message stringlengths 2 22.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 145 109k | cluster float64 9 9 | __index_level_0__ int64 290 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jzzhu has a big rectangular chocolate bar that consists of n Γ m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
* each cut should be straight (horizontal or vertical);
* each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
* each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 Γ 6 chocolate for 5 times.
<image>
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
Input
A single line contains three integers n, m, k (1 β€ n, m β€ 109; 1 β€ k β€ 2Β·109).
Output
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
Examples
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
Note
In the first sample, Jzzhu can cut the chocolate following the picture below:
<image>
In the second sample the optimal division looks like this:
<image>
In the third sample, it's impossible to cut a 2 Γ 3 chocolate 4 times.
Submitted Solution:
```
a = input().split(' ')
n = int(a[0])
m = int(a[1])
k = int(a[2])
if k>m+n-2:
print(-1)
else:
ans = 1
if k<n:
ans = max(ans, n//(k+1)*m)
if k<m:
ans = max(ans,m//(k+1)*n)
if k>=n:
ans = max( ans,m//(k-(n-1)+1) )
if k>=m:
ans = max( ans,n//(k-(m-1)+1) )
print (ans)
``` | instruction | 0 | 65,966 | 9 | 131,932 |
Yes | output | 1 | 65,966 | 9 | 131,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jzzhu has a big rectangular chocolate bar that consists of n Γ m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
* each cut should be straight (horizontal or vertical);
* each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
* each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 Γ 6 chocolate for 5 times.
<image>
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
Input
A single line contains three integers n, m, k (1 β€ n, m β€ 109; 1 β€ k β€ 2Β·109).
Output
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
Examples
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
Note
In the first sample, Jzzhu can cut the chocolate following the picture below:
<image>
In the second sample the optimal division looks like this:
<image>
In the third sample, it's impossible to cut a 2 Γ 3 chocolate 4 times.
Submitted Solution:
```
from sys import stdin as cin
from fractions import gcd
mod = 1000000007
n, m, k = map(int,cin.readline().split())
if k>(n+m-2):
print(-1)
else:
if n < m:
n,m = m,n
max_min_area = max((n*(m//(k+1))),(m*(n//(k+1))))
if (k+1) > m:
max_min_area = max(max_min_area,n//(k-m+2))
if (k+1) > n:
max_min_area = max(max_min_area,m//(k-n+2))
print(max_min_area)
``` | instruction | 0 | 65,967 | 9 | 131,934 |
Yes | output | 1 | 65,967 | 9 | 131,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jzzhu has a big rectangular chocolate bar that consists of n Γ m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
* each cut should be straight (horizontal or vertical);
* each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
* each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 Γ 6 chocolate for 5 times.
<image>
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
Input
A single line contains three integers n, m, k (1 β€ n, m β€ 109; 1 β€ k β€ 2Β·109).
Output
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
Examples
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
Note
In the first sample, Jzzhu can cut the chocolate following the picture below:
<image>
In the second sample the optimal division looks like this:
<image>
In the third sample, it's impossible to cut a 2 Γ 3 chocolate 4 times.
Submitted Solution:
```
import math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def printDivisors(n) :
A=[]
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
if (n / i == i) :
A.append(i)
else :
A.append(i)
A.append(n//i)
i = i + 1
return A
n,m,k= ilele()
z = n-1 + m-1
if k > z:
print(-1)
elif k == z:
print(1)
else:
mini = -1
A = printDivisors(n)
B = printDivisors(m)
#print(A,B)
for i in A:
l = i
b = k - ((n // l) - 1)
if b <= 0:
mini = max(mini,l*m)
else:
if b <= m-1:
r = max(1,m//(b+1))
mini = max(mini,l*r)
#print(mini)
for i in B:
r = i
b = k - ((m // r) - 1)
#print(i,b)
if b <= 0:
mini = max(mini,r*n)
else:
if b <= n-1:
l = max(1,n//(b+1))
mini = max(mini,l*r)
print(mini)
``` | instruction | 0 | 65,968 | 9 | 131,936 |
Yes | output | 1 | 65,968 | 9 | 131,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jzzhu has a big rectangular chocolate bar that consists of n Γ m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
* each cut should be straight (horizontal or vertical);
* each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
* each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 Γ 6 chocolate for 5 times.
<image>
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
Input
A single line contains three integers n, m, k (1 β€ n, m β€ 109; 1 β€ k β€ 2Β·109).
Output
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
Examples
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
Note
In the first sample, Jzzhu can cut the chocolate following the picture below:
<image>
In the second sample the optimal division looks like this:
<image>
In the third sample, it's impossible to cut a 2 Γ 3 chocolate 4 times.
Submitted Solution:
```
def split1(m,n,k):
if m%(k+1) == 0:
return n*(m//(k+1))
elif n%(k+1) == 0:
return m*(n//(k+1))
if k <= min(m,n):
return (m*n)//(max(m,n)//(k+1))
else:
k2 = k - (min(m,n) - 1)
return max(m,n) // (k2+1)
def split(*args):
r = split1(*args)
if r == 0: return -1
return r
args = input().split()
print(split(*map(int, args)))
``` | instruction | 0 | 65,969 | 9 | 131,938 |
No | output | 1 | 65,969 | 9 | 131,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jzzhu has a big rectangular chocolate bar that consists of n Γ m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
* each cut should be straight (horizontal or vertical);
* each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
* each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 Γ 6 chocolate for 5 times.
<image>
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
Input
A single line contains three integers n, m, k (1 β€ n, m β€ 109; 1 β€ k β€ 2Β·109).
Output
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
Examples
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
Note
In the first sample, Jzzhu can cut the chocolate following the picture below:
<image>
In the second sample the optimal division looks like this:
<image>
In the third sample, it's impossible to cut a 2 Γ 3 chocolate 4 times.
Submitted Solution:
```
a,b,k = map(int,input().split())
if a<b:
a,b=b,a
if k+2>a+b:
print(-1)
else:
area = a*b
if k>=a-1:
k-=(a-1)
area = b
print(area//(k+1))
else:
k = a//(k+1)
print(k*b)
``` | instruction | 0 | 65,970 | 9 | 131,940 |
No | output | 1 | 65,970 | 9 | 131,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jzzhu has a big rectangular chocolate bar that consists of n Γ m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
* each cut should be straight (horizontal or vertical);
* each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
* each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 Γ 6 chocolate for 5 times.
<image>
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
Input
A single line contains three integers n, m, k (1 β€ n, m β€ 109; 1 β€ k β€ 2Β·109).
Output
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
Examples
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
Note
In the first sample, Jzzhu can cut the chocolate following the picture below:
<image>
In the second sample the optimal division looks like this:
<image>
In the third sample, it's impossible to cut a 2 Γ 3 chocolate 4 times.
Submitted Solution:
```
a=input().split()
n,m,k=a
n,m,k=int(n),int(m),int(k)
if n-2+m>=k:
print(int(n*m/(k+1)))
else:
print(-1)
``` | instruction | 0 | 65,971 | 9 | 131,942 |
No | output | 1 | 65,971 | 9 | 131,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jzzhu has a big rectangular chocolate bar that consists of n Γ m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
* each cut should be straight (horizontal or vertical);
* each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
* each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 Γ 6 chocolate for 5 times.
<image>
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
Input
A single line contains three integers n, m, k (1 β€ n, m β€ 109; 1 β€ k β€ 2Β·109).
Output
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
Examples
Input
3 4 1
Output
6
Input
6 4 2
Output
8
Input
2 3 4
Output
-1
Note
In the first sample, Jzzhu can cut the chocolate following the picture below:
<image>
In the second sample the optimal division looks like this:
<image>
In the third sample, it's impossible to cut a 2 Γ 3 chocolate 4 times.
