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Provide tags and a correct Python 3 solution for this coding contest problem.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Tags: brute force, implementation
Correct Solution:
```
from collections import Counter
n = int(input())
s = input()
k = 0
for i in range(n-1):
for j in range(i+2, n+1, 2):
c = Counter(s[i:j])
k += (c['U']==c['D'] and c['R']==c['L'])
print(k)
```
| 87,100 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Tags: brute force, implementation
Correct Solution:
```
def f(i, j):
subs = s[i:j+1]
return subs.count("L") == subs.count("R") and subs.count("U") == subs.count("D")
n = int(input())
s = input()
r = 0
for i in range(n):
for j in range(i+1,n):
if f(i, j):
r += 1
print(r)
```
| 87,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Submitted Solution:
```
n = int(input())
string = input()
c = []
for char in string:
if char == 'U':
c.append(1)
if char == 'D':
c.append(-1)
if char == 'R':
c.append(2000)
if char == 'L':
c.append(-2000)
ans = 0
for k in range(2,len(c) + 1,2): #groups of odd lentgh can NOT sum to 0
for i in range(0,len(c) - k + 1):
#print(c[i:i + k])
t = sum(c[i:i + k])
if t == 0:
ans += 1
print(ans)
```
Yes
| 87,102 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Submitted Solution:
```
n = int(input())
c = input()
s = 0
for i in range(n):
for j in range(i + 1, n):
x = 0
y = 0
sub = c[i : j + 1]
for k in sub:
if k == "L":
x -= 1
elif k == "R":
x += 1
elif k == "U":
y += 1
else:
y -= 1
if x == 0 and y == 0:
s += 1
print(s)
```
Yes
| 87,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Submitted Solution:
```
n = int(input())
l = input()
ans = 0
for i in range(2, n + 1):
for j in range(n - i + 1):
st = l[j : j + i]
if st.count('L') == st.count('R') and st.count('U') == st.count('D'):
ans += 1
print(ans)
```
Yes
| 87,104 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Submitted Solution:
```
import collections
import math
n = int(input())
s = input()
ans = 0
for i in range(len(s)):
for j in range(i, len(s)):
t = s[i:j+1]
if t.count('U') == t.count('D') and t.count('L') == t.count('R'):
ans += 1
print(ans)
```
Yes
| 87,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Submitted Solution:
```
if __name__=='__main__':
n=int(input())
s=input()
i=0
ans=0
while(i<n):
r,l,u,d=0,0,0,0
g=i
while(g<n):
if(s[g]=='R'):
r+=1
elif(s[g]=='L'):
l+=1
elif(s[g]=='U'):
u+=1
elif(s[g]=='D'):
d+=1
if(d==u and l ==r):
print("i-> ",i," g-> ",g)
ans+=1
g+=1
i+=1
print(ans)
```
No
| 87,106 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Submitted Solution:
```
n = int(input())
s = input()
print(2)
```
No
| 87,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Submitted Solution:
```
#!/usr/bin/python3.5
n=int(input())
s=input()
i,e=0,0
while i<n:
k=i+1
while k<=n:
c,d=0,0
for j in range(i,k):
if s[j]=='U':
c+=1
elif s[j]=='D':
c-=1
elif s[j]=='R':
d+=1
else:
d-=1
if c==0 and c==0:
e+=1
k+=1
i+=1
print(e)
```
No
| 87,108 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.
Input
The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.
The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.
Output
Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.
Examples
Input
6
URLLDR
Output
2
Input
4
DLUU
Output
0
Input
7
RLRLRLR
Output
12
Note
In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.
Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.
Submitted Solution:
```
C=lambda x,y:x.count(y)
def P(s):
return C(s,"R")-C(s,"L")+1j*(C(s,"U")-C(s,"D"))
n=int(input())
s=input()
p=P(s)
c=0
for z in range(0,n):
for l in range(z+1,n+1):
print(s[z:l],p,P(s[z:l]),p+P(s[z:l]))
c+=p+P(s[z:l])==0
print(c)
```
No
| 87,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Tags: brute force, constructive algorithms, math
Correct Solution:
```
n,a,b,c,d = map(int, input().split())
num = 0
i = 0
while n > i:
i += 1
if i + b - c < 1 or i + b - c > n:
continue
if i + a - d < 1 or i + a - d > n:
continue
if i + a - d + b - c < 1 or i + a - d + b - c > n:
continue
num += 1
print(num * n)
```
| 87,110 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Tags: brute force, constructive algorithms, math
Correct Solution:
```
import sys
sys.stderr = sys.stdout
def painting(n, a, b, c, d):
lo = min(a+b, c+d, a+c, b+d)
hi = max(a+b, c+d, a+c, b+d)
delta = hi - lo
return max(n - delta, 0) * n
def main():
n, a, b, c, d = readinti()
print(painting(n, a, b, c, d))
##########
def readint():
return int(input())
def readinti():
return map(int, input().split())
def readintt():
return tuple(readinti())
def readintl():
return list(readinti())
def readinttl(k):
return [readintt() for _ in range(k)]
def readintll(k):
return [readintl() for _ in range(k)]
def log(*args, **kwargs):
print(*args, **kwargs, file=sys.__stderr__)
if __name__ == '__main__':
main()
```
| 87,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Tags: brute force, constructive algorithms, math
Correct Solution:
```
n, a, b, c, d = map(int, input().split())
k = 0
for u in range(1, n + 1):
v = u + b - c
if (v >= 1) and (v <= n):
z = a + v - d
if (z >= 1) and (z <= n):
y = c + z - b
if (y >= 1) and (y <= n):
u = d + y - a
if (u >= 1) and (u <= n):
k += 1
print(k * n)
```
| 87,112 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Tags: brute force, constructive algorithms, math
Correct Solution:
```
n, a, b, c, d = map(int, input().split())
lo = min(a+b, a+c, c+d, b+d)
hi = max(a+b, a+c, c+d, b+d)
ans = 0
for x in range(1, n+1):
ans += max(0, n-(hi-lo))
print(ans)
```
| 87,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Tags: brute force, constructive algorithms, math
Correct Solution:
```
from re import *
from sys import stderr
def readint():
return int(input())
def readfloat():
return float(input())
def readarray(N, foo=input):
return [foo() for i in range(N)]
def readlinearray(foo=int):
return map(foo, input().split())
def NOD(a, b):
while b:
a,b = b, a%b
return a
def gen_primes(max):
primes = [1]*(max+1)
for i in range(2, max+1):
if primes[i]:
for j in range(i+i, max+1, i):
primes[j] = 0
primes[0] = 0
return [x for x in range(max+1) if primes[x]]
def is_prime(N):
i = 3
if not(N % 2):
return 0
while i*i < N:
if not(N % i):
return 0
i += 3
return 1
n, a, b, c, d = readlinearray()
maxx = min(n, n - b + c, n - a + d, n - a + c - b + d)
minx = max(1, 1 - b + c, 1 - a + d, 1 - a + c - b + d)
if maxx < minx:
print(0)
else:
print((maxx - minx + 1) * n)
```
| 87,114 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Tags: brute force, constructive algorithms, math
Correct Solution:
```
import math as m
n,a,b,c,d=(map(int,input().split()))
ans=0
for i in range(n):
x=i+1
t=x+b-c
curr=1
if t>=1 and t<=n:
curr=1
else:
curr=0
t=x+a-d
if t>=1 and t<=n and curr!=0:
curr=1
else:
curr*=0
t=x+a+b-d-c
if t>=1 and t<=n and curr!=0:
curr=1
else:
curr*=0
ans+=curr*n
# print(x,curr)
print(ans)
```
| 87,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Tags: brute force, constructive algorithms, math
Correct Solution:
```
from collections import *
n,a,b,c,d = list(map(int,input().split()))
print((n-min(n,max(abs(a+b-c-d),abs(b-c),abs(a-d),abs(a+c-b-d))))*n)
```
| 87,116 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Tags: brute force, constructive algorithms, math
Correct Solution:
```
n, a, b, c, d = map(int, input().split())
count = 0
for x in range(1, n + 1):
w = x + c - b
if w >= 1 and w <= n:
y = a + w - d
if y >= 1 and y <= n:
z = a + x - d
if z >= 1 and z <= n:
count = count + 1
print(count * n)
```
| 87,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n, a, b, c, d = map(int, input().split())
ans = 0
for centro in range(1, n+1):
q1 = a + b + centro
q2 = a + c + centro
q3 = b + d + centro
q4 = c + d + centro
v = [q1, q2, q3, q4]
v.sort()
q_min = v[0]
q_max = v[-1]
if q_min + n < q_max + 1:
continue
elif q_min + n == q_max + 1:
ans += 1
else:
ans += (q_min + n) - (q_max + 1) + 1
print(ans)
```
Yes
| 87,118 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
import sys,os,io
import math,bisect,operator
inf,mod = float('inf'),10**9+7
# sys.setrecursionlimit(10 ** 6)
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,defaultdict
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
Neo = lambda : list(map(int,input().split()))
# test, = Neo()
n,a,b,c,d = Neo()
Ans = 0
for i in range(1,n+1):
p = i+a+b
q = p-a-c
r = p-b-d
s = p-c-d
if 1<=q<=n and 1<=r<=n and 1<=s<=n:
Ans += 1
print(n*Ans)
```
Yes
| 87,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n, a, b, c, d = map(int, input().split())
ans = n * (n - abs(a - d) - abs(b - c))
print(max(0, ans))
```
Yes
| 87,120 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n,a,b,c,d=tuple(map(int,input().split()))
ans=0
sum=0
def ch(c):
return (n>=c and c>0)
for i in range(1,n+1):
sum=i+a+b
x2=sum-a-c
x3=sum-c-d
x4=sum-b-d
if ch(i) and ch(x2) and ch(x3) and ch(x4):
ans+=1
print(ans*n)
```
Yes
| 87,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n,a,b,c,d=map(int,input().split())
print((max(a+b+1,b+c+1,c+d+1,a+d+1)-min(a+b+n,b+c+n,c+d+n,d+a+n)+1)*n)
```
No
| 87,122 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
nabcd = list(map(int, input().split()))
RANGE = range(1,nabcd[0])
a = nabcd[1]
b = nabcd[2]
c = nabcd[3]
d = nabcd[4]
dif1 = a-d
dif2 = b-c
dif = abs(dif1) + abs(dif2)
counter = nabcd[0]*(nabcd[0]-dif)
print(counter)
```
No
| 87,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n, a, b, c, d = map(int, input().split())
ans = 0
l = [a+b, a+c, b+d, d+c]
sum = (min(l)+n)-(max(l)+1)+1
ans = sum*n
print(ans)
```
No
| 87,124 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n, a, b, c, d = map(int, input().split())
ans = 0
if n >= 10 :
ans = 0
else :
S1 = set([a,b,c,d])
for i in range(1, n+1) :
s = a+b+i
j = s - a - c
k = s - b - d
l = s - c - d
if 1 <= j <=n and 1<= k <= n and 1<= l <= n :
S2 = set([i, j, k, l])
if len(S1 | S2) == n :
ans += n
elif len(S1 | S2) == n-1 :
ans += 1
print(ans)
```
No
| 87,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Tags: dp
Correct Solution:
```
import sys
from functools import lru_cache
from collections import defaultdict
possible = {
0: {'r'},
1: {'r', 'c'},
2: {'r', 'g'},
3: {'r', 'c', 'g'}
}
@lru_cache(maxsize=None)
def min_rest(i, prev_action):
if i == len(arr):
return 0
# At least one elem in the arr.
p = possible[arr[i]]
best = float("inf")
for act in p:
if act == 'r':
# Can always rest, no restrictions.
best = min(best, 1 + min_rest(i + 1, 'r'))
elif act != prev_action:
best = min(best, min_rest(i + 1, act))
return best
def iterative(arr):
DP = defaultdict(int)
for i in range(len(arr) - 1, -1, -1):
p = possible[arr[i]]
for prev_act in ('r', 'c', 'g'):
best = float("inf")
for act in p:
if act == 'r':
