda:int(input())
S=lambda:input()
r=range
def sm(arr , n):
ans = 0
for i in range(0, n):
ans = ans + arr[i]
return int(ans * math.pow(2, n - 1))
for t in r(IN()):
n=IN()
li=LI()
if n<3:
print(0)
continue
li.sort()
ans=li[-1]
diff=[0]*n
for i in range(1,n):
dif=li[i]-li[i-1]
diff[i]=dif
temp=0
for i in range(1,n):
temp+=((i)*(diff[i]))
ans-=temp
print(ans)
# smo=sum(diff)
# sma=sm(diff,len(diff))
# print(smo-sma)
``` | instruction | 0 | 54,208 | 3 | 108,416 |
Yes | output | 1 | 54,208 | 3 | 108,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, β¦, d_n (0 β€ d_i β€ 10^9) β the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
t=int(input())
while t>0:
n=int(input())
a=list(map(int,input().split()))
a.sort()
ans=max(a)
b=[]
for i in range(1,n):
term=a[i]-a[i-1]
b.append(term)
n-=1
maximum=(n+1)//2
#print("maximum",maximum)
#print('b',b,'n',n)
if(n%2==0):
for j in range(n):
#print("j",j)
p=min(j+1,n-j)
#print('p',p)
if(p<maximum):
subber=b[j]*(p*(p+1)+p*(n-2*p))
else:
subber=b[j]*(p*(p+1))
ans=ans-subber
#print("subber:",subber)
else:
for j in range(n):
#print("j",j)
p=min(j+1,n-j)
#print('p',p)
if(p<maximum):
subber=b[j]*(p*(p+1)+p*(n-2*p))
else:
subber=b[j]*p*p
#print("subber:",subber)
ans=ans-subber
print(ans)
t-=1
``` | instruction | 0 | 54,209 | 3 | 108,418 |
Yes | output | 1 | 54,209 | 3 | 108,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, β¦, d_n (0 β€ d_i β€ 10^9) β the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = sorted(map(int, input().split()))
if n < 3:
print(0)
else:
total = 0
for i in range(1,n-1):
total += i * a[-2-i]
for i in range(1,n-1):
total -= i * a[i+1]
print(total)
``` | instruction | 0 | 54,210 | 3 | 108,420 |
Yes | output | 1 | 54,210 | 3 | 108,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, β¦, d_n (0 β€ d_i β€ 10^9) β the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
# def
def main():
for _ in range(int(input())):
n = int(input())
d = sorted(list(map(int, input().split())))
if n == 1:
print(0)
continue
s = sum(d)
m = d[1]
f = n - 4
k = f + 1
r = 0
for i in range(2, n):
r += k * (d[i] - d[i - 1])
k += f
f -= 2
print(-(s - m) - r)
main()
``` | instruction | 0 | 54,211 | 3 | 108,422 |
Yes | output | 1 | 54,211 | 3 | 108,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, β¦, d_n (0 β€ d_i β€ 10^9) β the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import Counter
import bisect
for _ in range(int(input())):
n = int(input())
A = list(map(int, input().split()))
if n <= 2:
print(0)
else:
B = sorted(A)
tot = sum(B[2:])
print(-tot)
``` | instruction | 0 | 54,212 | 3 | 108,424 |
No | output | 1 | 54,212 | 3 | 108,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, β¦, d_n (0 β€ d_i β€ 10^9) β the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
from collections import defaultdict, Counter
from math import sqrt, log10, log2, log, gcd, floor, factorial
from bisect import bisect_left, bisect_right
from itertools import combinations, combinations_with_replacement
import sys, io, os
input = sys.stdin.readline
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# print=sys.stdout.write
# sys.setrecursionlimit(10000)
mod = 10 ** 9 + 7;inf = float('inf')
def get_list(): return [int(i) for i in input().split()]
yn = lambda a: print("YES" if a else "NO")
ceil = lambda a, b: (a + b - 1) // b
t=int(input())
def backward():
return i-1
def forward():
return n-i-1
for i in range(t):
n=int(input())
l=get_list()
l.sort()
counter = 0
suma = sum(l[:-2])
for i in range(2,n):
suma+=l[i-2]
counter-=(l[i]*backward())
print(counter+suma)
``` | instruction | 0 | 54,213 | 3 | 108,426 |
No | output | 1 | 54,213 | 3 | 108,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, β¦, d_n (0 β€ d_i β€ 10^9) β the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
import sys
IS_INTERACTIVE = False
input = input # type: ignore
# input: function
if not IS_INTERACTIVE:
*data, = sys.stdin.read().split("\n")[::-1]
def input(): # type: ignore
return data.pop()
def fprint(*args, **kwargs):
print(*args, **kwargs, flush=True)
def eprint(*args, **kwargs):
print(*args, **kwargs, file=sys.stderr)
for _ in range(int(input())):
n = int(input())
*arr, = map(int, input().split())
arr.sort()
if n == 1:
print(0)
else:
print(-sum(arr[2:]))
``` | instruction | 0 | 54,214 | 3 | 108,428 |
No | output | 1 | 54,214 | 3 | 108,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, β¦, d_n (0 β€ d_i β€ 10^9) β the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
t = int(input())
for c in range(t):
n = int(input())
a = [int(e) for e in input().split()]
a.sort()
ans = 0
for i in range(1, n):
d = a[i] - a[i - 1]
ans += d
for i in range(1, n):
d = a[i] - a[i - 1]
ans -= d * (n - 1)
print(ans)
``` | instruction | 0 | 54,215 | 3 | 108,430 |
No | output | 1 | 54,215 | 3 | 108,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000 | instruction | 0 | 54,264 | 3 | 108,528 |
Tags: dp, probabilities
Correct Solution:
```
##########################################################
from collections import Counter
def nCk(n, k):
res = 1
for i in range(1, k + 1):
res = res * (n - i + 1) // i
return res
import math
inf=10**20
n=int(input())
c=[]
dp=[0]*(n+1)
for i in range(n):
x,y, t,p = map(float, input().split())
c.append([int(x),int(y),int(t),p])
c.sort(key=lambda x:x[2])
for i in range(n):
dp[i]=c[i][3]
for j in range(i):
if (c[i][0]-c[j][0])**2+(c[i][1]-c[j][1])**2 <= (c[i][2]-c[j][2])**2:
dp[i]=max(dp[i],dp[j]+c[i][-1])
print(max(dp))
``` | output | 1 | 54,264 | 3 | 108,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000 | instruction | 0 | 54,265 | 3 | 108,530 |
Tags: dp, probabilities
Correct Solution:
```
if __name__ == '__main__':
n = int(input())
targets = [input() for _ in range(n)]
targets = [line.split() for line in targets]
targets = [(int(x), int(y), int(t), float(p)) for x, y, t, p in targets]
targets = sorted(targets, key=lambda x: x[2])
max_scores = [0] * n
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
if (targets[j][2] - targets[i][2])**2 >= (targets[j][0] - targets[i][0])**2 + (targets[j][1] - targets[i][1])**2:
max_scores[i] = max(max_scores[i], max_scores[j])
max_scores[i] += targets[i][3]
print(max(max_scores))
``` | output | 1 | 54,265 | 3 | 108,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000 | instruction | 0 | 54,266 | 3 | 108,532 |
Tags: dp, probabilities
Correct Solution:
```
def dist_square(p_1, p_2):
return (p_1[0] - p_2[0])**2 + (p_1[1] - p_2[1])**2
n = int(input())
events = []
for _ in range(n):
x, y, t, p = input().split()
x, y, t, p = int(x), int(y), int(t), float(p)
events += [(t, (x, y), p)]
events.sort()
A = [0 for _ in range(n)]
for i in range(n):
A[i] = 0
for j in range(i):
if dist_square(events[i][1], events[j][1]) <= \
(events[i][0] - events[j][0]) ** 2:
A[i] = max([A[i], A[j]])
A[i] += events[i][2]
print(max(A))
``` | output | 1 | 54,266 | 3 | 108,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000 | instruction | 0 | 54,267 | 3 | 108,534 |
Tags: dp, probabilities
Correct Solution:
```
n = int(input())
ts = sorted((list(map(float,input().strip().split())) for _ in range(n)), key=lambda x:x[2])
d = [0]*n
ans = 0
for i in range(n):
d[i] = ts[i][3]
for j in range(i):
if (ts[i][0]-ts[j][0])**2 + (ts[i][1]-ts[j][1])**2 <= (ts[i][2]-ts[j][2])**2:
d[i] = max(d[i], d[j]+ts[i][3])
ans = max(ans, d[i])
print('%.9f'%ans)
``` | output | 1 | 54,267 | 3 | 108,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000 | instruction | 0 | 54,268 | 3 | 108,536 |
Tags: dp, probabilities
Correct Solution:
```
def is_reachable(from_state, target):
return (from_state[0]-target[0])**2 + (from_state[1]-target[1])**2 <= (target[2] - from_state[2])**2
num_iter = int(input())
targets = sorted([list(map(float, input().strip().split(' '))) for dummy in range(0, num_iter)], key=lambda single_target: single_target[2])