Submitted Solution:
```
import sys
import time
import math
from functools import lru_cache
INF = 10 ** 18 + 3
EPS = 1e-10
MAX_CACHE = 10 ** 9
MOD = 1 << 30
# Decorators
def time_it(function, output=sys.stderr):
def wrapped(*args, **kwargs):
start = time.time()
res = function(*args, **kwargs)
elapsed_time = time.time() - start
print('"%s" took %f ms' % (function.__name__, elapsed_time * 1000),
file=output)
return res
return wrapped
@time_it
def main():
n, m, k = map(int, input().split())
n, m = sorted((n, m))
if k + 1 >= m:
res = n // (k + 2 - m)
else:
res = n * (m // (k + 1))
if res == 0:
res = -1
print(res)
def set_input(file):
global input
input = lambda: file.readline().strip()
def set_output(file):
global print
local_print = print
def print(*args, **kwargs):
kwargs["file"] = kwargs.get("file", file)
return local_print(*args, **kwargs)
if __name__ == '__main__':
set_input(open("input.txt", "r") if "MINE" in sys.argv else sys.stdin)
set_output(sys.stdout)
main()
``` | instruction | 0 | 65,972 | 9 | 131,944 |
No | output | 1 | 65,972 | 9 | 131,945 |
Provide a correct Python 3 solution for this coding contest problem.
A: Many Kinds of Apples
Problem Statement
Apple Farmer Mon has two kinds of tasks: "harvest apples" and "ship apples".
There are N different species of apples, and N distinguishable boxes. Apples are labeled by the species, and boxes are also labeled, from 1 to N. The i-th species of apples are stored in the i-th box.
For each i, the i-th box can store at most c_i apples, and it is initially empty (no apple exists).
Mon receives Q instructions from his boss Kukui, and Mon completely follows in order. Each instruction is either of two types below.
* "harvest apples": put d x-th apples into the x-th box.
* "ship apples": take d x-th apples out from the x-th box.
However, not all instructions are possible to carry out. Now we call an instruction which meets either of following conditions "impossible instruction":
* When Mon harvest apples, the amount of apples exceeds the capacity of that box.
* When Mon tries to ship apples, there are not enough apples to ship.
Your task is to detect the instruction which is impossible to carry out.
Input
Input is given in the following format.
N
c_1 c_2 $ \ cdots $ c_N
Q
t_1 x_1 d_1
t_2 x_2 d_2
$ \ vdots $
t_Q x_Q d_Q
In line 1, you are given the integer N, which indicates the number of species of apples.
In line 2, given c_i (1 \ leq i \ leq N) separated by whitespaces. C_i indicates the capacity of the i-th box.
In line 3, given Q, which indicates the number of instructions.
Instructions are given successive Q lines. T_i x_i d_i means what kind of instruction, which apple Mon handles in this instruction, how many apples Mon handles, respectively. If t_i is equal to 1, it means Mon does the task of "harvest apples" , else if t_i is equal to 2, it means Mon does the task of "ship apples".
Constraints
All input values ββare integers, and satisfy following constraints.
* 1 \ leq N \ leq 1,000
* 1 \ leq c_i \ leq 100,000 (1 \ leq i \ leq N)
* 1 \ leq Q \ leq 100,000
* t_i \ in \\ {1, 2 \\} (1 \ leq i \ leq Q)
* 1 \ leq x_i \ leq N (1 \ leq i \ leq Q)
* 1 \ leq d_i \ leq 100,000 (1 \ leq i \ leq Q)
Output
If there is "impossible instruction", output the index of the apples which have something to do with the first "impossible instruction".
Otherwise, output 0.
Sample Input 1
2
3 3
Four
1 1 2
one two Three
2 1 3
2 2 3
Sample Output 1
1
In this case, there are not enough apples to ship in the first box.
Sample Input 2
2
3 3
Four
1 1 3
2 1 2
one two Three
1 1 3
Sample Output 2
1
In this case, the amount of apples exceeds the capacity of the first box.
Sample Input 3
3
3 4 5
Four
1 1 3
one two Three
1 3 5
2 2 2
Sample Output 3
0
Sample Input 4
6
28 56 99 3 125 37
Ten
1 1 10
1 1 14
1 3 90
1 5 10
2 3 38
2 1 5
1 3 92
1 6 18
2 5 9
2 1 4
Sample Output 4
3
Example
Input
2
3 3
4
1 1 2
1 2 3
2 1 3
2 2 3
Output
1 | instruction | 0 | 66,420 | 9 | 132,840 |
"Correct Solution:
```
N = int(input())
D = [int(c) for c in input().split()]
C = [0]*(N)
Q = int(input())
for _ in range(0,Q):
t,x,d = map(int,input().split())
if t == 1:
C[x-1] = C[x-1] + d
if C[x-1] > D[x-1]:
print(x)
exit()
if t == 2:
C[x-1] = C[x-1] - d
if C[x-1] < 0:
print(x)
exit()
print(0)
``` | output | 1 | 66,420 | 9 | 132,841 |
Provide a correct Python 3 solution for this coding contest problem.
A: Many Kinds of Apples
Problem Statement
Apple Farmer Mon has two kinds of tasks: "harvest apples" and "ship apples".
There are N different species of apples, and N distinguishable boxes. Apples are labeled by the species, and boxes are also labeled, from 1 to N. The i-th species of apples are stored in the i-th box.
For each i, the i-th box can store at most c_i apples, and it is initially empty (no apple exists).
Mon receives Q instructions from his boss Kukui, and Mon completely follows in order. Each instruction is either of two types below.
* "harvest apples": put d x-th apples into the x-th box.
* "ship apples": take d x-th apples out from the x-th box.
However, not all instructions are possible to carry out. Now we call an instruction which meets either of following conditions "impossible instruction":
* When Mon harvest apples, the amount of apples exceeds the capacity of that box.
* When Mon tries to ship apples, there are not enough apples to ship.
Your task is to detect the instruction which is impossible to carry out.
Input
Input is given in the following format.
N
c_1 c_2 $ \ cdots $ c_N
Q
t_1 x_1 d_1
t_2 x_2 d_2
$ \ vdots $
t_Q x_Q d_Q
In line 1, you are given the integer N, which indicates the number of species of apples.
In line 2, given c_i (1 \ leq i \ leq N) separated by whitespaces. C_i indicates the capacity of the i-th box.
In line 3, given Q, which indicates the number of instructions.
Instructions are given successive Q lines. T_i x_i d_i means what kind of instruction, which apple Mon handles in this instruction, how many apples Mon handles, respectively. If t_i is equal to 1, it means Mon does the task of "harvest apples" , else if t_i is equal to 2, it means Mon does the task of "ship apples".
Constraints
All input values ββare integers, and satisfy following constraints.
* 1 \ leq N \ leq 1,000
* 1 \ leq c_i \ leq 100,000 (1 \ leq i \ leq N)
* 1 \ leq Q \ leq 100,000
* t_i \ in \\ {1, 2 \\} (1 \ leq i \ leq Q)
* 1 \ leq x_i \ leq N (1 \ leq i \ leq Q)
* 1 \ leq d_i \ leq 100,000 (1 \ leq i \ leq Q)
Output
If there is "impossible instruction", output the index of the apples which have something to do with the first "impossible instruction".
Otherwise, output 0.
Sample Input 1
2
3 3
Four
1 1 2
one two Three
2 1 3
2 2 3
Sample Output 1
1
In this case, there are not enough apples to ship in the first box.
Sample Input 2
2
3 3
Four
1 1 3
2 1 2
one two Three
1 1 3
Sample Output 2
1
In this case, the amount of apples exceeds the capacity of the first box.