# Can always rest, no restrictions.
best = min(best, 1 + DP[('r', i + 1)])
elif act != prev_act:
best = min(best, DP[act, i + 1])
DP[(prev_act, i)] = best
return DP[('r', 0)]
if __name__ == "__main__":
mat = []
for e, line in enumerate(sys.stdin.readlines()):
if e == 0:
continue
arr = list(map(int, line.strip().split()))
# print(min_rest(0, 'r'))
print(iterative(arr))
```
| 87,126 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Tags: dp
Correct Solution:
```
days = int(input())
lst = list(map(int,input().split()))
rest, gym, cont = [0], [0], [0]
if lst[0] in (1,3):
cont = [1]
if lst[0] in (2,3):
gym = [1]
for i in range(1,days):
rest.append(max(rest[-1],gym[-1],cont[-1]))
if lst[i] == 3:
cont.append(max(rest[-2],gym[-1])+1)
gym.append(max(rest[-2],cont[-2])+1)
elif lst[i] == 1:
cont.append(max(rest[-2],gym[-1])+1)
elif lst[i] == 2:
gym.append(max(rest[-2],cont[-1])+1)
#print(rest,gym,cont)
print(days-max(rest[-1],gym[-1],cont[-1]))
```
| 87,127 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Tags: dp
Correct Solution:
```
n = int(input())
a = list(map(int, input().split(' ')))
dp = [[0 for i in range(3)] for _ in range(102)]
for i in range(1, n + 1):
if a[i - 1] == 0:
dp[i][0] = min(dp[i - 1]) + 1
dp[i][1] = dp[i - 1][1] + 1
dp[i][2] = dp[i - 1][2] + 1
elif a[i - 1] == 1:
dp[i][0] = min(dp[i - 1]) + 1
dp[i][1] = min(dp[i - 1][0], dp[i - 1][2])
dp[i][2] = dp[i- 1][2] + 1
elif a[i - 1] == 2:
dp[i][0] = min(dp[i - 1]) + 1
dp[i][1] = dp[i - 1][1] + 1
dp[i][2] = min(dp[i - 1][0], dp[i - 1][1])
elif a[i - 1] == 3:
dp[i][0] = min(dp[i - 1]) + 1
dp[i][2] = min(dp[i - 1][0], dp[i - 1][1])
dp[i][1] = min(dp[i - 1][0], dp[i - 1][2])
print(min(dp[n]))
```
| 87,128 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Tags: dp
Correct Solution:
```
a = int(input())
days = [int(x) for x in input().split()]
dp = [[666, 666, 666] for i in range(a)]
if days[0] == 3:
dp[0][0] = 0
dp[0][1] = 0
dp[0][2] = 0
else:
dp[0][0], dp[0][days[0]] = 0, 0
dp[0][0] += 1
for x in range(1, a):
dp[x][0] = min(dp[x - 1][:]) + 1
if days[x] == 3:
dp[x][1] = min(dp[x - 1][0], dp[x - 1][2])
dp[x][2] = min(dp[x - 1][0], dp[x - 1][1])
elif days[x] == 2:
dp[x][2] = min(dp[x - 1][0], dp[x - 1][1])
elif days[x] == 1:
dp[x][1] = min(dp[x - 1][0], dp[x - 1][2])
print(min(dp[a - 1][:]))
```
| 87,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Tags: dp
Correct Solution:
```
n = int(input())
arr = list(map(lambda x: int(x), input().split(" ")))
dp = [[None] * n for _ in range(3)]
first = arr[0]
dp[0][0] = 1
if first == 1:
dp[1][0] = 0
if first == 2:
dp[2][0] = 0
if first == 3:
dp[1][0] = 0
dp[2][0] = 0
for c in range(1, n):
val = arr[c]
cand = [dp[0][c-1], dp[1][c-1], dp[2][c-1]]
cand = list(filter(lambda x: x != None, cand))
dp[0][c] = 1+min(cand)
if val == 1 or val == 3:
cand = [dp[0][c-1], dp[2][c-1]]
cand = list(filter(lambda x: x != None, cand))
dp[1][c] = min(cand)
if val == 2 or val == 3:
cand = [dp[0][c-1], dp[1][c-1]]
cand = list(filter(lambda x: x != None, cand))
dp[2][c] = min(cand)
cand = [dp[0][-1], dp[1][-1], dp[2][-1]]
#print(dp)
print(min(filter(lambda x: x != None, cand)))
```
| 87,130 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Tags: dp
Correct Solution:
```
def get_ints(string):
return list(map(int, string.split()))
def get_input():
n = int(input())
xs = get_ints(input())
return n, xs
def foo(xs, res, last):
# print(xs)
if len(xs) is []:
return res
for i, x in enumerate(xs):
if x == 0:
res += 1
last = 0
elif x == 1 and last == 1:
res += 1
last = 0
elif x == 1:
last = 1
elif x == 2 and last == 2:
res += 1
last = 0
elif x == 2:
last = 2
else:
if last == 1:
last = 2
elif last == 2:
last = 1
elif i < len(xs)-1 and xs[i+1] == 0:
last = 1
else:
return min(foo(xs[i+1:], res, 1), foo(xs[i+1:], res, 2))
return res
def main():
n, xs = get_input()
res = foo(xs, 0, 0)
print(res)
return res
if __name__ == '__main__':
main()
```
| 87,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Tags: dp
Correct Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
dp = [[0]*(n+1), [0]*(n+1), [0]*(n+1)]
INF = 1e9
for i in range(1, n + 1):
dp[0][i] = min(dp[0][i - 1], dp[1][i - 1], dp[2][i - 1]) + 1
if a[i - 1] == 0:
dp[1][i] = INF
dp[2][i] = INF
elif a[i - 1] == 1:
dp[1][i] = min(dp[0][i - 1], dp[2][i - 1])
dp[2][i] = INF
elif a[i - 1] == 2:
dp[1][i] = INF
dp[2][i] = min(dp[0][i - 1], dp[1][i - 1])
elif a[i - 1] == 3:
dp[1][i] = min(dp[0][i - 1], dp[2][i - 1])
dp[2][i] = min(dp[0][i - 1], dp[1][i - 1])
print(min(dp[0][n], dp[1][n], dp[2][n]))
```
| 87,132 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Tags: dp
Correct Solution:
```
n = int(input())
ai = list(map(int,input().split()))
for i in range(n):
if i < n-1:
if ai[i] % 3 != 0 and ai[i] == ai[i+1]:
ai[i+1] = 0
elif ai[i+1] == 3 and ai[i] % 3 != 0:
ai[i+1] = 3 - ai[i]
ai[i] = int(ai[i] > 0)
print(n - sum(ai))
```
| 87,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
import math
n = int(input())
arr = list(map(int,input().split()))
dp = [[float("inf") for _ in range(n+1)] for j in range(3)]
for i in range(3):
dp[i][0] = 0
for idx in range(1,n+1):
# print(f"idx = {idx} day = {arr[idx-1]}")
# print('\n'.join([''.join(['{:4}'.format(item) for item in row])
# for row in dp]))
day = arr[idx-1]
dp[0][idx] = min([dp[0][idx-1],dp[1][idx-1],dp[2][idx-1]])+1
if day == 1 or day ==3:
dp[1][idx] = min(dp[0][idx-1],dp[2][idx-1])
if day == 2 or day == 3:
dp[2][idx] = min(dp[0][idx-1],dp[1][idx-1])
# print("------------------------------------------------")
# print('\n'.join([''.join(['{:4}'.format(item) for item in row])
# for row in dp]))
#print(dp)
ans = float("inf")
for i in range(3):
ans = min(ans,dp[i][n])
print(ans)
```
Yes
| 87,134 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n = int(input())
list1 = [int(i)for i in input().split()]
dp = []
for x in range(n):
dp.append([10**9,10**9,10**9])
dp[0][0]=1
if list1[0]==0:
dp[0][1]=1
dp[0][2]=1
elif list1[0]==1:
dp[0][1]=0
dp[0][2]=1
elif list1[0]==2:
dp[0][1]=1
dp[0][2]=0
else:
dp[0][1]=0
dp[0][2]=0
for x in range(1,n):
dp[x][0]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
if list1[x]==0:
dp[x][1]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
dp[x][2]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
elif list1[x]==1:
dp[x][1]=min([dp[x-1][0],dp[x-1][2]])
dp[x][2]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
elif list1[x]==2:
dp[x][1]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
dp[x][2]=min([dp[x-1][0],dp[x-1][1]])
else:
dp[x][1]=min([dp[x-1][0],dp[x-1][2]])
dp[x][2]=min([dp[x-1][0],dp[x-1][1]])
print(min([dp[n-1][0],dp[n-1][1],dp[n-1][2]]))
```
Yes
| 87,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n = int(input())
nums = input()
nums = [int(i) for i in nums.split()]
MAXVAL = 999
dp = [[MAXVAL for i in range(3)] for i in range(n)]
if nums[0] == 0:
dp[0][0] = 1
elif nums[0] == 1:
dp[0][1] = 0
elif nums[0] == 2:
dp[0][2] = 0
else:
dp[0][1] = 0 # if both, CONTEST
dp[0][2] = 0 # if both, GYM
#print(dp)
for i in range(1,n):
if nums[i] == 0:
dp[i][0] = min(dp[i-1]) + 1
elif nums[i] == 1:
# Do contest
dp[i][1] = min(dp[i-1][0], dp[i-1][2]) + 0
# Rest
dp[i][0] = min(dp[i-1][0], dp[i-1][1], dp[i-1][2]) + 1
elif nums[i] == 2:
# Do gym
dp[i][2] = min(dp[i-1][0], dp[i-1][1]) + 0
# Rest
dp[i][0] = min(dp[i-1][0], dp[i-1][1], dp[i-1][2]) + 1
else:
# Rest
dp[i][0] = min(dp[i-1][0], dp[i-1][1], dp[i-1][2]) + 1
# Do gym
dp[i][2] = min(dp[i-1][0], dp[i-1][1]) + 0
# Do contest
dp[i][1] = min(dp[i-1][0], dp[i-1][2]) + 0
#print(dp)
ans = min(dp[n-1])
print(ans)
```
Yes
| 87,136 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
z,zz=input,lambda:list(map(int,z().split()))
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in stdin.readline().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from re import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if (x%i==0 and x!=2) or x==0:return 0
return 1
def dfs(u,visit,graph):
visit[u]=True
for i in graph[u]:
if not visit[i]:
dfs(i,visit,graph)
###########################---Test-Case---#################################
"""
"""
###########################---START-CODING---##############################
n=int(z())
arr=zzz()
dp=[[10**9]*3 for _ in range(n+1)]
dp[0][0]=0
for i in range(n):
k=arr[i]
dp[i+1][0]=min(dp[i][2],dp[i][1],dp[i][0])+1
if k==1:
dp[i+1][1]=min(dp[i][0],dp[i][2])
if k==2:
dp[i+1][2]=min(dp[i][0],dp[i][1])
if k==3:
dp[i+1][1]=min(dp[i][0],dp[i][2])
dp[i+1][2]=min(dp[i][0],dp[i][1])
print(min(dp[n]))
```
Yes
| 87,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n = int(input())
c = [int(x) for x in input().split(' ')]
a = [(2, -1)]
b = [(1, -1)]
for i in range(n):
if c[i] == 0: pass
elif c[i] == 1 or c[i] == 2:
if a[-1][0] != c[i]: a.append((c[i], i))
if b[-1][0] != c[i]: b.append((c[i], i))
else:
a.append((2, i) if a[-1][0] == 1 else (1, i))
b.append((2, i) if b[-1][0] == 1 else (1, i))
#print(a, b)
print( n - max(len(a), len(b)) + 1)
```
No
| 87,138 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n=int(input())
s=input().split()
aflag=0
bflag=0
ans=0
for i in range(n):
if s[i]=='0':
ans+=1;aflag=0;bflag=0
elif s[i]=='1':
if aflag==1:
ans+=1;aflag=0;bflag=0
else:
aflag=1;bflag=0
elif s[i]=='2':
if bflag==1:
ans+=1;aflag=0;bflag=0
else:
bflag=1;aflag=0
else:
if aflag==1:
bflag=1;aflag=0
else:
aflag=1;bflag=0
print(ans)
```
No
| 87,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
n = val()
l = li()
dp = [[0 for i in range(2)] for j in range(n)]
dp[0][0] = 1 if l[0]>1 else 0
dp[0][1] = 1 if l[0]&1 else 0
ans = 0
if dp[0][0] == 0 and dp[0][1] == 0:ans = 1
currmax = ans
for i in range(1,n):
if not dp[i-1][0] and l[i] > 1:dp[i][0] = 1
if not dp[i-1][1] and l[i]&1:dp[i][1] = 1
if sum(dp[i]) == 0:ans += 1
else:ans = 0
currmax = max(ans,currmax)
print(currmax)
```
No
| 87,140 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
count = 0
prev = 0
f=0
for i in range(n):
if a[i] == 0:
count+=1
prev = 0
if a[i] == 1 or a[i] == 2:
if prev == a[i]:
count+=1
prev = 0
prev = a[i]
if a[i] == 3:
if prev != 3 and prev != 0:
prev = 3-prev
else:
prev = 3
print(count)
```
No
| 87,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is the following puzzle popular among nuclear physicists.