states = []
for i, target in enumerate(targets):
states.append(target[3])
for j in range(i):
if is_reachable(targets[j], target):
score_estimate = states[j] + target[3]
if (states[i] < score_estimate):
states[i] = score_estimate
max_score = max(states)
print(max_score)
``` | output | 1 | 54,268 | 3 | 108,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000 | instruction | 0 | 54,269 | 3 | 108,538 |
Tags: dp, probabilities
Correct Solution:
```
import sys
import math
sys.setrecursionlimit(100000)
def solve():
n, = rv()
targets = list()
for i in range(n):
x, y, t, p, = input().split()
x = int(x)
y = int(y)
t = int(t)
p = float(p)
targets.append((t, x, y, p))
targets.sort()
mem = [[-2] * (n + 2) for _ in range(n + 1)]
res = dp(0, -1, targets, mem)
print(res)
def dp(index, previndex, targets, mem):
if index == len(targets): return 0
if mem[index][previndex + 1] > -1: return mem[index][previndex + 1]
x, y = targets[previndex][1], targets[previndex][2]
distance = 0 if previndex == -1 else math.sqrt((x - targets[index][1]) * (x - targets[index][1]) + (y - targets[index][2]) * (y - targets[index][2]))
dont = dp(index + 1, previndex, targets, mem)
take = 0
if previndex == -1 or distance <= targets[index][0] - targets[previndex][0]:
take = targets[index][3] + dp(index + 1, index, targets, mem)
mem[index][previndex + 1] = max(take, dont)
return max(take, dont)
def prt(l): return print(''.join(l))
def rv(): return map(int, input().split())
def rl(n): return [list(map(int, input().split())) for _ in range(n)]
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
``` | output | 1 | 54,269 | 3 | 108,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000 | instruction | 0 | 54,270 | 3 | 108,540 |
Tags: dp, probabilities
Correct Solution:
```
import heapq
def is_reachable(from_state, target):
return (from_state[0]-target[0])**2 + (from_state[1]-target[1])**2 <= (target[2] - from_state[2])**2
num_iter = int(input())
targets = sorted([list(map(float, input().strip().split(' '))) for dummy in range(0, num_iter)], key=lambda single_target: single_target[2])
states = []
for i, target in enumerate(targets):
score_estimate = target[3]
sorted_states = heapq.nlargest(len(states), states)
for j in range(len(sorted_states)):
if is_reachable(sorted_states[j][1], target):
score_estimate += sorted_states[j][1][3]
break;
target[3] = score_estimate
heapq.heappush(states, (score_estimate, target))
max_score = heapq.nlargest(1, states)[0][1][3]
print(max_score)
``` | output | 1 | 54,270 | 3 | 108,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000 | instruction | 0 | 54,271 | 3 | 108,542 |
Tags: dp, probabilities
Correct Solution:
```
F = lambda: map(float, input().split())
n = int(input())
arr = []
for i in range(n):
x, y, t, s = F()
arr.append([int(x), int(y), int(t), s])
arr.sort(key=lambda x: x[2])
dp = [0] * n
for i in range(n):
dp[i] = arr[i][-1]
for j in range(i):
if (arr[i][0] - arr[j][0]) ** 2 + (arr[i][1] - arr[j][1]) ** 2 <= (arr[i][2] - arr[j][2]) ** 2:
dp[i] = max(dp[i], dp[j] + arr[i][-1])
print(max(dp))
``` | output | 1 | 54,271 | 3 | 108,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000
Submitted Solution:
```
class Target:
def __init__(self, x, y, t, p):
self.x = x
self.y = y
self.t = t
self.p = p
self.best = 0
n = int(input())
targets = n * [ None ]
for i in range(n):
parts = input().split()
x, y, t = map(int, parts[:3])
p = float(parts[3])
targets[i] = Target(x, y, t, p)
targets.sort(key=lambda target: target.t)
best = 0
for i, a in enumerate(targets):
a.best += a.p
best = max(best, a.best)
for j in range(i + 1, n):
b = targets[j]
if (b.t - a.t) ** 2 < (b.x - a.x) ** 2 + (b.y - a.y) ** 2:
continue
b.best = max(b.best, a.best)
print(best)
``` | instruction | 0 | 54,272 | 3 | 108,544 |
Yes | output | 1 | 54,272 | 3 | 108,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000