Sample Input 3
3
3 4 5
Four
1 1 3
one two Three
1 3 5
2 2 2
Sample Output 3
0
Sample Input 4
6
28 56 99 3 125 37
Ten
1 1 10
1 1 14
1 3 90
1 5 10
2 3 38
2 1 5
1 3 92
1 6 18
2 5 9
2 1 4
Sample Output 4
3
Example
Input
2
3 3
4
1 1 2
1 2 3
2 1 3
2 2 3
Output
1 | instruction | 0 | 66,421 | 9 | 132,842 |
"Correct Solution:
```
n = int(input())
c = list(map(int,input().split()))
b = [ 0 for _ in range(n) ]
q = int(input())
r = 0
for _ in range(q):
[t,x,d] = map(int,input().split())
if t == 1: # εη©«
b[x-1] += d
if b[x-1] > c[x-1]:
r = x
break
elif t == 2: # εΊθ·
b[x-1] -= d
if b[x-1] < 0:
r = x
break
print(r)
``` | output | 1 | 66,421 | 9 | 132,843 |
Provide a correct Python 3 solution for this coding contest problem.
A: Many Kinds of Apples
Problem Statement
Apple Farmer Mon has two kinds of tasks: "harvest apples" and "ship apples".
There are N different species of apples, and N distinguishable boxes. Apples are labeled by the species, and boxes are also labeled, from 1 to N. The i-th species of apples are stored in the i-th box.
For each i, the i-th box can store at most c_i apples, and it is initially empty (no apple exists).
Mon receives Q instructions from his boss Kukui, and Mon completely follows in order. Each instruction is either of two types below.
* "harvest apples": put d x-th apples into the x-th box.
* "ship apples": take d x-th apples out from the x-th box.
However, not all instructions are possible to carry out. Now we call an instruction which meets either of following conditions "impossible instruction":
* When Mon harvest apples, the amount of apples exceeds the capacity of that box.
* When Mon tries to ship apples, there are not enough apples to ship.
Your task is to detect the instruction which is impossible to carry out.
Input
Input is given in the following format.
N
c_1 c_2 $ \ cdots $ c_N
Q
t_1 x_1 d_1
t_2 x_2 d_2
$ \ vdots $
t_Q x_Q d_Q
In line 1, you are given the integer N, which indicates the number of species of apples.
In line 2, given c_i (1 \ leq i \ leq N) separated by whitespaces. C_i indicates the capacity of the i-th box.
In line 3, given Q, which indicates the number of instructions.
Instructions are given successive Q lines. T_i x_i d_i means what kind of instruction, which apple Mon handles in this instruction, how many apples Mon handles, respectively. If t_i is equal to 1, it means Mon does the task of "harvest apples" , else if t_i is equal to 2, it means Mon does the task of "ship apples".
Constraints
All input values ββare integers, and satisfy following constraints.
* 1 \ leq N \ leq 1,000
* 1 \ leq c_i \ leq 100,000 (1 \ leq i \ leq N)
* 1 \ leq Q \ leq 100,000
* t_i \ in \\ {1, 2 \\} (1 \ leq i \ leq Q)
* 1 \ leq x_i \ leq N (1 \ leq i \ leq Q)
* 1 \ leq d_i \ leq 100,000 (1 \ leq i \ leq Q)
Output
If there is "impossible instruction", output the index of the apples which have something to do with the first "impossible instruction".
Otherwise, output 0.
Sample Input 1
2
3 3
Four
1 1 2
one two Three
2 1 3
2 2 3
Sample Output 1
1
In this case, there are not enough apples to ship in the first box.
Sample Input 2
2
3 3
Four
1 1 3
2 1 2
one two Three
1 1 3
Sample Output 2
1
In this case, the amount of apples exceeds the capacity of the first box.
Sample Input 3
3
3 4 5
Four
1 1 3
one two Three
1 3 5
2 2 2
Sample Output 3
0
Sample Input 4
6
28 56 99 3 125 37
Ten
1 1 10
1 1 14
1 3 90
1 5 10
2 3 38
2 1 5
1 3 92
1 6 18
2 5 9
2 1 4
Sample Output 4
3
Example
Input
2
3 3
4
1 1 2
1 2 3
2 1 3
2 2 3
Output
1 | instruction | 0 | 66,422 | 9 | 132,844 |
"Correct Solution:
```
n = int(input())
C = list(map(int, input().split()))
q = int(input())
data = []
for _ in range(q):
t, x, d = map(int, input().split())
data.append([t, x, d])
apple = [0] * n
for D in data:
t, x, d = D
if t == 1:
apple[x-1] += d
if apple[x-1] > C[x-1]:
print(x)
break
else:
apple[x-1] -= d
if apple[x-1] < 0:
print(x)
break
else:
print(0)
``` | output | 1 | 66,422 | 9 | 132,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14 | instruction | 0 | 66,698 | 9 | 133,396 |
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
h, res = 0, -1
for _ in range(n):
i = int(input())
res += abs(i - h) + 2
h = i
print(res)
``` | output | 1 | 66,698 | 9 | 133,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14 | instruction | 0 | 66,699 | 9 | 133,398 |
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
a = []
for _ in range(n):
a.append(int(input()))
t = a[0]
for i in range(n-1):
if a[i+1] >= a[i]:
t += a[i+1]-a[i]+1
else:
t += a[i]-a[i+1]+1
print(t+n)
``` | output | 1 | 66,699 | 9 | 133,399 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14 | instruction | 0 | 66,700 | 9 | 133,400 |
Tags: greedy, implementation
Correct Solution:
```
l = [int(input()) for i in range(int(input()))]
print(sum([abs(l[i]-l[i-1]) if i>0 else l[i] for i in range(len(l))])+(2*len(l)-1))
``` | output | 1 | 66,700 | 9 | 133,401 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14 | instruction | 0 | 66,701 | 9 | 133,402 |
Tags: greedy, implementation
Correct Solution:
```
n=int(input())
ar=[]
for x in range(n):
ar.append(int(input()))
sum=n+n-1
h=0
for i in range(n):
sum+=abs(ar[i]-h)
h=ar[i]
# sum+=ar[-1]
print(sum)
``` | output | 1 | 66,701 | 9 | 133,403 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14 | instruction | 0 | 66,702 | 9 | 133,404 |
Tags: greedy, implementation
Correct Solution:
```
#
# 24.06.2021
#
# CF 162 B
n = int (input ())
hp = int (input ())
k = hp + 1
for i in range (1, n) :
hc = int (input ())
if ( hp <= hc ) :
k += hc - hp + 2
else :
k += hp - hc + 2
hp = hc
print (k)
``` | output | 1 | 66,702 | 9 | 133,405 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14 | instruction | 0 | 66,703 | 9 | 133,406 |
Tags: greedy, implementation
Correct Solution:
```
n=int(input())
l=[int(input()) for _ in range(n)]
ans=n+l[0]
for i in range(1,n):
ans+=abs(l[i-1]-l[i])+1
print(ans)
``` | output | 1 | 66,703 | 9 | 133,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14 | instruction | 0 | 66,704 | 9 | 133,408 |
Tags: greedy, implementation
Correct Solution:
```
n = int(input())
curr=0
summ=0
for i in range(n):
new= int(input())