A reactor contains a set of n atoms of some chemical elements. We shall understand the phrase "atomic number" as the number of this atom's element in the periodic table of the chemical elements.
You are allowed to take any two different atoms and fuse a new one from them. That results in a new atom, whose number is equal to the sum of the numbers of original atoms. The fusion operation can be performed several times.
The aim is getting a new pregiven set of k atoms.
The puzzle's difficulty is that it is only allowed to fuse two atoms into one, it is not allowed to split an atom into several atoms. You are suggested to try to solve the puzzle.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 17). The second line contains space-separated symbols of elements of n atoms, which are available from the start. The third line contains space-separated symbols of elements of k atoms which need to be the result of the fusion. The symbols of the elements coincide with the symbols from the periodic table of the chemical elements. The atomic numbers do not exceed 100 (elements possessing larger numbers are highly unstable). Some atoms can have identical numbers (that is, there can be several atoms of the same element). The sum of numbers of initial atoms is equal to the sum of numbers of the atoms that need to be synthesized.
Output
If it is impossible to synthesize the required atoms, print "NO" without the quotes. Otherwise, print on the first line «YES», and on the next k lines print the way of synthesizing each of k atoms as equations. Each equation has the following form: "x1+x2+...+xt->yi", where xj is the symbol of the element of some atom from the original set, and yi is the symbol of the element of some atom from the resulting set. Each atom from the input data should occur in the output data exactly one time. The order of summands in the equations, as well as the output order does not matter. If there are several solutions, print any of them. For a better understanding of the output format, see the samples.
Examples
Input
10 3
Mn Co Li Mg C P F Zn Sc K
Sn Pt Y
Output
YES
Mn+C+K->Sn
Co+Zn+Sc->Pt
Li+Mg+P+F->Y
Input
2 1
H H
He
Output
YES
H+H->He
Input
2 2
Bk Fm
Cf Es
Output
NO
Note
The reactions from the first example possess the following form (the atomic number is written below and to the left of the element):
<image>
<image>
<image>
To find a periodic table of the chemical elements, you may use your favorite search engine.
The pretest set contains each of the first 100 elements of the periodic table at least once. You can use that information to check for misprints.
Tags: bitmasks, dp
Correct Solution:
```
numbersDict = {
"1": "H",
"2": "He",
"3": "Li",
"4": "Be",
"5": "B",
"6": "C",
"7": "N",
"8": "O",
"9": "F",
"10": "Ne",
"11": "Na",
"12": "Mg",
"13": "Al",
"14": "Si",
"15": "P",
"16": "S",
"17": "Cl",
"18": "Ar",
"19": "K",
"20": "Ca",
"21": "Sc",
"22": "Ti",
"23": "V",
"24": "Cr",
"25": "Mn",
"26": "Fe",
"27": "Co",
"28": "Ni",
"29": "Cu",
"30": "Zn",
"31": "Ga",
"32": "Ge",
"33": "As",
"34": "Se",
"35": "Br",
"36": "Kr",
"37": "Rb",
"38": "Sr",
"39": "Y",
"40": "Zr",
"41": "Nb",
"42": "Mo",
"43": "Tc",
"44": "Ru",
"45": "Rh",
"46": "Pd",
"47": "Ag",
"48": "Cd",
"49": "In",
"50": "Sn",
"51": "Sb",
"52": "Te",
"53": "I",
"54": "Xe",
"55": "Cs",
"56": "Ba",
"57": "La",
"58": "Ce",
"59": "Pr",
"60": "Nd",
"61": "Pm",
"62": "Sm",
"63": "Eu",
"64": "Gd",
"65": "Tb",
"66": "Dy",
"67": "Ho",
"68": "Er",
"69": "Tm",
"70": "Yb",
"71": "Lu",
"72": "Hf",
"73": "Ta",
"74": "W",
"75": "Re",
"76": "Os",
"77": "Ir",
"78": "Pt",
"79": "Au",
"80": "Hg",
"81": "Tl",
"82": "Pb",
"83": "Bi",
"84": "Po",
"85": "At",
"86": "Rn",
"87": "Fr",
"88": "Ra",
"89": "Ac",
"90": "Th",
"91": "Pa",
"92": "U",
"93": "Np",
"94": "Pu",
"95": "Am",
"96": "Cm",
"97": "Bk",
"98": "Cf",
"99": "Es",
"100": "Fm"
}
lettersDict = {
"H": "1",
"He": "2",
"Li": "3",
"Be": "4",
"B": "5",
"C": "6",
"N": "7",
"O": "8",
"F": "9",
"Ne": "10",
"Na": "11",
"Mg": "12",
"Al": "13",
"Si": "14",
"P": "15",
"S": "16",
"Cl": "17",
"Ar": "18",
"K": "19",
"Ca": "20",
"Sc": "21",
"Ti": "22",
"V": "23",
"Cr": "24",
"Mn": "25",
"Fe": "26",
"Co": "27",
"Ni": "28",
"Cu": "29",
"Zn": "30",
"Ga": "31",
"Ge": "32",
"As": "33",
"Se": "34",
"Br": "35",
"Kr": "36",
"Rb": "37",
"Sr": "38",
"Y": "39",
"Zr": "40",
"Nb": "41",
"Mo": "42",
"Tc": "43",
"Ru": "44",
"Rh": "45",
"Pd": "46",
"Ag": "47",
"Cd": "48",
"In": "49",
"Sn": "50",
"Sb": "51",
"Te": "52",
"I": "53",
"Xe": "54",
"Cs": "55",
"Ba": "56",
"La": "57",
"Ce": "58",
"Pr": "59",
"Nd": "60",
"Pm": "61",
"Sm": "62",
"Eu": "63",
"Gd": "64",
"Tb": "65",
"Dy": "66",
"Ho": "67",
"Er": "68",
"Tm": "69",
"Yb": "70",
"Lu": "71",
"Hf": "72",
"Ta": "73",
"W": "74",
"Re": "75",
"Os": "76",
"Ir": "77",
"Pt": "78",
"Au": "79",
"Hg": "80",
"Tl": "81",
"Pb": "82",
"Bi": "83",
"Po": "84",
"At": "85",
"Rn": "86",
"Fr": "87",
"Ra": "88",
"Ac": "89",
"Th": "90",
"Pa": "91",
"U": "92",
"Np": "93",
"Pu": "94",
"Am": "95",
"Cm": "96",
"Bk": "97",
"Cf": "98",
"Es": "99",
"Fm": "100"
}
_ = input() # Supposed to be n, k but we do not need them
atoms = input().split(" ")
outAtoms = input().split(" ")
atoms = sorted(list(map(lambda x: int(lettersDict[x]), atoms)))
outAtoms = sorted(list(map(lambda x: int(lettersDict[x]), outAtoms)))
sumAtoms = 0
def testIfPossible():
atomsx = atoms.copy()
outAtomsx = outAtoms.copy()
for i in range(len(atoms) - 1, -1, -1):
if atomsx[i] > outAtomsx[-1]:
atomsx.pop()
if sum(outAtomsx) > sum(atomsx):
print("NO")
exit()
testIfPossible()
for at in atoms:
sumAtoms += at
outAtom = 0
for at in outAtoms:
outAtom += at
def dfs(i: int, currentSum: int, arr: [int], searchSum: int) -> [[int]]:
if i >= len(arr) or currentSum + arr[i] > searchSum:
return []
totalRes = []
# we take
res = dfs(i + 1, currentSum + arr[i], arr, searchSum)
totalRes += [[i] + a for a in res]
# we don't take
res = dfs(i + 1, currentSum, arr, searchSum)
totalRes += [a for a in res]
if currentSum + arr[i] == searchSum:
totalRes.append([i])
return totalRes
allCombos = [[set(x) for x in dfs(0, 0, atoms, out)] for out in outAtoms]
currentSet = set()
stack = []
resultFound = False
def dfs2(i: int):
global resultFound
global stack
global currentSet
if i >= len(allCombos):
resultFound = True
return
for set in allCombos[i]:
if not set & currentSet:
stack.append(set)
currentSet = currentSet | set
dfs2(i + 1)
if resultFound:
break
stack.pop()
currentSet = currentSet - set
isAnyEmpty = False
for comb in allCombos:
if not comb:
isAnyEmpty = True
if not isAnyEmpty:
dfs2(0)
if resultFound:
print("YES")
res = list(map(lambda x: "+".join(list(map(lambda y: numbersDict[f"{atoms[y]}"],list(x)))), stack))
outs = list(map(lambda x: numbersDict[f"{x}"], outAtoms))
tot = list(map(lambda x: x[0] + "->" + x[1], zip(res, outs)))
for t in tot:
print(t)
else:
print("NO")
```
| 87,142 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is the following puzzle popular among nuclear physicists.
A reactor contains a set of n atoms of some chemical elements. We shall understand the phrase "atomic number" as the number of this atom's element in the periodic table of the chemical elements.
You are allowed to take any two different atoms and fuse a new one from them. That results in a new atom, whose number is equal to the sum of the numbers of original atoms. The fusion operation can be performed several times.
The aim is getting a new pregiven set of k atoms.
The puzzle's difficulty is that it is only allowed to fuse two atoms into one, it is not allowed to split an atom into several atoms. You are suggested to try to solve the puzzle.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 17). The second line contains space-separated symbols of elements of n atoms, which are available from the start. The third line contains space-separated symbols of elements of k atoms which need to be the result of the fusion. The symbols of the elements coincide with the symbols from the periodic table of the chemical elements. The atomic numbers do not exceed 100 (elements possessing larger numbers are highly unstable). Some atoms can have identical numbers (that is, there can be several atoms of the same element). The sum of numbers of initial atoms is equal to the sum of numbers of the atoms that need to be synthesized.
Output
If it is impossible to synthesize the required atoms, print "NO" without the quotes. Otherwise, print on the first line «YES», and on the next k lines print the way of synthesizing each of k atoms as equations. Each equation has the following form: "x1+x2+...+xt->yi", where xj is the symbol of the element of some atom from the original set, and yi is the symbol of the element of some atom from the resulting set. Each atom from the input data should occur in the output data exactly one time. The order of summands in the equations, as well as the output order does not matter. If there are several solutions, print any of them. For a better understanding of the output format, see the samples.
Examples
Input
10 3
Mn Co Li Mg C P F Zn Sc K
Sn Pt Y
Output
YES
Mn+C+K->Sn
Co+Zn+Sc->Pt
Li+Mg+P+F->Y
Input
2 1
H H
He
Output
YES
H+H->He
Input
2 2
Bk Fm
Cf Es
Output
NO
Note
The reactions from the first example possess the following form (the atomic number is written below and to the left of the element):
<image>
<image>
<image>
To find a periodic table of the chemical elements, you may use your favorite search engine.
The pretest set contains each of the first 100 elements of the periodic table at least once. You can use that information to check for misprints.