Submitted Solution:
```
n=int(input())
b=[]
for i in range(n):
x,y,t,p=map(float,input().split())
b.append([t,x,y,p])
b.sort()
dp=[0 for j in range(n)]
dp[0]=b[0][3]
ans=0
for i in range(1,n):
t,x,y,p=b[i]
j=i-1
while(j>=0):
t1,x1,y1,p1=b[j]
if ((x1-x)**2+(y1-y)**2)**0.5<=(t-t1):
dp[i]=max(dp[i],dp[j])
j+=-1
dp[i]+=p
print(max(dp))
``` | instruction | 0 | 54,273 | 3 | 108,546 |
Yes | output | 1 | 54,273 | 3 | 108,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000
Submitted Solution:
```
#!/usr/bin/python3 -SOO
n = int(input())
ts = sorted((list(map(float,input().strip().split())) for _ in range(n)), key=lambda x:x[2])
d = [0]*n
ans = 0
for i in range(n):
d[i] = ts[i][3]
for j in range(i):
if (ts[i][0]-ts[j][0])**2 + (ts[i][1]-ts[j][1])**2 <= (ts[i][2]-ts[j][2])**2:
d[i] = max(d[i], d[j]+ts[i][3])
ans = max(ans, d[i])
print('%.9f'%ans)
``` | instruction | 0 | 54,274 | 3 | 108,548 |
Yes | output | 1 | 54,274 | 3 | 108,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000
Submitted Solution:
```
I = lambda : map(float , input().split())
n = int(input())
li = sorted ((list(I()) for i in range(n) ) ,key = lambda x : x[2] )
#print(li)
d = {}
ans=0
for i in range (n) :
x1 = li[i][0] ; y1 = li[i][1] ; t1 = li[i][2] ; p1 = li[i][3] ;
#print(p1)
d[i] = p1
for j in range (i) :
x = li[j][0] ; y = li[j][1] ; t = li[j][2] ; p = li[j][3] ;
if ((y1-y)**2 + (x1-x)**2 ) <= (t1-t)**2 :
d[i] = max(d[i],d[j]+p1)
ans = max(ans,d[i])
print(ans)
``` | instruction | 0 | 54,275 | 3 | 108,550 |
Yes | output | 1 | 54,275 | 3 | 108,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000
Submitted Solution:
```
#!/usr/bin/python3 -SOO
n = int(input())
ts = [[0,0,0,0]] + sorted(filter(lambda x:x[2]>=0, (list(map(float,input().strip().split())) for _ in range(n))), key=lambda x:x[2])
d = [0]*(n+1)
ans = 0
for i in range(1,n+1):
for j in range(i):
if (ts[i][0]-ts[j][0])**2 + (ts[i][1]-ts[j][1])**2 <= (ts[i][2]-ts[j][2])**2:
d[i] = max(d[i], d[j]+ts[i][3])
ans = max(ans, d[i])
print('%.9f'%ans)
``` | instruction | 0 | 54,276 | 3 | 108,552 |
No | output | 1 | 54,276 | 3 | 108,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000
Submitted Solution:
```
if __name__ == '__main__':
n = int(input())
targets = [input() for _ in range(n)]
targets = [line.split() for line in targets]
targets = [(int(x), int(y), int(t), float(p)) for x, y, t, p in targets]
targets = sorted(targets, key=lambda x: x[2])
max_scores = [0] * n
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
if targets[j][2] - targets[i][2] >= (abs(targets[j][0] - targets[i][0]) + abs(targets[j][1] - targets[i][1])):
max_scores[i] = max(max_scores[i], max_scores[j])
max_scores[i] += targets[i][3]
print(max(max_scores))
``` | instruction | 0 | 54,277 | 3 | 108,554 |
No | output | 1 | 54,277 | 3 | 108,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000
Submitted Solution:
```
I = lambda : map(float , input().split())
n = int(input())
li = sorted ((list(I()) for i in range(n) ) ,key = lambda x : x[2] )
#print(li)
d = {}
ans=0
for i in range (n) :
x1 = li[i][0] ; y1 = li[i][1] ; t1 = li[i][2] ; p1 = li[i][3] ;
#print(p1)
d[i] = p1
for j in range (i) :
x = li[j][0] ; y = li[j][1] ; t = li[j][2] ; p = li[j][3] ;
if ((y1-y)**2 + (x1-x)**2 ) <= (t1-t)**2 :
d[i] = max(d[i],d[j]+p1)
else :
d[i] = max(d[i],d[j])
ans = max(ans,d[i])
print(ans)
``` | instruction | 0 | 54,278 | 3 | 108,556 |
No | output | 1 | 54,278 | 3 | 108,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize β big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
Input
The first line contains integer n (1 β€ n β€ 1000) β amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti β integers, - 1000 β€ xi, yi β€ 1000, 0 β€ ti β€ 109, real number pi is given with no more than 6 digits after the decimal point, 0 β€ pi β€ 1). No two targets may be at the same point.