summ+=abs(new-curr)
curr=new
summ+=2
summ-=1
print(summ)
``` | output | 1 | 66,704 | 9 | 133,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14
Submitted Solution:
```
n=int(input())
j=0
l=0
for i in range(n):
L=int(input())
l+=int(i>0)+abs(L-j)+1
j=L
print(l)
``` | instruction | 0 | 66,706 | 9 | 133,412 |
Yes | output | 1 | 66,706 | 9 | 133,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14
Submitted Solution:
```
n=int(input())
h=[]
for _ in range(n):
h.append(int(input()))
# print(h)
ans=h[0]+1
for i in range(1,n):
if h[i]>=h[i-1]:
ans+=h[i]-h[i-1]+1
else:
ans+=h[i-1]-h[i]+1
print(ans+n-1)
``` | instruction | 0 | 66,707 | 9 | 133,414 |
Yes | output | 1 | 66,707 | 9 | 133,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14
Submitted Solution:
```
n=int(input())
a=1
s=0
for _ in range(n):
h=int(input())
s=s+abs(a-h)+2
a=h
print(s)
``` | instruction | 0 | 66,708 | 9 | 133,416 |
Yes | output | 1 | 66,708 | 9 | 133,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14
Submitted Solution:
```
n = int(input())
lst = list()
for i in range(n):
s = int(input())
lst.append(s)
result, temp = 2 * n - 1 + lst[0], 0
for i in range(1, n):
temp = lst[i] - lst[i - 1]
result += abs(temp)
print(result)
``` | instruction | 0 | 66,709 | 9 | 133,418 |
Yes | output | 1 | 66,709 | 9 | 133,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14
Submitted Solution:
```
n=int(input())
arr=[]
for i in range(0,n):
arr.append(int(input()))
m=min(arr)
ans=(n)+2*(sum(arr)-n)+n-1
print(ans)
``` | instruction | 0 | 66,710 | 9 | 133,420 |
No | output | 1 | 66,710 | 9 | 133,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14
Submitted Solution:
```
n = int(input())
first_tree = int(input())
cur_height = first_tree
steps = first_tree+1
c = 0
while c < n-1:
next_tree = int(input())
steps += 2 # jump and eat
# go up
if cur_height < next_tree:
steps += (next_tree-cur_height)
# go down
elif cur_height > next_tree:
steps += (cur_height-next_tree)
c+=1
print(steps)
``` | instruction | 0 | 66,711 | 9 | 133,422 |
No | output | 1 | 66,711 | 9 | 133,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14
Submitted Solution:
```
n=int(input())
s=0
x=[0]
for i in range(1,n):
x.append(int(input()))
s+=abs(x[i-1]-x[i])+1
print(s+n)
``` | instruction | 0 | 66,712 | 9 | 133,424 |
No | output | 1 | 66,712 | 9 | 133,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
* Walk up or down one unit on a tree.
* Eat a nut on the top of the current tree.
* Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 β€ i β€ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
Input
The first line contains an integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 β€ hi β€ 104) β the height of the tree with the number i.
Output
Print a single integer β the minimal time required to eat all nuts in seconds.
Examples
Input
2
1
2
Output
5
Input
5
2
1
2
1
1
Output
14
Submitted Solution:
```
n = int(input())
li = []
for i in range(n):
li.append(int(input()))
sum = li[0] + 1
for i in range(n-1):
if li[i] > li[i+1]:
sum += min(li[i], li[i+1]) + 2
else:
diff = li[i+1] - li[i]
sum += diff+2
#print(sum)
print(sum)
``` | instruction | 0 | 66,713 | 9 | 133,426 |
No | output | 1 | 66,713 | 9 | 133,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample. | instruction | 0 | 66,816 | 9 | 133,632 |
Tags: brute force, dp
Correct Solution:
```
t = int(input())
dp = [[[0 for i in range(51)] for j in range(31)] for k in range(31)]
def cost(n, m, k):
if (dp[n][m][k] or k == 0 or n * m == k): return dp[n][m][k]
c = 10**9
for i in range(1, n // 2 + 1):
for j in range(k + 1):
c = min(c, cost(i, m, j) + cost(n - i, m, k - j) + m * m)
for i in range(1, m // 2 + 1):
for j in range(k + 1):
c = min(c, cost(n, i, j) + cost(n, m - i, k - j) + n * n)
dp[n][m][k] = c
return c
for _ in range(t):
n, m, k = map(int, input().split())
print(cost(n, m, k))
``` | output | 1 | 66,816 | 9 | 133,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample. | instruction | 0 | 66,817 | 9 | 133,634 |
Tags: brute force, dp
Correct Solution:
```
import sys
input=sys.stdin.readline
def main():
ans=[]
memo=[[[-1 for _ in range(51)] for _ in range(31)] for _ in range(31)]
def solve(n, m , k) :
if n*m == k or k==0: return 0
if memo[n][m][k] > -1 : return memo[n][m][k]
if memo[m][n][k] > -1 : memo[n][m][k]=memo[m][n][k] ; return memo[n][m][k]
r=float('inf')
for i in range(k+1):
for j in range(1,max(m,n)):
if m > j :
r=min(r,n**2+solve(j,n,i)+solve(m-j,n,k-i))
if n > j :
r=min(r,m**2+solve(m,j,i)+solve(m,n-j,k-i))
memo[n][m][k] = r
return r
for _ in range(int(input())):
n,m,k = map(int,input().split())
ans.append(str(solve(n,m,k)))
print('\n'.join(ans))
main()
``` | output | 1 | 66,817 | 9 | 133,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample. | instruction | 0 | 66,818 | 9 | 133,636 |
Tags: brute force, dp
Correct Solution:
```
import sys
input = sys.stdin.readline
d={}
testnumber = int(input())
def calc(n, m, k):
if k <= 0 or k == m*n:
return 0
if k > n*m:
return 1000_000_000
global d
if n < m:
n, m = m, n
if k > (m*n - m):
return m*m + 1
if k < m:
return m*m + 1
if k % m == 0:
return m*m
if (n, m, k) in d:
return d[ (n, m, k)]
d[ (n, m, k) ] = min( calc2(n, m, k), calc2(m, n, k) )
return d[ (n, m, k) ]
def calc2(n, m, k):
m2 = m*m
ans = m2*2 + 1
for i in range(1, n):
if i*m >= k:
ans = min(ans, m2 + calc(m, i, k) )
else:
ans = min(ans, m2 + calc(m, n-i, k - i*m))
return ans
for ntest in range(testnumber):
n, m, k = map( int, input().split() )
if k == n*m:
print(0)
continue
print( calc(n, m, k) )
``` | output | 1 | 66,818 | 9 | 133,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample. | instruction | 0 | 66,819 | 9 | 133,638 |
Tags: brute force, dp
Correct Solution:
```
t = int(input())
dp = [[[0 for i in range(51)] for j in range(31)] for k in range(31)]
def cost(n, m, k):
if (dp[n][m][k] or k == 0 or n * m == k): return dp[n][m][k]
c = 10**9
for i in range(1, n // 2 + 1):
for j in range(k + 1):
c = min(c, cost(n - i, m, k - j) + cost(i, m, j) + m * m)
for i in range(1, m // 2 + 1):
for j in range(k + 1):
c = min(c, cost(n, m - i, k - j) + cost(n, i, j) + n * n)
dp[n][m][k] = c
return c
for _ in range(t):