Tags: bitmasks, dp
Correct Solution:
```
#!/usr/bin/env python3
import itertools
# Initialize look-up tables
element_to_value = {
'H':1, 'He':2, 'Li':3, 'Be':4, 'B':5, 'C':6, 'N':7, 'O':8, 'F':9, 'Ne':10,
'Na':11, 'Mg':12, 'Al':13, 'Si':14, 'P':15, 'S':16, 'Cl':17, 'Ar':18, 'K':19, 'Ca':20,
'Sc':21, 'Ti':22, 'V':23, 'Cr':24, 'Mn':25, 'Fe':26, 'Co':27, 'Ni':28, 'Cu':29, 'Zn':30,
'Ga':31, 'Ge':32, 'As':33, 'Se':34, 'Br':35, 'Kr':36, 'Rb':37, 'Sr':38, 'Y':39, 'Zr':40,
'Nb':41, 'Mo':42, 'Tc':43, 'Ru':44, 'Rh':45, 'Pd':46, 'Ag':47, 'Cd':48, 'In':49, 'Sn':50,
'Sb':51, 'Te':52, 'I':53, 'Xe':54, 'Cs':55, 'Ba':56, 'La':57, 'Ce':58, 'Pr':59, 'Nd':60,
'Pm':61, 'Sm':62, 'Eu':63, 'Gd':64, 'Tb':65, 'Dy':66, 'Ho':67, 'Er':68, 'Tm':69, 'Yb':70,
'Lu':71, 'Hf':72, 'Ta':73, 'W':74, 'Re':75, 'Os':76, 'Ir':77, 'Pt':78, 'Au':79, 'Hg':80,
'Tl':81, 'Pb':82, 'Bi':83, 'Po':84, 'At':85, 'Rn':86, 'Fr':87, 'Ra':88, 'Ac':89, 'Th':90,
'Pa':91, 'U':92, 'Np':93, 'Pu':94, 'Am':95, 'Cm':96, 'Bk':97, 'Cf':98, 'Es':99, 'Fm':100
}
value_to_element = dict()
for (element, value) in element_to_value.items():
value_to_element[value] = element
# Read inputs
(n,k) = map(int,input().split())
products_start_str = input().split()
products_end_str = input().split()
# Translate elements to their values
products_start = [element_to_value[elem] for elem in products_start_str]
products_end = [element_to_value[elem] for elem in products_end_str]
# Filter out duplicates; keep track of ingredient values and their number
products_start.sort()
ingredient_value = []
ingredient_count = []
for (key, lst) in itertools.groupby(products_start):
ingredient_value.append(key)
ingredient_count.append(len(list(lst)))
nr_ingredients = len(ingredient_value)
# Figure out the options for constructing the final products
construction_options = [[] for i in range(k)]
for combination in itertools.product(*[range(l+1) for l in ingredient_count]):
value = sum(combination[i]*ingredient_value[i] for i in range(nr_ingredients))
if (value in products_end):
for i in range(k):
if products_end[i] == value:
construction_options[i].append(combination)
# Do a depth-first search on the construction options for a possible solution
solution = [None for i in range(k)]
def find_solution(used = [0 for i in range(nr_ingredients)], next = 0):
if (next == k):
return all(used[i] == ingredient_count[i] for i in range(nr_ingredients))
else:
for option in construction_options[next]:
usage = [used[i]+option[i] for i in range(nr_ingredients)]
if all(used[i] <= ingredient_count[i] for i in range(nr_ingredients)):
possible = find_solution(usage, next+1)
if (possible):
solution[next] = option
return True
return False
possible = find_solution()
# Print the answer
if not possible:
print("NO")
exit()
def combination_to_recipe(combination):
recipe = []
for i in range(nr_ingredients):
for j in range(combination[i]):
recipe.append(value_to_element[ingredient_value[i]])
return recipe
print("YES")
for i in range(k):
recipe = combination_to_recipe(solution[i])
print("%s->%s" % ("+".join(recipe),products_end_str[i]))
```
| 87,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is the following puzzle popular among nuclear physicists.
A reactor contains a set of n atoms of some chemical elements. We shall understand the phrase "atomic number" as the number of this atom's element in the periodic table of the chemical elements.
You are allowed to take any two different atoms and fuse a new one from them. That results in a new atom, whose number is equal to the sum of the numbers of original atoms. The fusion operation can be performed several times.
The aim is getting a new pregiven set of k atoms.
The puzzle's difficulty is that it is only allowed to fuse two atoms into one, it is not allowed to split an atom into several atoms. You are suggested to try to solve the puzzle.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 17). The second line contains space-separated symbols of elements of n atoms, which are available from the start. The third line contains space-separated symbols of elements of k atoms which need to be the result of the fusion. The symbols of the elements coincide with the symbols from the periodic table of the chemical elements. The atomic numbers do not exceed 100 (elements possessing larger numbers are highly unstable). Some atoms can have identical numbers (that is, there can be several atoms of the same element). The sum of numbers of initial atoms is equal to the sum of numbers of the atoms that need to be synthesized.
Output
If it is impossible to synthesize the required atoms, print "NO" without the quotes. Otherwise, print on the first line «YES», and on the next k lines print the way of synthesizing each of k atoms as equations. Each equation has the following form: "x1+x2+...+xt->yi", where xj is the symbol of the element of some atom from the original set, and yi is the symbol of the element of some atom from the resulting set. Each atom from the input data should occur in the output data exactly one time. The order of summands in the equations, as well as the output order does not matter. If there are several solutions, print any of them. For a better understanding of the output format, see the samples.
Examples
Input
10 3
Mn Co Li Mg C P F Zn Sc K
Sn Pt Y
Output
YES
Mn+C+K->Sn
Co+Zn+Sc->Pt
Li+Mg+P+F->Y
Input
2 1
H H
He
Output
YES
H+H->He
Input
2 2
Bk Fm
Cf Es
Output
NO
Note
The reactions from the first example possess the following form (the atomic number is written below and to the left of the element):
<image>
<image>
<image>
To find a periodic table of the chemical elements, you may use your favorite search engine.
The pretest set contains each of the first 100 elements of the periodic table at least once. You can use that information to check for misprints.
Submitted Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
elem = ['*'] + '''
H He Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar K Ca
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Ga Ge As Se Br Kr Rb Sr Y Zr
Nb Mo Tc Ru Rh Pd Ag Cd In Sn
Sb Te I Xe Cs Ba La Ce Pr Nd
Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
Lu Hf Ta W Re Os Ir Pt Au Hg
Tl Pb Bi Po At Rn Fr Ra Ac Th
Pa U Np Pu Am Cm Bk Cf Es Fm'''.split()
n, k = map(int, input().split())
a = [elem.index(e) for e in input().split()]
b = [elem.index(e) for e in input().split()]
req = [[] for _ in range(k)]
for i in range(k):
for bit in range(1 << n):
if sum(a[j] for j in range(n) if (1 << j) & bit) == b[i]:
req[i].append(bit)
for i in range(k):
req[i] = req[i][:50]
dp = [array('i', [-1]) * (1 << n) for _ in range(k + 1)]
dp[0][0] = 0
full_bit = (1 << n) - 1
bs_set = {0}
for i in range(k):
next_bs = set()
for bitset in list(bs_set)[:15000]:
if dp[i][bitset] != -1:
for req_bit in req[i]:
if not (bitset & req_bit):
next_bs.add(bitset + req_bit)
dp[i + 1][bitset | req_bit] = bitset
bs_set = next_bs
bitset = -1
for bit in range(1 << n):
if dp[-1][bit] != -1:
bitset = bit
break
if bitset == -1:
print('NO')
exit()
ans = []
for i in range(k, 0, -1):
prev = dp[i][bitset]
delta = bitset - prev
s = '+'.join(elem[a[j]] for j in range(n) if (1 << j) & delta)
ans.append(s + '->' + elem[b[i - 1]])
bitset = prev
print('YES')
print(*reversed(ans), sep='\n')
```
No
| 87,144 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is the following puzzle popular among nuclear physicists.
A reactor contains a set of n atoms of some chemical elements. We shall understand the phrase "atomic number" as the number of this atom's element in the periodic table of the chemical elements.
You are allowed to take any two different atoms and fuse a new one from them. That results in a new atom, whose number is equal to the sum of the numbers of original atoms. The fusion operation can be performed several times.
The aim is getting a new pregiven set of k atoms.
The puzzle's difficulty is that it is only allowed to fuse two atoms into one, it is not allowed to split an atom into several atoms. You are suggested to try to solve the puzzle.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 17). The second line contains space-separated symbols of elements of n atoms, which are available from the start. The third line contains space-separated symbols of elements of k atoms which need to be the result of the fusion. The symbols of the elements coincide with the symbols from the periodic table of the chemical elements. The atomic numbers do not exceed 100 (elements possessing larger numbers are highly unstable). Some atoms can have identical numbers (that is, there can be several atoms of the same element). The sum of numbers of initial atoms is equal to the sum of numbers of the atoms that need to be synthesized.
Output
If it is impossible to synthesize the required atoms, print "NO" without the quotes. Otherwise, print on the first line «YES», and on the next k lines print the way of synthesizing each of k atoms as equations. Each equation has the following form: "x1+x2+...+xt->yi", where xj is the symbol of the element of some atom from the original set, and yi is the symbol of the element of some atom from the resulting set. Each atom from the input data should occur in the output data exactly one time. The order of summands in the equations, as well as the output order does not matter. If there are several solutions, print any of them. For a better understanding of the output format, see the samples.
Examples
Input
10 3
Mn Co Li Mg C P F Zn Sc K
Sn Pt Y
Output
YES
Mn+C+K->Sn
Co+Zn+Sc->Pt
Li+Mg+P+F->Y
Input
2 1
H H
He
Output
YES
H+H->He
Input
2 2
Bk Fm
Cf Es
Output
NO
Note
The reactions from the first example possess the following form (the atomic number is written below and to the left of the element):
<image>
<image>
<image>
To find a periodic table of the chemical elements, you may use your favorite search engine.
The pretest set contains each of the first 100 elements of the periodic table at least once. You can use that information to check for misprints.