Output
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
Examples
Input
1
0 0 0 0.5
Output
0.5000000000
Input
2
0 0 0 0.6
5 0 5 0.7
Output
1.3000000000
Submitted Solution:
```
#---------------------------------------------
import sys
# sys.stdin = open('../input.txt', 'r'); sys.stdout = open('../output.txt', 'w')
mod = 1000000007
get_arr = lambda: list(map(int, input().split()))
get_int = lambda: int(input())
get_ints = lambda: map(int, input().split())
get_str = lambda: input()
get_strs = lambda: input().split()
#--------------------------------------------
cache = {}
def dist(x1, y1, x2, y2):
return ((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1))**0.5
def fn(i):
if(i==n): return 0
if i in cache: return cache[i]
ans = A[i][3]
for j in range(i+1, n):
time_left = A[j][0] - A[i][0]
time_need = dist(A[i][1], A[i][2], A[j][1], A[j][2])
if(time_need <= time_left):
ans = max(ans, A[i][3]+fn(j))
ans = max(ans, fn(i+1))
cache[i] = ans
return ans
n = get_int()
A = []
for i in range(n):
x, y, t, p = map(float, input().split())
A.append([t, x, y, p])
A = sorted(A)
print('%.12f'%fn(0))
``` | instruction | 0 | 54,279 | 3 | 108,558 |
No | output | 1 | 54,279 | 3 | 108,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The New Vasjuki village is stretched along the motorway and that's why every house on it is characterized by its shift relative to some fixed point β the xi coordinate. The village consists of n houses, the i-th house is located in the point with coordinates of xi.
TELE3, a cellular communication provider planned to locate three base stations so as to provide every house in the village with cellular communication. The base station having power d located in the point t provides with communication all the houses on the segment [t - d, t + d] (including boundaries).
To simplify the integration (and simply not to mix anything up) all the three stations are planned to possess the equal power of d. Which minimal value of d is enough to provide all the houses in the village with cellular communication.
Input
The first line contains an integer n (1 β€ n β€ 2Β·105) which represents the number of houses in the village. The second line contains the coordinates of houses β the sequence x1, x2, ..., xn of integer numbers (1 β€ xi β€ 109). It is possible that two or more houses are located on one point. The coordinates are given in a arbitrary order.
Output
Print the required minimal power d. In the second line print three numbers β the possible coordinates of the base stations' location. Print the coordinates with 6 digits after the decimal point. The positions of the stations can be any from 0 to 2Β·109 inclusively. It is accepted for the base stations to have matching coordinates. If there are many solutions, print any of them.
Examples
Input
4
1 2 3 4
Output
0.500000
1.500000 2.500000 3.500000
Input
3
10 20 30
Output
0
10.000000 20.000000 30.000000
Input
5
10003 10004 10001 10002 1
Output
0.500000
1.000000 10001.500000 10003.500000
Submitted Solution:
```
import os, sys
from io import IOBase, BytesIO
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill():
pass
return super(FastIO, self).read()
def readline(self):
while self.newlines == 0:
s = self._fill()
self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
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")
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self, a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = "\n"
def solve():
n = int(input())
homes = list(map(int, input().split()))
homes.sort()
def can_cover(d_i):
# print(d_i)
total_posts = 3
curr_cover = 0
ptr = 0
res = []
while ptr < len(homes):
if curr_cover < homes[ptr]:
if total_posts > 0:
total_posts -= 1
curr_cover = (homes[ptr] + (2 * d_i))
res.append(float("{:.6f}".format(min(homes[ptr] + d_i, homes[-1]))))
else:
break
ptr += 1
if ptr < len(homes):
return False, []
else:
return True, res
lo = 0
hi = 4 * (10 ** 8)
res = -1
mx = 100
while lo < hi and mx:
mid = lo + (hi - lo) / 2
r, pos = can_cover(mid)
if r:
res = mid
hi = mid
else:
lo = mid
mx -= 1
cout << ("%.6f" % res) << endl
cout << " ".join(map(str, can_cover(res)[1]))<<endl
def main():
solve()
if __name__ == "__main__":
main()
``` | instruction | 0 | 54,365 | 3 | 108,730 |
No | output | 1 | 54,365 | 3 | 108,731 |
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