n, m, k = map(int, input().split())
print(cost(n, m, k))
# mem = [[[0 for i in range(51)] for j in range(31)] for k in range(31)]
# def f(n, m, k):
# if mem[n][m][k]:
# return mem[n][m][k]
# if (n*m == k) or (k == 0):
# return 0
# cost = 10**9
# for x in range(1, n//2 + 1):
# for z in range(k+1):
# cost = min(cost, m*m + f(n-x, m, k-z) + f(x, m, z))
# for y in range(1, m//2 + 1):
# for z in range(k+1):
# cost = min(cost, n*n + f(n, m-y, k-z) + f(n, y, z))
# mem[n][m][k] = cost
# return cost
# t = int(input())
# for i in range(t):
# n, m, k = map(int, input().split())
# print(f(n, m, k))
``` | output | 1 | 66,819 | 9 | 133,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample. | instruction | 0 | 66,820 | 9 | 133,640 |
Tags: brute force, dp
Correct Solution:
```
d = [0] * 49011
def g(n, m, k):
t = 1e9
for i in range(1, m // 2 + 1):
for j in range(k + 1):
t = min(t, f(n, m - i, k - j) + f(n, i, j))
return n * n + t
def f(n, m, k):
if n > m: n, m = m, n
k = min(k, n * m - k)
if k == 0: return 0
if k < 0: return 1e9
q = n + 31 * m + 961 * k
if d[q] == 0: d[q] = min(g(n, m, k), g(m, n, k))
return d[q]
for q in range(int(input())):
n, m, k = map(int, input().split())
print(f(n, m, k))
``` | output | 1 | 66,820 | 9 | 133,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample. | instruction | 0 | 66,821 | 9 | 133,642 |
Tags: brute force, dp
Correct Solution:
```
import sys
#import random
#import bisect
from collections import deque
#sys.setrecursionlimit(10**6)
from queue import PriorityQueue
from math import gcd
from math import log
from math import ceil
from math import pi
input_ = lambda: sys.stdin.readline().strip("\r\n")
ii = lambda : int(input_())
il = lambda : list(map(int, input_().split()))
ilf = lambda : list(map(float, input_().split()))
ip = lambda : input_()
fi = lambda : float(input_())
ap = lambda ab,bc,cd : ab[bc].append(cd)
li = lambda : list(input_())
pr = lambda x : print(x)
prinT = lambda x : print(x)
f = lambda : sys.stdout.flush()
mod = 10**9 + 7
dp = [[[0 for k in range (51)] for j in range (31)] for i in range (31)]
def calc (n,m,k) :
#print(n,m,k)
if (dp[n][m][k] != 0) or (n*m == k) or (k == 0) :
return dp[n][m][k]
ans = 10**9
for i in range (1,n//2 + 1) :
for j in range (k+1) :
#print(i,j,'a')
if (i*m >= (k-j)) and ((n-i)*m >= j) :
ans = min(ans, m*m + calc(i,m,k-j) + calc(n-i,m,j))
for i in range (1,m//2 + 1) :
for j in range (k+1) :
#print(i,j,'b')
if (i*n >= (k-j)) and ((m-i)*n >= j) :
ans = min(ans, n*n + calc(n,i,k-j) + calc(n,m-i,j))
dp[n][m][k] = ans
return ans
for _ in range (ii()) :
n,m,k = il()
print(calc(n,m,k))
``` | output | 1 | 66,821 | 9 | 133,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample. | instruction | 0 | 66,822 | 9 | 133,644 |
Tags: brute force, dp
Correct Solution:
```
import sys
# sys.stdin = open('ivo.in')
mem = []
for i in range(32):
mem.append([[-1] * 52 for u in range(32)])
def solve(x, y, z):
if x > y:
mem[x][y][z] = solve(y, x, z)
return mem[x][y][z]
if x * y == z or z == 0:
mem[x][y][z] = 0
return 0
if x * y < z:
mem[x][y][z] = -2
return -2
res = -2
for i in range(1, x//2 + 1):
for eaten in range(z + 1):
t1 = mem[i][y][eaten] if mem[i][y][eaten] != -1 else solve(i, y, eaten)
if t1 == -2:
continue
t2 = mem[x - i][y][z - eaten] if mem[x - i][y][z - eaten] != -1 else solve(x - i, y, z - eaten)
if t2 == -2:
continue
if res == -2 or res > t1 + t2 + y * y:
res = t1 + t2 + y * y
for j in range(1, y//2 + 1):
for eaten in range(z + 1):
t1 = mem[x][j][eaten] if mem[x][j][eaten] != -1 else solve(x, j, eaten)
if t1 == -2:
continue
t2 = mem[x][y - j][z - eaten] if mem[x][y - j][z - eaten] != -1 else solve(x, y - j, z - eaten)
if t2 == -2:
continue
if res == -2 or res > t1 + t2 + x * x:
res = t1 + t2 + x * x
mem[x][y][z] = res
return mem[x][y][z]
t = int(sys.stdin.readline())
for it in range(t):
n, m, k = map(int, sys.stdin.readline().split())
print(solve(n, m, k))
``` | output | 1 | 66,822 | 9 | 133,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample. | instruction | 0 | 66,823 | 9 | 133,646 |
Tags: brute force, dp
Correct Solution:
```
d=[[[0 for i in range(51)] for j in range(31)] for k in range(31)]
for i in range(31):
d.append([])
for j in range(31):
d[i].append([])
for k in range(50):
d[i][j].append(0)
def rec(n,m,k):
global d
if n*m==k or k==0:
return 0
if d[n][m][k]>0:
return d[n][m][k]
if n*m<k:
return 10**10
cost=10**10
for i in range(1,n//2+1):
for j in range(k+1):
cost=min(cost,m*m+rec(n-i,m,k-j)+rec(i,m,j))
for i in range(1,m//2+1):
for j in range(k+1):
cost=min(cost,n*n+rec(n,m-i,k-j)+rec(n,i,j))
d[n][m][k]=cost
return cost
for i in range(int(input())):
a,b,c=map(int,input().split())
print(rec(a,b,c))
``` | output | 1 | 66,823 | 9 | 133,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.
Submitted Solution:
```
d=[[[0 for i in range(51)] for j in range(31)] for k in range(31)]
def rec(n,m,k):
global d
if n*m==k or k==0:
return 0
if d[n][m][k]>0:
return d[n][m][k]
if n*m<k:
return 10**10
cost=10**10
for i in range(1,n//2+1):
for j in range(k+1):
cost=min(cost,m*m+rec(n-i,m,k-j)+rec(i,m,j))
for i in range(1,m//2+1):
for j in range(k+1):
cost=min(cost,n*n+rec(n,m-i,k-j)+rec(n,i,j))
d[n][m][k]=cost
return cost
for i in range(int(input())):
a,b,c=map(int,input().split())
print(rec(a,b,c))
``` | instruction | 0 | 66,824 | 9 | 133,648 |
Yes | output | 1 | 66,824 | 9 | 133,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.
Submitted Solution:
```
d = [[[0 for i in range(51)] for j in range(31)] for k in range(31)]
def rec(n, m, k):
global d
if n*m == k or k == 0:
return 0
if d[n][m][k] > 0:
return d[n][m][k]
if n*m<k:
return 10**10
cost = 10**10
for i in range(1, n // 2 + 1):
for j in range(k+1):
cost = min(cost, m*m + rec(n-i, m, k-j) + rec(i, m, j))
for i in range(1, m // 2 + 1):
for j in range(k+1):
cost = min(cost, n*n + rec(n, m-i, k-j) + rec(n, i, j))
d[n][m][k] = cost
return cost
p = []
t = int(input())
for i in range(t):
n, m, k = map(int, input().split())
p.append(rec(n, m, k))
print('\n'.join(str(x) for x in p))
``` | instruction | 0 | 66,825 | 9 | 133,650 |
Yes | output | 1 | 66,825 | 9 | 133,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.