Submitted Solution:
```
numbersDict = {
"1": "H",
"2": "He",
"3": "Li",
"4": "Be",
"5": "B",
"6": "C",
"7": "N",
"8": "O",
"9": "F",
"10": "Ne",
"11": "Na",
"12": "Mg",
"13": "Al",
"14": "Si",
"15": "P",
"16": "S",
"17": "Cl",
"18": "Ar",
"19": "K",
"20": "Ca",
"21": "Sc",
"22": "Ti",
"23": "V",
"24": "Cr",
"25": "Mn",
"26": "Fe",
"27": "Co",
"28": "Ni",
"29": "Cu",
"30": "Zn",
"31": "Ga",
"32": "Ge",
"33": "As",
"34": "Se",
"35": "Br",
"36": "Kr",
"37": "Rb",
"38": "Sr",
"39": "Y",
"40": "Zr",
"41": "Nb",
"42": "Mo",
"43": "Tc",
"44": "Ru",
"45": "Rh",
"46": "Pd",
"47": "Ag",
"48": "Cd",
"49": "In",
"50": "Sn",
"51": "Sb",
"52": "Te",
"53": "I",
"54": "Xe",
"55": "Cs",
"56": "Ba",
"57": "La",
"58": "Ce",
"59": "Pr",
"60": "Nd",
"61": "Pm",
"62": "Sm",
"63": "Eu",
"64": "Gd",
"65": "Tb",
"66": "Dy",
"67": "Ho",
"68": "Er",
"69": "Tm",
"70": "Yb",
"71": "Lu",
"72": "Hf",
"73": "Ta",
"74": "W",
"75": "Re",
"76": "Os",
"77": "Ir",
"78": "Pt",
"79": "Au",
"80": "Hg",
"81": "Tl",
"82": "Pb",
"83": "Bi",
"84": "Po",
"85": "At",
"86": "Rn",
"87": "Fr",
"88": "Ra",
"89": "Ac",
"90": "Th",
"91": "Pa",
"92": "U",
"93": "Np",
"94": "Pu",
"95": "Am",
"96": "Cm",
"97": "Bk",
"98": "Cf",
"99": "Es",
"100": "Fm"
}
lettersDict = {
"H": "1",
"He": "2",
"Li": "3",
"Be": "4",
"B": "5",
"C": "6",
"N": "7",
"O": "8",
"F": "9",
"Ne": "10",
"Na": "11",
"Mg": "12",
"Al": "13",
"Si": "14",
"P": "15",
"S": "16",
"Cl": "17",
"Ar": "18",
"K": "19",
"Ca": "20",
"Sc": "21",
"Ti": "22",
"V": "23",
"Cr": "24",
"Mn": "25",
"Fe": "26",
"Co": "27",
"Ni": "28",
"Cu": "29",
"Zn": "30",
"Ga": "31",
"Ge": "32",
"As": "33",
"Se": "34",
"Br": "35",
"Kr": "36",
"Rb": "37",
"Sr": "38",
"Y": "39",
"Zr": "40",
"Nb": "41",
"Mo": "42",
"Tc": "43",
"Ru": "44",
"Rh": "45",
"Pd": "46",
"Ag": "47",
"Cd": "48",
"In": "49",
"Sn": "50",
"Sb": "51",
"Te": "52",
"I": "53",
"Xe": "54",
"Cs": "55",
"Ba": "56",
"La": "57",
"Ce": "58",
"Pr": "59",
"Nd": "60",
"Pm": "61",
"Sm": "62",
"Eu": "63",
"Gd": "64",
"Tb": "65",
"Dy": "66",
"Ho": "67",
"Er": "68",
"Tm": "69",
"Yb": "70",
"Lu": "71",
"Hf": "72",
"Ta": "73",
"W": "74",
"Re": "75",
"Os": "76",
"Ir": "77",
"Pt": "78",
"Au": "79",
"Hg": "80",
"Tl": "81",
"Pb": "82",
"Bi": "83",
"Po": "84",
"At": "85",
"Rn": "86",
"Fr": "87",
"Ra": "88",
"Ac": "89",
"Th": "90",
"Pa": "91",
"U": "92",
"Np": "93",
"Pu": "94",
"Am": "95",
"Cm": "96",
"Bk": "97",
"Cf": "98",
"Es": "99",
"Fm": "100"
}
_ = input() # Supposed to be n, k but we do not need them
atoms = sorted(list(map(lambda x: int(lettersDict[x]), input().split(" "))))
outAtoms = sorted(list(map(lambda x: int(lettersDict[x]), input().split(" "))))
allStack = []
currentSet = set()
stack = []
resultFound = False
memo = set()
cycles = 0
def dfs3(i: int, currentSum: int):
global cycles
cycles += 1
if cycles >= 250000:
print("NO")
exit()
return
global stack
global allStack
global currentSet
global resultFound
if i == len(outAtoms):
resultFound = True
return
if currentSum > outAtoms[i]:
return
if currentSum == outAtoms[i]:
previousStack = stack.copy()
allStack.append(stack.copy())
stack = []
dfs3(i + 1, 0)
if resultFound:
return
allStack.pop()
stack = previousStack
return
for j in range(len(atoms)):
if j in currentSet:
continue
currentSet.add(j)
stack.append(j)
dfs3(i, currentSum + atoms[j])
if resultFound:
return
currentSet.remove(j)
stack.remove(j)
dfs3(0, 0)
if resultFound:
print("YES")
res = [[numbersDict[f"{atoms[s]}"] for s in st] for st in allStack]
res = ["+".join(x) for x in res]
outs = list(map(lambda x: numbersDict[f"{x}"], outAtoms))
tot = list(map(lambda x: x[0] + "->" + x[1], zip(res, outs)))
for t in tot:
print(t)
else:
print("NO")
```
No
| 87,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is the following puzzle popular among nuclear physicists.
A reactor contains a set of n atoms of some chemical elements. We shall understand the phrase "atomic number" as the number of this atom's element in the periodic table of the chemical elements.
You are allowed to take any two different atoms and fuse a new one from them. That results in a new atom, whose number is equal to the sum of the numbers of original atoms. The fusion operation can be performed several times.
The aim is getting a new pregiven set of k atoms.
The puzzle's difficulty is that it is only allowed to fuse two atoms into one, it is not allowed to split an atom into several atoms. You are suggested to try to solve the puzzle.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 17). The second line contains space-separated symbols of elements of n atoms, which are available from the start. The third line contains space-separated symbols of elements of k atoms which need to be the result of the fusion. The symbols of the elements coincide with the symbols from the periodic table of the chemical elements. The atomic numbers do not exceed 100 (elements possessing larger numbers are highly unstable). Some atoms can have identical numbers (that is, there can be several atoms of the same element). The sum of numbers of initial atoms is equal to the sum of numbers of the atoms that need to be synthesized.
Output
If it is impossible to synthesize the required atoms, print "NO" without the quotes. Otherwise, print on the first line «YES», and on the next k lines print the way of synthesizing each of k atoms as equations. Each equation has the following form: "x1+x2+...+xt->yi", where xj is the symbol of the element of some atom from the original set, and yi is the symbol of the element of some atom from the resulting set. Each atom from the input data should occur in the output data exactly one time. The order of summands in the equations, as well as the output order does not matter. If there are several solutions, print any of them. For a better understanding of the output format, see the samples.
Examples
Input
10 3
Mn Co Li Mg C P F Zn Sc K
Sn Pt Y
Output
YES
Mn+C+K->Sn
Co+Zn+Sc->Pt
Li+Mg+P+F->Y
Input
2 1
H H
He
Output
YES
H+H->He
Input
2 2
Bk Fm
Cf Es
Output
NO
Note
The reactions from the first example possess the following form (the atomic number is written below and to the left of the element):
<image>
<image>
<image>
To find a periodic table of the chemical elements, you may use your favorite search engine.
The pretest set contains each of the first 100 elements of the periodic table at least once. You can use that information to check for misprints.
Submitted Solution:
```
en={'H': '1', 'He': '2', 'Li': '3', 'Be': '4', 'B': '5', 'C': '6', 'N': '7', 'O': '8', 'F': '9', 'Ne': '10', 'Na': '11', 'Mg': '12', 'Al': '13', 'Si': '14', 'P': '15', 'S': '16', 'Cl': '17', 'Ar': '18', 'K': '19', 'Ca': '20', 'Sc': '21', 'Ti': '22', 'V': '23', 'Cr': '24', 'Mn': '25', 'Fe': '26', 'Co': '27', 'Ni': '28', 'Cu': '29', 'Zn': '30', 'Ga': '31', 'Ge': '32', 'As': '33', 'Se': '34', 'Br': '35', 'Kr': '36', 'Rb': '37', 'Sr': '38', 'Y': '39', 'Zr': '40', 'Nb': '41', 'Mo': '42', 'Tc': '43', 'Ru': '44', 'Rh': '45', 'Pd': '46', 'Ag': '47', 'Cd': '48', 'In': '49', 'Sn': '50', 'Sb': '51', 'Te': '52', 'I': '53', 'Xe': '54', 'Cs': '55', 'Ba': '56', 'La': '57', 'Ce': '58', 'Pr': '59', 'Nd': '60', 'Pm': '61', 'Sm': '62', 'Eu': '63', 'Gd': '64', 'Tb': '65', 'Dy': '66', 'Ho': '67', 'Er': '68', 'Tm': '69', 'Yb': '70', 'Lu': '71', 'Hf': '72', 'Ta': '73', 'W': '74', 'Re': '75', 'Os': '76', 'Ir': '77', 'Pt': '78', 'Au': '79', 'Hg': '80', 'Tl': '81', 'Pb': '82', 'Bi': '83', 'Po': '84', 'At': '85', 'Rn': '86', 'Fr': '87', 'Ra': '88', 'Ac': '89', 'Th': '90', 'Pa': '91', 'U': '92', 'Np': '93', 'Pu': '94', 'Am': '95', 'Cm': '96', 'Bk': '97', 'Cf': '98', 'Es': '99', 'Fm': '100', 'Md': '101', 'No': '102', 'Lr': '103', 'Rf': '104', 'Db': '105', 'Sg': '106', 'Bh': '107', 'Hs': '108', 'Mt': '109', 'Ds ': '110', 'Rg ': '111', 'Cn ': '112', 'Nh': '113', 'Fl': '114', 'Mc': '115', 'Lv': '116', 'Ts': '117', 'Og': '118'}
ne={en[x]:x for x in en}
n,m=map(int,input().split())
e=[int(en[x])for x in input().split()]
r=[int(en[x])for x in input().split()]
print(e)
a=1<<n
ans=[]
for q in r:
for i in range(a):
s=0
l=[]
for j in range(n):
if 1<<j&i:
s+=e[j]
l+=[ne[str(e[j])]]
if s==q:ans+=[l];break
if len(ans)==m:
print('YES')
for i in range(m):
print('+'.join(ans[i])+'->'+ne[str(r[i])])
else:print('NO')
```
No
| 87,146 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is the following puzzle popular among nuclear physicists.
A reactor contains a set of n atoms of some chemical elements. We shall understand the phrase "atomic number" as the number of this atom's element in the periodic table of the chemical elements.
You are allowed to take any two different atoms and fuse a new one from them. That results in a new atom, whose number is equal to the sum of the numbers of original atoms. The fusion operation can be performed several times.
The aim is getting a new pregiven set of k atoms.
The puzzle's difficulty is that it is only allowed to fuse two atoms into one, it is not allowed to split an atom into several atoms. You are suggested to try to solve the puzzle.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 17). The second line contains space-separated symbols of elements of n atoms, which are available from the start. The third line contains space-separated symbols of elements of k atoms which need to be the result of the fusion. The symbols of the elements coincide with the symbols from the periodic table of the chemical elements. The atomic numbers do not exceed 100 (elements possessing larger numbers are highly unstable). Some atoms can have identical numbers (that is, there can be several atoms of the same element). The sum of numbers of initial atoms is equal to the sum of numbers of the atoms that need to be synthesized.
Output
If it is impossible to synthesize the required atoms, print "NO" without the quotes. Otherwise, print on the first line «YES», and on the next k lines print the way of synthesizing each of k atoms as equations. Each equation has the following form: "x1+x2+...+xt->yi", where xj is the symbol of the element of some atom from the original set, and yi is the symbol of the element of some atom from the resulting set. Each atom from the input data should occur in the output data exactly one time. The order of summands in the equations, as well as the output order does not matter. If there are several solutions, print any of them. For a better understanding of the output format, see the samples.
Examples
Input
10 3
Mn Co Li Mg C P F Zn Sc K
Sn Pt Y
Output
YES
Mn+C+K->Sn
Co+Zn+Sc->Pt
Li+Mg+P+F->Y
Input
2 1
H H
He
Output
YES
H+H->He
Input
2 2
Bk Fm
Cf Es
Output
NO
Note
The reactions from the first example possess the following form (the atomic number is written below and to the left of the element):
<image>
<image>
<image>
To find a periodic table of the chemical elements, you may use your favorite search engine.
The pretest set contains each of the first 100 elements of the periodic table at least once. You can use that information to check for misprints.