Submitted Solution:
```
d = [[[0 for i in range(51)] for j in range(31)] for k in range(31)]
def rec(n, m, k):
global d
if n*m == k or k == 0:
return 0
if d[n][m][k] > 0:
return d[n][m][k]
if n*m<k:
return 10**10
cost = 10**10
for i in range(1, n // 2 + 1):
for j in range(k+1):
cost = min(cost, m*m + rec(n-i, m, k-j) + rec(i, m, j))
for i in range(1, m // 2 + 1):
for j in range(k+1):
cost = min(cost, n*n + rec(n, m-i, k-j) + rec(n, i, j))
d[n][m][k] = cost
return cost
p = []
t = int(input())
for i in range(t):
n, m, k = map(int, input().split())
#p.append(rec(n, m, k))
print(rec(n, m, k))
#print('\n'.join(str(x) for x in p))
``` | instruction | 0 | 66,826 | 9 | 133,652 |
Yes | output | 1 | 66,826 | 9 | 133,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <=key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
dp=[[[-1 for i in range(51)] for j in range(31)] for t in range(31)]
for ik in range(int(input())):
n,m,k=map(int,input().split())
def find(n,m,k):
if k==0:
return 0
if dp[n][m][k]!=-1:
return dp[n][m][k]
if n*m<k:
return 10**9
if n*m==k:
return 0
ans=10**9
if n==1 or m==1:
t=1
if n*m==k:
t=0
dp[n][m][k]=t
return t
for i in range(m-1):
for j in range(k+1):
ans=min(ans,find(n,i+1,j)+find(n,m-1-i,k-j)+n*n)
for i in range(n-1):
for j in range(k+1):
ans=min(ans,find(i+1,m,j)+find(n-1-i,m,k-j)+m*m)
dp[n][m][k]=ans
return ans
print(find(n,m,k))
``` | instruction | 0 | 66,827 | 9 | 133,654 |
Yes | output | 1 | 66,827 | 9 | 133,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.
Submitted Solution:
```
t=int(input())
p=[0]*t
for i in range(t):
n,m,k=map(int,input().split())
k=min(k,m*n-k)
if n>m:
n,m=m,n
while n*m!=k and k!=0:
p[i]+=n**2
m=k/n
if m!=int(m):
m=int(m+1)
if n>m:
n,m=m,n
k=min(k,m*n-k)
for i in p:
print(int(i))
``` | instruction | 0 | 66,828 | 9 | 133,656 |
No | output | 1 | 66,828 | 9 | 133,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.
Submitted Solution:
```
t = int(input())
din = dict()
def rec(n, m, k):
if n * m == k:
return 0
if n < 2 or m < 2:
return 1
if n in din:
if m in din[n]:
if k in din[n][m]:
return din[n][m][k]
r1 = n * n
if k % n != 0:
r1 += rec(n, m - k // n - 1, k % n)
r2 = m * m
if k % m != 0:
r2 += rec(n - k // m - 1, m, k % m)
res = min(r1, r2)
if not n in din:
din[n] = dict()
if not m in din[n]:
din[n][m] = dict()
din[n][m][k] = res
return res
for it in range(t):
n, m, k = [int(i) for i in input().split()]
print(rec(n, m, k))
``` | instruction | 0 | 66,829 | 9 | 133,658 |
No | output | 1 | 66,829 | 9 | 133,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.
Submitted Solution:
```
t=int(input())
p=[0]*t
for i in range(t):
n,m,k=map(int,input().split())
if i==-1:
print(n,m,k)
k=min(k,m*n-k)
if n>m:
n,m=m,n
jj=-1
if k%m==0:
jj=m**2
while k!=0:
p[i]+=n**2
mm=m
m=k/n
if m!=int(m):
m=int(m+1)
if mm*n-m*n<k and k%m!=0:
k=k-(mm*n-m*n)
if n>m:
n,m=m,n
k=min(k,m*n-k)
p[i]=int(p[i])
if jj!=-1:
p[i]=min(p[i],jj)
for i in p:
print(i)
``` | instruction | 0 | 66,830 | 9 | 133,660 |
No | output | 1 | 66,830 | 9 | 133,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a rectangular chocolate bar consisting of n Γ m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 Γ 3 unit squares then you can break it horizontally and get two 1 Γ 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 Γ 1 and 2 Γ 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining nΒ·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 β€ t β€ 40910) β the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 β€ n, m β€ 30, 1 β€ k β€ min(nΒ·m, 50)) β the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
Input
4
2 2 1
2 2 3
2 2 2
2 2 4
Output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
* to split 2 Γ 2 bar into two pieces of 2 Γ 1 (cost is 22 = 4),
* to split the resulting 2 Γ 1 into two 1 Γ 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.
Submitted Solution:
```
t=int(input())
p=[0]*t
for i in range(t):
n,m,k=map(int,input().split())
if i==-1:
print(n,m,k)
k=min(k,m*n-k)
if n>m:
n,m=m,n
jj=-1
if k%m==0:
jj=m**2
while k!=0:
p[i]+=n**2
m=k/n
if m!=int(m):
m=int(m+1)
if n>m:
n,m=m,n
k=min(k,m*n-k)
p[i]=int(p[i])
if jj!=-1:
p[i]=min(p[i],jj)
for i in p:
print(i)
``` | instruction | 0 | 66,831 | 9 | 133,662 |
No | output | 1 | 66,831 | 9 | 133,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification:
* The first box stores toys with fun values in range of (-β,-1].
* The second box stores toys with fun values in range of (-1, 0).
* The third box stores toys with fun values in range of (0, 1).
* The fourth box stores toys with fun value in range of [1, β).
Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.
While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.
As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.
Input
The first line has an integer T (1 β€ T β€ 5 β
10^4), the number of test cases.
Every case contains a line with four space-separated integers A B C D (0 β€ A, B, C, D β€ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.
Output
For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.
For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.
Example
Input
2
1 2 0 1
0 1 0 0
Output
Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak
Note
For the first case, here is a scenario where the first box is the special box:
* The first box had toys with fun values \{-3\}.
* The second box had toys with fun values \{ -0.5, -0.5 \}
* The fourth box had toys with fun values \{ 3 \}
The sewing sequence:
1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box.
2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box.
3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box.
Here is a scenario where the second box ends up being the special box:
* The first box had toys with fun values \{-3\}
* The second box had toys with fun values \{ -0.33, -0.25 \}.
* The fourth box had toys with fun values \{ 3 \}.
The sewing sequence:
1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box.
2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box.
3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box.
There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy. | instruction | 0 | 67,482 | 9 | 134,964 |
Tags: constructive algorithms
Correct Solution:
```
def main():
for _ in range(int(input())):
a, b, c, d = [int(x) for x in input().split()]
is_negative_sign = (a + b) & 1
can_be_large = a or d
can_be_small = b or c
if is_negative_sign:
print('Ya' if can_be_large else 'Tidak', end = ' ')
print('Ya' if can_be_small else 'Tidak', end = ' ')
print('Tidak Tidak')
else:
print('Tidak Tidak', end = ' ')
print('Ya' if can_be_small else 'Tidak', end = ' ')
print('Ya' if can_be_large else 'Tidak')
main()
``` | output | 1 | 67,482 | 9 | 134,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification:
* The first box stores toys with fun values in range of (-β,-1].
* The second box stores toys with fun values in range of (-1, 0).
* The third box stores toys with fun values in range of (0, 1).
* The fourth box stores toys with fun value in range of [1, β).
Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.
While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.
As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.
Input
The first line has an integer T (1 β€ T β€ 5 β
10^4), the number of test cases.
Every case contains a line with four space-separated integers A B C D (0 β€ A, B, C, D β€ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.
Output
For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.
For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.
Example
Input
2
1 2 0 1
0 1 0 0
Output
Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak
Note
For the first case, here is a scenario where the first box is the special box:
* The first box had toys with fun values \{-3\}.
* The second box had toys with fun values \{ -0.5, -0.5 \}
* The fourth box had toys with fun values \{ 3 \}
The sewing sequence:
1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box.
2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box.
3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box.
Here is a scenario where the second box ends up being the special box:
* The first box had toys with fun values \{-3\}
* The second box had toys with fun values \{ -0.33, -0.25 \}.
* The fourth box had toys with fun values \{ 3 \}.
The sewing sequence:
1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box.
2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box.
3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box.
There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy. | instruction | 0 | 67,483 | 9 | 134,966 |
Tags: constructive algorithms
Correct Solution:
```
import sys,functools,collections,bisect,math,heapq
input = sys.stdin.readline
#print = sys.stdout.write
sys.setrecursionlimit(300000)
mod = 10**9 + 7
t = int(input())
for _ in range(t):
a,b,c,d = map(int,input().strip().split())
ans = []
if (a+b)%2:
if a or d:
ans.append('Ya')
else:
ans.append('Tidak')
if b or c:
ans.append('Ya')
else:
ans.append('Tidak')
ans.append('Tidak')
ans.append('Tidak')
else:
ans.append('Tidak')
ans.append('Tidak')
if b or c:
ans.append('Ya')
else:
ans.append('Tidak')
if a or d:
ans.append('Ya')
else:
ans.append('Tidak')
print(' '.join(ans))
``` | output | 1 | 67,483 | 9 | 134,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification:
* The first box stores toys with fun values in range of (-β,-1].
* The second box stores toys with fun values in range of (-1, 0).
* The third box stores toys with fun values in range of (0, 1).
* The fourth box stores toys with fun value in range of [1, β).
Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.
While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.
As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.
Input
The first line has an integer T (1 β€ T β€ 5 β
10^4), the number of test cases.
Every case contains a line with four space-separated integers A B C D (0 β€ A, B, C, D β€ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.
Output
For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.
For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.
Example
Input
2
1 2 0 1
0 1 0 0
Output
Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak
Note
For the first case, here is a scenario where the first box is the special box:
* The first box had toys with fun values \{-3\}.
* The second box had toys with fun values \{ -0.5, -0.5 \}
* The fourth box had toys with fun values \{ 3 \}
The sewing sequence:
1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box.
2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box.
3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box.
Here is a scenario where the second box ends up being the special box:
* The first box had toys with fun values \{-3\}
* The second box had toys with fun values \{ -0.33, -0.25 \}.
* The fourth box had toys with fun values \{ 3 \}.
The sewing sequence:
1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box.
2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box.
3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box.
There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy. | instruction | 0 | 67,484 | 9 | 134,968 |
Tags: constructive algorithms
Correct Solution:
```
t=int(input())
while t>0:
t-=1
a,b,c,d = map(int,input().split(" "))
l1=["Tidak"]*4
if (a+b)%2!=0:
if a+d>0:
l1[0]="Ya"
if b+c>0:
l1[1]="Ya"
else:
if a+d>0:
l1[3]="Ya"
if b+c>0:
l1[2]="Ya"
for i in l1:
print(i,end=" ")
print()
``` | output | 1 | 67,484 | 9 | 134,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification:
* The first box stores toys with fun values in range of (-β,-1].
* The second box stores toys with fun values in range of (-1, 0).
* The third box stores toys with fun values in range of (0, 1).
* The fourth box stores toys with fun value in range of [1, β).
Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.
While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.
As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.
Input
The first line has an integer T (1 β€ T β€ 5 β
10^4), the number of test cases.
Every case contains a line with four space-separated integers A B C D (0 β€ A, B, C, D β€ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.
Output
For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.
For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.
Example
Input
2
1 2 0 1
0 1 0 0
Output
Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak
Note
For the first case, here is a scenario where the first box is the special box:
* The first box had toys with fun values \{-3\}.
* The second box had toys with fun values \{ -0.5, -0.5 \}
* The fourth box had toys with fun values \{ 3 \}
The sewing sequence:
1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box.
2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box.
3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box.
Here is a scenario where the second box ends up being the special box:
* The first box had toys with fun values \{-3\}
* The second box had toys with fun values \{ -0.33, -0.25 \}.
* The fourth box had toys with fun values \{ 3 \}.
The sewing sequence:
1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box.
2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box.
3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box.
There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy. | instruction | 0 | 67,485 | 9 | 134,970 |
Tags: constructive algorithms
Correct Solution:
```
import sys, heapq, collections
tc = int(sys.stdin.readline())
for _ in range(tc):
a,b,c,d = map(int, sys.stdin.readline().split())
res = [-1,-1,1,1]
if a % 2 == 0:
res[0] = 1
if b % 2 == 0:
res[1] = 1
ans = 1
for i in res:
ans *= i
temp = ['Tidak'] * 4
if ans == -1:
if a != 0:
temp[0] = 'Ya'
if b != 0:
temp[1] = 'Ya'
if c != 0:
temp[1] = 'Ya'
if d != 0:
temp[0] = 'Ya'
else:
if a != 0:
temp[3] = 'Ya'
if b != 0:
temp[2] = 'Ya'
if c != 0:
temp[2] = 'Ya'
if d != 0:
temp[3] = 'Ya'
print(' '.join(temp))
``` | output | 1 | 67,485 | 9 | 134,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification:
* The first box stores toys with fun values in range of (-β,-1].
* The second box stores toys with fun values in range of (-1, 0).
* The third box stores toys with fun values in range of (0, 1).
* The fourth box stores toys with fun value in range of [1, β).
Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.
While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.
As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.
Input
The first line has an integer T (1 β€ T β€ 5 β
10^4), the number of test cases.
Every case contains a line with four space-separated integers A B C D (0 β€ A, B, C, D β€ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.
Output
For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.
For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.
Example
Input
2
1 2 0 1
0 1 0 0
Output
Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak
Note
For the first case, here is a scenario where the first box is the special box:
* The first box had toys with fun values \{-3\}.
* The second box had toys with fun values \{ -0.5, -0.5 \}
* The fourth box had toys with fun values \{ 3 \}
The sewing sequence:
1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box.
2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box.
3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box.
Here is a scenario where the second box ends up being the special box:
* The first box had toys with fun values \{-3\}
* The second box had toys with fun values \{ -0.33, -0.25 \}.
* The fourth box had toys with fun values \{ 3 \}.
The sewing sequence:
1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box.
2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box.
3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box.
There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy. | instruction | 0 | 67,486 | 9 | 134,972 |
Tags: constructive algorithms
Correct Solution:
```
import math
t = int(input())
for _ in range(t):
abcd = input().split()
a = int(abcd[0])
b = int(abcd[1])
c = int(abcd[2])
d = int(abcd[3])
if a == 0 and d == 0:
if b % 2 == 0:
print("Tidak Tidak Ya Tidak")
else:
print("Tidak Ya Tidak Tidak")
elif b == 0 and c == 0:
if a % 2 == 0:
print("Tidak Tidak Tidak Ya")
else:
print("Ya Tidak Tidak Tidak")
else:
if (a+b) % 2 == 0:
print("Tidak Tidak Ya Ya")
else:
print("Ya Ya Tidak Tidak")
``` | output | 1 | 67,486 | 9 | 134,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification:
* The first box stores toys with fun values in range of (-β,-1].
* The second box stores toys with fun values in range of (-1, 0).
* The third box stores toys with fun values in range of (0, 1).
* The fourth box stores toys with fun value in range of [1, β).
Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.
While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.
As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.
Input
The first line has an integer T (1 β€ T β€ 5 β
10^4), the number of test cases.
Every case contains a line with four space-separated integers A B C D (0 β€ A, B, C, D β€ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.
Output
For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.
For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.
Example
Input
2
1 2 0 1
0 1 0 0
Output
Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak
Note
For the first case, here is a scenario where the first box is the special box:
* The first box had toys with fun values \{-3\}.