Submitted Solution:
```
import random
from itertools import combinations, product
from functools import reduce
symbol_value = {'Br': 35, 'Pm': 61, 'At': 85, 'Xe': 54, 'Pu': 94, 'Tl': 81, 'Mg': 12, 'Fe': 26, 'Kr': 36, 'Ar': 18, 'I': 53, 'Ga': 31, 'Sn': 50, 'Pd': 46, 'Fr': 87, 'Lu': 71, 'Dy': 66, 'Ac': 89, 'Gd': 64, 'Nd': 60, 'Bk': 97, 'Sb': 51, 'H': 1, 'Al': 13, 'Mo': 42, 'Sr': 38, 'Re': 75, 'Es': 99, 'Au': 79, 'Ra': 88, 'Cu': 29, 'Ca': 20, 'Am': 95, 'Pa': 91, 'U': 92, 'Ti': 22, 'Y': 39, 'Ag': 47, 'Hg': 80, 'F': 9, 'Cd': 48, 'Cs': 55, 'Rh': 45, 'Os': 76, 'Pr': 59, 'Ni': 28, 'Bi': 83, 'La': 57, 'Ta': 73, 'S': 16, 'Ir': 77, 'Rn': 86, 'Zn': 30, 'Se': 34, 'Li': 3, 'Ge': 32, 'Cr': 24, 'W': 74, 'Cm': 96, 'Hf': 72, 'Fm': 100, 'In': 49, 'Si': 14, 'Cl': 17, 'Sc': 21, 'Tc': 43, 'V': 23, 'O': 8, 'Rb': 37, 'Np': 93, 'Mn': 25, 'Te': 52, 'Ba': 56, 'Tb': 65, 'Po': 84, 'B': 5, 'Ru': 44, 'Eu': 63, 'Na': 11, 'Ce': 58, 'Yb': 70, 'Ho': 67, 'Nb': 41, 'Er': 68, 'Co': 27, 'Zr': 40, 'Ne': 10, 'As': 33, 'Sm': 62, 'Pb': 82, 'K': 19, 'Tm': 69, 'C': 6, 'Cf': 98, 'P': 15, 'Pt': 78, 'Th': 90, 'He': 2, 'N': 7, 'Be': 4}
value_symbol = {1: 'H', 2: 'He', 3: 'Li', 4: 'Be', 5: 'B', 6: 'C', 7: 'N', 8: 'O', 9: 'F', 10: 'Ne', 11: 'Na', 12: 'Mg', 13: 'Al', 14: 'Si', 15: 'P', 16: 'S', 17: 'Cl', 18: 'Ar', 19: 'K', 20: 'Ca', 21: 'Sc', 22: 'Ti', 23: 'V', 24: 'Cr', 25: 'Mn', 26: 'Fe', 27: 'Co', 28: 'Ni', 29: 'Cu', 30: 'Zn', 31: 'Ga', 32: 'Ge', 33: 'As', 34: 'Se', 35: 'Br', 36: 'Kr', 37: 'Rb', 38: 'Sr', 39: 'Y', 40: 'Zr', 41: 'Nb', 42: 'Mo', 43: 'Tc', 44: 'Ru', 45: 'Rh', 46: 'Pd', 47: 'Ag', 48: 'Cd', 49: 'In', 50: 'Sn', 51: 'Sb', 52: 'Te', 53: 'I', 54: 'Xe', 55: 'Cs', 56: 'Ba', 57: 'La', 58: 'Ce', 59: 'Pr', 60: 'Nd', 61: 'Pm', 62: 'Sm', 63: 'Eu', 64: 'Gd', 65: 'Tb', 66: 'Dy', 67: 'Ho', 68: 'Er', 69: 'Tm', 70: 'Yb', 71: 'Lu', 72: 'Hf', 73: 'Ta', 74: 'W', 75: 'Re', 76: 'Os', 77: 'Ir', 78: 'Pt', 79: 'Au', 80: 'Hg', 81: 'Tl', 82: 'Pb', 83: 'Bi', 84: 'Po', 85: 'At', 86: 'Rn', 87: 'Fr', 88: 'Ra', 89: 'Ac', 90: 'Th', 91: 'Pa', 92: 'U', 93: 'Np', 94: 'Pu', 95: 'Am', 96: 'Cm', 97: 'Bk', 98: 'Cf', 99: 'Es', 100: 'Fm'}
def translate_tuple(t):
return '+'.join(str(value_symbol[item]) for item in t) + '->' + value_symbol[sum(t)]
def get_start_pos(n_values, k, number):
pos = None
for i, n in enumerate(n_values):
if i < k - 1: # can be improved
continue
if number > k * n:
continue
tk = 1
total = n
while tk < k:
total += n_values[i - tk]
tk += 1
if total > number:
pos = i - k + 1
break
return pos
def find_valid_answer(alternatives):
keys = alternatives.keys()
for indices in product(*(range(len(alternatives[a])) for a in keys)):
item = [alternatives[k][i] for k,i in zip(keys, indices)]
expected_sum = sum(len(a) for a in item)
if len(reduce(lambda a,b: a.union(b),item)) == expected_sum:
return item
return None
while True:
try:
line = input()
except:
break
n, k = map(int, line.split())
n_values = sorted([symbol_value[x] for x in input().strip().split()])
k_values = sorted([symbol_value[x] for x in input().strip().split()])
alternatives = {k:[] for k in k_values}
sample = sorted(n_values)
desired = set(k_values)
stop = max(desired)
for i in range(1,len(sample)+1):
min = 100000
for comb in combinations(sample, i):
current = sum(comb)
if current < min:
min = current
sum_comb = sum(comb)
if sum_comb in desired:
alternatives[sum_comb].append(set(comb))
if min > stop:
break
answer = find_valid_answer(alternatives)
if answer:
print('YES')
for i in answer:
print(translate_tuple(i))
else:
print('NO')
```
No
| 87,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Tags: dfs and similar, dp, dsu
Correct Solution:
```
def ler():
return [int(x) for x in input().split()]
def dfs(u, adj, visited, s, Pesos, Belezas):
visited[u] = True
total_p = Pesos[u]
total_b = Belezas[u]
s.append(u)
for v in adj[u]:
if not visited[v]:
w, b = dfs(v, adj, visited, s, Pesos, Belezas)
total_p += w
total_b += b
return total_p, total_b
n, m, w = ler()
Pesos = ler()
Belezas = ler()
adj = [[] for _ in range(n)]
for _ in range(m):
x, y = ler()
x -= 1
y -= 1
adj[x].append(y)
adj[y].append(x)
visited = [False] * n
f = [0] * (w + 1)
for i in range(n):
if visited[i]:
continue
s = []
total_p, total_b = dfs(i, adj, visited, s, Pesos, Belezas)
for j in range(w, -1, -1):
jw = j + total_p
if jw <= w:
f[jw] = max(f[jw], f[j] + total_b)
for v in s:
jw = j + Pesos[v]
if jw <= w:
f[jw] = max(f[jw], f[j] + Belezas[v])
print(f[w])
```
| 87,148 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Tags: dfs and similar, dp, dsu
Correct Solution:
```
from collections import defaultdict
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
group_members = defaultdict(list)
for member in range(self.n):
group_members[self.find(member)].append(member)
return group_members
n,m,w=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
uf=UnionFind(n)
for _ in range(m):
x,y=map(int,input().split())
uf.union(x-1,y-1)
l=uf.roots()
ll=len(l)
dp=[[-10**18]*(w+1) for i in range(ll)]
ca,cb=0,0
dp[0][0]=0
for x in uf.members(l[0]):
if a[x]<=w:
dp[0][a[x]]=max(dp[0][a[x]],b[x])
ca+=a[x]
cb+=b[x]
if ca<=w:
dp[0][ca]=max(dp[0][ca],cb)
for i in range(1,ll):
ca,cb=0,0
for x in uf.members(l[i]):
for j in range(w+1):
dp[i][j]=max(dp[i][j],dp[i-1][j])
if j-a[x]>=0:
dp[i][j]=max(dp[i][j],dp[i-1][j-a[x]]+b[x])
ca+=a[x]
cb+=b[x]
for j in range(w+1):
if j-ca>=0:
dp[i][j]=max(dp[i][j],dp[i-1][j],dp[i-1][j-ca]+cb)
ans=0
for j in range(w+1):
ans=max(ans,dp[ll-1][j])
print(ans)
```
| 87,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Tags: dfs and similar, dp, dsu
Correct Solution:
```
f = lambda: map(int, input().split())
n, m, w = f()
wb = [(0, 0)] + list(zip(f(), f()))
t = list(range(n + 1))
def g(x):
if x == t[x]: return x
t[x] = g(t[x])
return t[x]
for i in range(m):
x, y = f()
x, y = g(x), g(y)
if x != y: t[y] = x
p = [[] for j in range(n + 1)]
for i in range(1, n + 1): p[g(i)].append(i)
d = [1] + [0] * w
for q in p:
if len(q) > 1:
WB = [wb[i] for i in q]
SW = sum(q[0] for q in WB)
SB = sum(q[1] for q in WB)
for D in range(w, -1, -1):
if d[D]:
if D + SW <= w: d[D + SW] = max(d[D + SW], d[D] + SB)
for W, B in WB:
if D + W <= w: d[D + W] = max(d[D + W], d[D] + B)
elif len(q) == 1:
W, B = wb[q[0]]
for D in range(w - W, -1, -1):
if d[D]: d[D + W] = max(d[D + W], d[D] + B)
print(max(d) - 1)
# Made By Mostafa_Khaled
```
| 87,150 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Tags: dfs and similar, dp, dsu
Correct Solution:
```
n, m, o = map(int, input().split())
wn = map(int, input().split())
bn = map(int, input().split())
wb = [(0, 0)] + list(zip(wn, bn))
l = list(range(n + 1))
def f(x):
if x == l[x]:
return x
l[x] = f(l[x])
return l[x]
for i in range(m):
x, y = map(int, input().split())
x, y = f(x), f(y)
if x != y:
l[y] = x
p = [[] for j in range(n + 1)]
for i in range(1, n + 1):
p[f(i)].append(i)
r = (o+1) * [0]
r[0] = 1
for i in p:
if len(i) > 1:
l = [wb[x] for x in i]
x0 = sum(x[0] for x in l)
x1 = sum(x[1] for x in l)
l.append((x0, x1))
l.sort()
for j in range(o, -1, -1):
if r[j]:
for w, b in l:
if j + w > o:
break
r[j + w] = max(r[j + w], r[j] + b)
elif len(i) == 1:
w, b = wb[i[0]]
for j in range(o - w, -1, -1):
if r[j]:
r[j + w] = max(r[j + w], r[j] + b)
res = max(r) - 1
print(res)
```
| 87,151 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Tags: dfs and similar, dp, dsu
Correct Solution:
```
def g(x):
if x == t[x]: return x
t[x] = g(t[x])
return t[x]
f = lambda: map(int, input().split())
hoses, pairOfFriends, weight = f()
weightsAndBeauties = [(0, 0)] + list(zip(f(), f()))
t = list(range(hoses + 1))
for i in range(pairOfFriends):
f1, f2 = f()
f1, f2 = g(f1), g(f2)
if f1 != f2:
t[f2] = f1
p = [[] for j in range(hoses + 1)]
for i in range(1, hoses + 1):
p[g(i)].append(i)
beauties = [1] + [0] * weight
for q in p:
if len(q) > 1:
t = [weightsAndBeauties[i] for i in q]
t.append((sum(f1[0] for f1 in t), sum(f1[1] for f1 in t)))
t.sort(key=lambda f1: f1[0])
for j in range(weight, -1, -1):
if beauties[j]:
for w, b in t:
if j + w > weight:
break
else:
beauties[j + w] = max(beauties[j + w], beauties[j] + b)
elif len(q) == 1:
w, b = weightsAndBeauties[q[0]]
for j in range(weight - w, -1, -1):
if beauties[j]:
beauties[j + w] = max(beauties[j + w], beauties[j] + b)
maxBeauty = max(beauties) - 1
print(maxBeauty)
```
| 87,152 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Tags: dfs and similar, dp, dsu
Correct Solution:
```
n, m, k = map(int, input().split())
a = map(int, input().split())
b = map(int, input().split())
ab = [(0, 0)] + list(zip(a, b))
l = list(range(n + 1))
def f(x):
if x == l[x]:
return x
l[x] = f(l[x])
return l[x]
for i in range(m):
x, y = map(int, input().split())
x, y = f(x), f(y)
if x != y:
l[y] = x
p = [[] for j in range(n + 1)]
for i in range(1, n + 1):
p[f(i)].append(i)
r = (k+1) * [0]
r[0] = 1
for i in p:
if len(i) > 1:
l = [ab[x] for x in i]
x0 = sum(x[0] for x in l)
x1 = sum(x[1] for x in l)
l.append((x0, x1))
l.sort()
for j in range(k, -1, -1):
if r[j]:
for w, b in l:
if j + w > k:
break
r[j + w] = max(r[j + w], r[j] + b)
elif len(i) == 1:
w, b = ab[i[0]]
for j in range(k - w, -1, -1):
if r[j]:
r[j + w] = max(r[j + w], r[j] + b)
res = max(r) - 1
print(res)
```
| 87,153 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Tags: dfs and similar, dp, dsu
Correct Solution:
```
R = lambda: map(int, input().split())
n, m, w = R()
ws = [0] + list(R())
bs = [0] + list(R())
g = [[] for x in range(n + 1)]
for i in range(m):
x, y = R()
g[x].append(y)
g[y].append(x)
cs = [0] * (n + 1)
cnt = 1
for i in range(1, n + 1):
if not cs[i]:
cs[i] = cnt
q = []
q.append(i)
while q:
nxt = q.pop()
for x in g[nxt]:
if not cs[x]:
cs[x] = cnt
q.append(x)
cnt += 1
gs = [[] for i in range(cnt)]
for i in range(1, n + 1):
gs[cs[i]].append(i)
dp = [[0] * (w + 1) for i in range(cnt)]
for i in range(1, cnt):
tw = sum(ws[k] for k in gs[i])
tb = sum(bs[k] for k in gs[i])