* The second box had toys with fun values \{ -0.5, -0.5 \}
* The fourth box had toys with fun values \{ 3 \}
The sewing sequence:
1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box.
2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box.
3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box.
Here is a scenario where the second box ends up being the special box:
* The first box had toys with fun values \{-3\}
* The second box had toys with fun values \{ -0.33, -0.25 \}.
* The fourth box had toys with fun values \{ 3 \}.
The sewing sequence:
1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box.
2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box.
3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box.
There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy. | instruction | 0 | 67,487 | 9 | 134,974 |
Tags: constructive algorithms
Correct Solution:
```
for k in range(int(input())):
a, b, c, d = list(map(int, input().split()))
s = ''
if (a+b)%2==1 and (a > 0 or d > 0):
s += 'Ya '
else:
s += 'Tidak '
if (a+b)%2 == 1 and (b > 0 or c > 0):
s += 'Ya '
else:
s += 'Tidak '
if (a+b)%2 == 0 and (b > 0 or c > 0):
s += 'Ya '
else:
s += 'Tidak '
if (a > 0 and (a+b)%2 == 0) or (d > 0 and (a+b)%2 == 0):
s += 'Ya'
else:
s += 'Tidak'
print(s)
``` | output | 1 | 67,487 | 9 | 134,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification:
* The first box stores toys with fun values in range of (-β,-1].
* The second box stores toys with fun values in range of (-1, 0).
* The third box stores toys with fun values in range of (0, 1).
* The fourth box stores toys with fun value in range of [1, β).
Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.
While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.
As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.
Input
The first line has an integer T (1 β€ T β€ 5 β
10^4), the number of test cases.
Every case contains a line with four space-separated integers A B C D (0 β€ A, B, C, D β€ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.
Output
For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.
For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.
Example
Input
2
1 2 0 1
0 1 0 0
Output
Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak
Note
For the first case, here is a scenario where the first box is the special box:
* The first box had toys with fun values \{-3\}.
* The second box had toys with fun values \{ -0.5, -0.5 \}
* The fourth box had toys with fun values \{ 3 \}
The sewing sequence:
1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box.
2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box.
3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box.
Here is a scenario where the second box ends up being the special box:
* The first box had toys with fun values \{-3\}
* The second box had toys with fun values \{ -0.33, -0.25 \}.
* The fourth box had toys with fun values \{ 3 \}.
The sewing sequence:
1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box.
2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box.
3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box.
There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy. | instruction | 0 | 67,488 | 9 | 134,976 |
Tags: constructive algorithms
Correct Solution:
```
# _
#####################################################################################################################
def main(nTestCases):
tuple(print(specialBox_sPosition(*testCase_1425h())) for i in range(nTestCases))
def testCase_1425h():
return map(int, input().split())
def specialBox_sPosition(nToysB1, nToysB2, nToysB3, nToysB4):
if (nToysB1 + nToysB2)%2:
if nToysB1 and nToysB2 or (nToysB1 and nToysB3) or (nToysB2 and nToysB4):
return 'Ya Ya Tidak Tidak'
elif nToysB1:
return 'Ya Tidak Tidak Tidak'
return 'Tidak Ya Tidak Tidak'
if nToysB3 and nToysB4 or (nToysB1 and nToysB3) or (nToysB2 and nToysB4) or (nToysB1 and nToysB2):
return 'Tidak Tidak Ya Ya'
elif nToysB4 or nToysB1:
return 'Tidak Tidak Tidak Ya'
return 'Tidak Tidak Ya Tidak'
if __name__ == '__main__':
main(int(input()))
``` | output | 1 | 67,488 | 9 | 134,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification:
* The first box stores toys with fun values in range of (-β,-1].
* The second box stores toys with fun values in range of (-1, 0).
* The third box stores toys with fun values in range of (0, 1).
* The fourth box stores toys with fun value in range of [1, β).
Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.
While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.
As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.
Input
The first line has an integer T (1 β€ T β€ 5 β
10^4), the number of test cases.
Every case contains a line with four space-separated integers A B C D (0 β€ A, B, C, D β€ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.
Output
For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.
For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.
Example
Input
2
1 2 0 1
0 1 0 0
Output
Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak
Note
For the first case, here is a scenario where the first box is the special box:
* The first box had toys with fun values \{-3\}.
* The second box had toys with fun values \{ -0.5, -0.5 \}
* The fourth box had toys with fun values \{ 3 \}
The sewing sequence:
1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box.
2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box.
3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box.
Here is a scenario where the second box ends up being the special box:
* The first box had toys with fun values \{-3\}
* The second box had toys with fun values \{ -0.33, -0.25 \}.
* The fourth box had toys with fun values \{ 3 \}.
The sewing sequence:
1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box.
2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box.
3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box.
There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy. | instruction | 0 | 67,489 | 9 | 134,978 |
Tags: constructive algorithms
Correct Solution:
```
for _ in range(int(input())):
l=list(map(int,input().split()))
a=[0]*4
if l[0]+l[3]!=0:a[0]=a[3]=1
if l[1]+l[2]!=0:a[1]=a[2]=1
if (l[0]+l[1])%2==0:a[0]=a[1]=0
else:a[2]=a[3]=0
for x in a:
print(["Tidak","Ya"][x>0],end=' ')
print('')
``` | output | 1 | 67,489 | 9 | 134,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification:
* The first box stores toys with fun values in range of (-β,-1].
* The second box stores toys with fun values in range of (-1, 0).
* The third box stores toys with fun values in range of (0, 1).
* The fourth box stores toys with fun value in range of [1, β).
Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.
While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.
As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.
Input
The first line has an integer T (1 β€ T β€ 5 β
10^4), the number of test cases.
Every case contains a line with four space-separated integers A B C D (0 β€ A, B, C, D β€ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.
Output
For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.
For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.
Example
Input
2
1 2 0 1
0 1 0 0
Output
Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak
Note
For the first case, here is a scenario where the first box is the special box:
* The first box had toys with fun values \{-3\}.
* The second box had toys with fun values \{ -0.5, -0.5 \}
* The fourth box had toys with fun values \{ 3 \}
The sewing sequence:
1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box.
2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box.
3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box.
Here is a scenario where the second box ends up being the special box:
* The first box had toys with fun values \{-3\}
* The second box had toys with fun values \{ -0.33, -0.25 \}.
* The fourth box had toys with fun values \{ 3 \}.
The sewing sequence:
1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box.
2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box.
3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box.
There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy.
Submitted Solution:
```
t=int(input())
for i in range(t):
a,b,c,d=map(int,input().split())
if a+b+c+d==1:
lst=[a,b,c,d]
for i in lst:
if i==0:
print("Tidak",end=" ")
else:
print("Ya",end=" ")
print()
else:
s1=""
s2=""
s3=""
s4=""
a1=a+b
if a1%2==0:
if b+c>0 and a+d>0:
s1+="Tidak"
s2+="Tidak"
s3+="Ya"
s4+="Ya"
elif b+c==0:
s1+="Tidak"
s2+="Tidak"
s3+="Tidak"
s4+="Ya"
else:
s1+="Tidak"
s2+="Tidak"
s3+="Ya"
s4+="Tidak"
else:
if b+c>0 and a+d>0:
s1+="Ya"
s2+="Ya"
s3+="Tidak"
s4+="Tidak"
elif b+c==0:
s1+="Ya"
s2+="Tidak"
s3+="Tidak"
s4+="Tidak"
else:
s1+="Tidak"
s2+="Ya"
s3+="Tidak"
s4+="Tidak"
print(s1,s2,s3,s4)
#if bx2 to be superbox
``` | instruction | 0 | 67,490 | 9 | 134,980 |
Yes | output | 1 | 67,490 | 9 | 134,981 |
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