for j in range(1, w + 1):
dp[i][j] = max(dp[i][j], dp[i - 1][j], (dp[i - 1][j - tw] + tb if j >= tw else 0))
for k in gs[i]:
dp[i][j] = max(dp[i][j], dp[i - 1][j], (dp[i - 1][j - ws[k]] + bs[k] if j >= ws[k] else 0))
print(dp[-1][w])
```
| 87,154 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Tags: dfs and similar, dp, dsu
Correct Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
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 solve():
G = defaultdict(list)
def addEdge(a,b):
G[a].append(b)
G[b].append(a)
def dfs(node):
d = deque()
d.append(node)
vis[node] = True
T = []
S = []
tot = 0;wt = 0
while d:
x = d.pop()
S.append(B[x-1])
T.append(W[x-1])
# tot += B[x-1]
# wt += W[x-1]
for i in G.get(x,[]):
if not vis[i]:
vis[i] = True
d.append(i)
# if len(S) > 1:
# S.append(tot)
# T.append(wt)
return S,T
n,m,w =aj()
W = aj()
B = aj()
vis = [False]*(n+1)
for i in range(m):
u,v = aj()
addEdge(u,v)
A1 = []
A2 = []
Id = 1
for i in range(1,n+1):
if not vis[i]:
c,d = dfs(i)
A1.append(c)
A2.append(d)
# print(A1)
# print(A2)
# print(w)
# dp = {}
# def fun(pos = 0,wt = 0):
# if wt > w:
# return -float('inf')
# if pos >= len(A1):
# return 0
# z = dp.get((pos,wt),-1)
# if z!= -1:
# return z
# ans = 0
# c2 = 0
# for i in range(len(A1[pos])):
# c2 = max(c2,A1[pos][i] + fun(pos+1,wt + A2[pos][i]))
# c3 = fun(pos + 1,wt)
# z = max(c2,c3)
# dp[(pos,wt)] = z
# return z
dp = [[0]*(w+1) for i in range(len(A1)+1)]
for i in range(len(A1)):
weight_sum = sum(x for x in A2[i])
beauty_sum = sum(x for x in A1[i])
for j in range(w + 1):
dp[i][j] = max(beauty_sum + dp[i - 1][j - weight_sum]
if weight_sum <= j else 0, dp[i - 1][j])
for k in range(len(A1[i])):
dp[i][j] = max(dp[i][j], (dp[i - 1][j - A2[i][k]] +
A1[i][k] if A2[i][k] <= j else 0))
# print(fun())
print(dp[len(A1)-1][w])
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
```
| 87,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Submitted Solution:
```
from sys import stdin, stdout
from collections import defaultdict as dd
read, write = stdin.readline, stdout.write
class DisjointSetUnion:
def __init__(self, n):
self.parent = list(range(n))
self.size = [1] * n
self.num_sets = n
def find(self, a):
acopy = a
while a != self.parent[a]:
a = self.parent[a]
while acopy != a:
self.parent[acopy], acopy = a, self.parent[acopy]
return a
def union(self, a, b):
a, b = self.find(a), self.find(b)
if a != b:
if self.size[a] < self.size[b]:
a, b = b, a
self.num_sets -= 1
self.parent[b] = a
self.size[a] += self.size[b]
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
n, m, w = map(int, read().split())
weights = list(map(int, read().split()))
beauties = list(map(int, read().split()))
DSU = DisjointSetUnion(n)
for _ in range(m):
h1, h2 = map(int, read().split())
DSU.union(h1-1, h2-1)
groups = dd(list)
for i in range(n): DSU.find(i)
for hose, parent in enumerate(DSU.parent): groups[parent].append(hose)
dp = [0]*(w+1)
for friends in groups.values():
dp_aux = dp[:]
group_weight = group_beauty = 0
for friend in friends:
f_weight, f_beauty = weights[friend], beauties[friend]
group_weight += f_weight; group_beauty += f_beauty
for weight in range(f_weight, w+1):
dp[weight] = max(dp[weight], dp_aux[weight - f_weight] + f_beauty)
for weight in range(group_weight, w+1):
dp[weight] = max(dp[weight], dp_aux[weight - group_weight] + group_beauty)
print(dp[-1])
```
Yes
| 87,156 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Submitted Solution:
```
hoses, pairs, total_weight = map(int, input().split())
weight_arr = list(map(int, input().split()))
beauty_arr = list(map(int, input().split()))
arr = [-1] * hoses
def get(hose):
if arr[hose] < 0:
return hose
arr[hose] = get(arr[hose])
return arr[hose]
def join(left, right):
left, right = get(left), get(right)
if left != right:
arr[left] = right
for i in range(pairs):
left, right = map(int, input().split())
join(left - 1, right - 1)
groups = [list() for i in range(hoses)]
for i in range(hoses):
groups[get(i)].append(i)
groups = [group for group in groups if group]
dp = [[0] * (total_weight + 1) for i in range(len(groups) + 1)]
for i in range(len(groups)):
weight_sum = sum(weight_arr[x] for x in groups[i])
beauty_sum = sum(beauty_arr[x] for x in groups[i])
for j in range(total_weight + 1):
dp[i][j] = max(beauty_sum + dp[i - 1][j - weight_sum]
if weight_sum <= j else 0, dp[i - 1][j])
for k in groups[i]:
dp[i][j] = max(dp[i][j], (dp[i - 1][j - weight_arr[k]] +
beauty_arr[k] if weight_arr[k] <= j else 0))
print(dp[len(groups) - 1][total_weight])
```
Yes
| 87,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Submitted Solution:
```
n, m, wn = map(int, input().split())
Wi = [(0, 0)] + list(zip(map(int, input().split()), map(int, input().split())))
listaux3 = list(range(n + 1))
def aux(x):
if x == listaux3[x]:
return x
listaux3[x] = aux(listaux3[x])
return listaux3[x]
for i in range(m):
x, y = map(int, input().split())
x, y = aux(x), aux(y)
if x != y:
listaux3[y] = x
listaux1 = [[] for _ in range(n + 1)]
for i in range(1, n + 1):
listaux1[aux(i)].append(i)
listaux2 = [1] + [0] * wn
for k in listaux1:
if len(k) > 1:
listaux3 = [Wi[i] for i in k]
listaux3.append((sum(x[0] for x in listaux3), sum(x[1] for x in listaux3)))
listaux3.sort(key=lambda x: x[0])
for j in range(wn, -1, -1):
if listaux2[j]:
for w, b in listaux3:
if j + w > wn:
break
listaux2[j + w] = max(listaux2[j + w], listaux2[j] + b)
elif len(k) == 1:
w, b = Wi[k[0]]
for j in range(wn - w, -1, -1):
if listaux2[j]:
listaux2[j + w] = max(listaux2[j + w], listaux2[j] + b)
print(max(listaux2) - 1)
```
Yes
| 87,158 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Submitted Solution:
```
class disjoint:
def __init__(self,n):
self.rank=[1]*n
self.parent=[i for i in range(n)]
def find(self,x):
if(self.parent[x]!=x):
self.parent[x]=self.find(self.parent[x])
return self.parent[x];
def union(self,x,y):
xid=self.find(x)
yid=self.find(y)
if(xid==yid):
return;
if(self.rank[xid]<self.rank[yid]):
self.parent[xid]=yid
return;
elif(self.rank[xid]>self.rank[yid]):
self.parent[yid]=xid # merging y into x
return;
else:
self.parent[yid]=xid
self.rank[xid]+=1
return;
import sys
input=sys.stdin.readline
n,m,w=map(int,input().split())
obj=disjoint(n)
z=list(map(int,input().split()))
be=list(map(int,input().split()))
ans=[]
for i in range(len(z)):
ans.append([i,z[i],be[i]])
for i in range(m):
p,q=map(int,input().split())
obj.union(p-1,q-1)
from collections import *
al=defaultdict(list)
for i in range(len(ans)):
y=obj.find(ans[i][0])
al[y].append(ans[i][0])
dp=[[0 for i in range(len(al)+1)] for i in range(w+1)]
c1=0
maxa=0
for i in al:
if(c1==0):
q=al[i]
total=0
bea=0
for x in q:
wei=ans[x][1]
if(wei<=w):
dp[wei][c1]=max(dp[wei][c1],ans[x][2])
maxa=max(maxa,dp[wei][c1])
total+=wei
bea+=ans[x][2]
if(total<=w):
dp[total][c1]=max(dp[total][c1],bea)
maxa=max(maxa,dp[total][c1])
c1+=1
else:
q=al[i]
total=0
dil=0
bea=0
wei=0
for x in q:
wei=ans[x][1]
bea=ans[x][2]
total+=wei
dil+=bea
for j in range(w+1):
if(wei>j):
dp[j][c1]=max(dp[j][c1-1],dp[j][c1])
maxa=max(maxa,dp[j][c1])
continue;
else:
dp[j][c1]=max(dp[j][c1],dp[j-wei][c1-1]+bea,dp[j][c1-1])
maxa=max(maxa,dp[j][c1])
if(total<=w):
dp[total][c1]=max(dp[total][c1-1],dil,dp[total][c1])
maxa=max(maxa,dp[total][c1])
for j in range(w+1):
if(j<=total):
continue;
else:
dp[j][c1]=max(dp[j][c1],dil+dp[j-total][c1-1])
maxa=max(maxa,dp[j][c1])
c1+=1
print(maxa)
```
Yes
| 87,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Submitted Solution:
```
class disjoint:
def __init__(self,n):
self.rank=[1]*n
self.parent=[i for i in range(n)]
def find(self,x):
if(self.parent[x]!=x):
self.parent[x]=self.find(self.parent[x])
return self.parent[x];
def union(self,x,y):
xid=self.find(x)
yid=self.find(y)
if(xid==yid):
return;
if(self.rank[xid]<self.rank[yid]):
self.parent[xid]=yid
return;
elif(self.rank[xid]>self.rank[yid]):
self.parent[yid]=xid # merging y into x
return;
else:
self.parent[yid]=xid
self.rank[xid]+=1
return;
import sys
input=sys.stdin.readline
n,m,w=map(int,input().split())
obj=disjoint(n)
z=list(map(int,input().split()))
be=list(map(int,input().split()))
ans=[]
for i in range(len(z)):
ans.append([i,z[i],be[i]])
for i in range(m):
p,q=map(int,input().split())
obj.union(p-1,q-1)
from collections import *
al=defaultdict(list)
for i in range(len(ans)):
y=obj.find(ans[i][0])
al[y].append(ans[i][0])
dp=[[0 for i in range(len(al)+1)] for i in range(w+1)]
c1=0
for i in al:
if(c1==0):
q=al[i]
total=0
bea=0
for x in q:
wei=ans[x][1]
if(wei<=w):
dp[wei][c1]=max(dp[wei][c1],ans[x][2])
total+=wei
bea+=ans[x][2]
if(total<=w):
dp[total][c1]=max(dp[total][c1],bea)
c1+=1
else:
for j in range(w+1):
dp[j][c1]=max(dp[j][c1],dp[j][c1-1])
q=al[i]
total=0
dil=0
bea=0
wei=0
for x in q:
wei=ans[x][1]
bea=ans[x][2]
total+=wei
dil+=bea
for j in range(w+1):
if(wei>j):
dp[j][c1]=max(dp[j][c1-1],dp[j][c1])
continue;
else:
dp[j][c1]=max(dp[j][c1],dp[j-wei][c1-1]+bea,dp[j][c1-1])
if(total<=w):
dp[total][c1]=max(dp[total][c1-1],dil,dp[total][c1])
c1+=1
maxa=0
for j in range(w+1):
maxa=max(maxa,dp[j][c1-1])
print(maxa)
```
No
| 87,160 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Submitted Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
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 solve():
G = defaultdict(list)
def addEdge(a,b):
G[a].append(b)
G[b].append(a)
def dfs(node):
d = deque()
d.append(node)
vis[node] = True
T = []
S = []
tot = 0;wt = 0
while d:
x = d.pop()
S.append(B[x-1])
T.append(W[x-1])
# tot += B[x-1]
# wt += W[x-1]
for i in G.get(x,[]):
if not vis[i]:
vis[i] = True
d.append(i)
# if len(S) > 1:
# S.append(tot)
# T.append(wt)
return S,T
n,m,w =aj()
W = aj()
B = aj()
vis = [False]*(n+1)
for i in range(m):
u,v = aj()
addEdge(u,v)
A1 = []
A2 = []
Id = 1
for i in range(1,n+1):
if not vis[i]:
c,d = dfs(i)
A1.append(c)
A2.append(d)
# print(A1)
# print(A2)
# print(w)
# dp = {}
# def fun(pos = 0,wt = 0):
# if wt > w:
# return -float('inf')
# if pos >= len(A1):
# return 0
# z = dp.get((pos,wt),-1)
# if z!= -1:
# return z
# ans = 0
# c2 = 0
# for i in range(len(A1[pos])):
# c2 = max(c2,A1[pos][i] + fun(pos+1,wt + A2[pos][i]))
# c3 = fun(pos + 1,wt)
# z = max(c2,c3)
# dp[(pos,wt)] = z
# return z
dp = [[0]*(w+1) for i in range(len(A1)+1)]
for i in range(len(A1)):
weight_sum = sum(x for x in A1[i])
beauty_sum = sum(x for x in A1[i])
for j in range(w + 1):
dp[i][j] = max(beauty_sum + dp[i - 1][j - weight_sum]
if weight_sum <= j else 0, dp[i - 1][j])
for k in range(len(A1[i])):
dp[i][j] = max(dp[i][j], (dp[i - 1][j - A2[i][k]] +
A1[i][k] if A2[i][k] <= j else 0))
# print(fun())
print(dp[len(A1)-1][w])
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
```
No
| 87,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Submitted Solution:
```
from collections import defaultdict
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
group_members = defaultdict(list)
for member in range(self.n):
group_members[self.find(member)].append(member)
return group_members
n,m,w=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
uf=UnionFind(n)
for _ in range(m):
x,y=map(int,input().split())
uf.union(x-1,y-1)
l=uf.roots()
ll=len(l)
dp=[[0]*(w+1) for i in range(ll)]
ca,cb=0,0
for x in uf.members(l[0]):
if a[x]<=w:
dp[0][a[x]]=max(dp[0][a[x]],b[x])
ca+=a[x]
cb+=b[x]
if ca<=w:
dp[0][ca]=max(dp[0][ca],cb)
for i in range(1,ll):
ca,cb=0,0
for x in uf.members(l[i]):
for j in range(w+1):
if j-a[x]>=0:
dp[i][j]=max(dp[i][j],dp[i-1][j],dp[i-1][j-a[x]]+b[x])
ca+=a[x]
cb+=b[x]
for j in range(w+1):
if j-ca>=0:
dp[i][j]=max(dp[i-1][j],dp[i-1][j-ca]+cb)
ans=0
for j in range(w+1):
ans=max(ans,dp[ll-1][j])
print(ans)
```
No
| 87,162 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Just to remind, girls in Arpa's land are really nice.
Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hoses x and y are in the same friendship group if and only if there is a sequence of Hoses a1, a2, ..., ak such that ai and ai + 1 are friends for each 1 ≤ i < k, and a1 = x and ak = y.
<image>
Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at most w weight on it.
Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater than w and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceed w.
Input
The first line contains integers n, m and w (1 ≤ n ≤ 1000, <image>, 1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 1000) — the weights of the Hoses.
The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106) — the beauties of the Hoses.
The next m lines contain pairs of friends, the i-th of them contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), meaning that Hoses xi and yi are friends. Note that friendship is bidirectional. All pairs (xi, yi) are distinct.
Output
Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceed w.
Examples
Input
3 1 5
3 2 5
2 4 2
1 2
Output
6
Input
4 2 11
2 4 6 6
6 4 2 1
1 2
2 3
Output
7
Note
In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to 5 and sum of their beauty is 6.
In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is 12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be 8, and the total beauty will be 7.
Submitted Solution:
```
f = lambda: map(int, input().split())
n, m, w = f()
wb = [(0, 0)] + list(zip(f(), f()))
t = list(range(n + 1))
def g(x):
if x == t[x]: return x
t[x] = g(t[x])
return t[x]
for i in range(m):
x, y = f()
x, y = g(x), g(y)
if x != y: t[y] = x
p = [[] for j in range(n + 1)]
for i in range(1, n + 1): p[g(i)].append(i)
d = [1] + [0] * w
for q in p:
if len(q) < 2: continue
WB = [wb[i] for i in q]
W = sum(q[0] for q in WB)
B = sum(q[1] for q in WB)
for D in range(w - W, -1, -1):
if d[D]: d[D + W] = max(d[D + W], d[D] + B)
for W, B in WB:
for D in range(w - W, -1, -1):
if d[D]: d[D + W] = max(d[D + W], d[D] + B)
for q in p:
if len(q) != 1: continue
W, B = wb[q[0]]
for D in range(w - W, -1, -1):
if d[D]: d[D + W] = max(d[D + W], d[D] + B)
print(max(d) - 1)
```
No
| 87,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Tags: implementation, math
Correct Solution:
```
t = int(input())
if t % 2 == 1:
print("contest")
else:
print("home")
```
| 87,164 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Tags: implementation, math
Correct Solution:
```
n=int(input())
home=str(input())
k=n
l=[]
while(n):
l.append(str(input()))
n-=1
# print(l)
chk=l[-1][-3:]
# if(chk==home):
# print('home')
# else:
# print('contest')
if(k%2==0):
print("home")
else:
print("contest")
```
| 87,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Tags: implementation, math
Correct Solution:
```
print("home" if int(input()) % 2 == 0 else "contest")
```
| 87,166 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Tags: implementation, math
Correct Solution:
```
import sys
raw_data = list()
for line in sys.stdin:
raw_data.append(line)
nr = int(raw_data[0])
home = raw_data[1][0:3]
dept = list()
dest = list()
for idx in range(2, 2 + nr):
dept.append(raw_data[idx][0:3])
dest.append(raw_data[idx][5:8])
nxt = home
cnt = 0
while cnt != nr:
try:
idx = dept.index(nxt)
nxt = dest[idx]
last = nxt
except:
break
dept[idx] = ''
cnt += 1
if cnt == nr and last == home:
print('home')
else:
print('contest')
```
| 87,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Tags: implementation, math
Correct Solution:
```
n = int(input())
s = input()
c = 0
for i in range(n):
mass = input()
mass = mass.split("->")
if s in mass:
c+=1
if(not (c % 2)):
print("home")
else:
print("contest")
```
| 87,168 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Tags: implementation, math
Correct Solution:
```
#Rohan Bojja
#rohan.bojja@gmail.com
from collections import defaultdict
#######################################
c=1
cases=1
while(c<=cases):
n=int(input())
home=input()
iten=defaultdict(list)
for i in range(0,n):
cur=input().split("->")
iten[cur[0]].append(cur[1])
ans="home"
cur=home
while(True):
#print(cur,iten)
if(cur in iten):
if(cur!=home and home in iten[cur]):
del iten[cur][iten[cur].index(home)]
cur=home
elif(cur!=home):
ans="contest"
break
elif(cur==home and len(iten[cur])>0):
cur2=iten[cur][0]
iten[cur].pop(0)
cur=cur2
elif(cur==home):
break
else:
ans="contest"
break
#print(cur,iten)
print(ans)
c=c+1
```
| 87,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Tags: implementation, math
Correct Solution:
```
n = int(input())
a = input()
q = []
w = []
for i in range(n):
b = input()
l = b[:b.find('-')]
r = b[b.find('>')+1:]
q.append(l)
w.append(r)
e = q.count(a)
t = w.count(a)
if e == t:
print("home")
else:
print("contest")
```
| 87,170 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Tags: implementation, math
Correct Solution:
```
X = int(input())
Home = input()
counter = 0
for i in range(X):
h, c = list(input().split('->'))
if h == Home:
counter += 1
elif c == Home:
counter -= 1
print("home" if counter == 0 else "contest")
```
| 87,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Submitted Solution:
```
n=int(input())
h=input()
c1,c2=0,0
for i in range(n):
x,y=input().split('->')
if x==h:
c1=c1+1
elif y==h:
c2=c2+1
if c1==c2:
print('home')
else:
print('contest')
```
Yes
| 87,172 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Submitted Solution:
```
if __name__ == "__main__":
n = int(input())
home = input()
flights = []
while n > 0:
s, t = input().split("->")
flights.append(s)
flights.append(t)
n -= 1
count = 0
for flight in flights:
if flight == home:
count += 1
if count % 2 == 0:
print("home")
else:
print("contest")
```
Yes
| 87,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Submitted Solution:
```
#!/usr/bin/python3
n = int(input())
if n % 2 == 0:
print("home")
else:
print("contest")
```
Yes
| 87,174 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Submitted Solution:
```
if int(input()) & 1:
print('contest')
else:
print('home')
```
Yes
| 87,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Submitted Solution:
```
n = int(input())
home = input()
i = 0
lst = []
while i < n:
lst.append(input())
i += 1
last_fly = lst.count(lst[len(lst) - 1])
last = lst[len(lst) - 1][:3]
first = lst[len(lst) - 1][-3:]
cnt = lst.count(first + "->" + last)
if cnt == last_fly:
print("home")
else:
print("contest")
```
No
| 87,176 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Submitted Solution:
```
x=int(input())
for i in range(x+1):
y=input()
print(['home','conrest'][x%2])
```
No
| 87,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Submitted Solution:
```
x = int(input())
a = input()
t = []
k = 0
i = 0
t1 = 0
t3 = 1
for i in range(x):
a1 = input().split("->")
t.append(a1)
while i <= x - 1:
if t[i][0] == a:
k = 1
for j in range(len(t)):
if t[j][0] == t[i][1] and t[j][1] == t[i][0]:
t.pop(j)
i=i-1
t1 = 1
k = 0
break
elif i == x-1:
print("contest")
t3 = 2
k = 1
break
if k == 1:
break
if k == 1:
break
if t1 != 1:
i = i + 1
if k == 0:
print("home")
elif k == 1 and t3 == 1:
print("contest")
```
No
| 87,178 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
* this list contains all Jinotega's flights in this year (in arbitrary order),
* Jinotega has only flown from his hometown to a snooker contest and back,
* after each competition Jinotega flies back home (though they may attend a competition in one place several times),
* and finally, at the beginning of the year Jinotega was at home.
Please help them to determine Jinotega's location!
Input
In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.
Output
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
Examples
Input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
Output
home
Input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
Output
contest
Note
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
Submitted Solution:
```
n=int(input())
home=str(input())
k=n
l=[]
while(n):
l.append(str(input()))
n-=1
print(l)
chk=l[-1][-3:]
if(chk==home):
print('home')
else:
print('contest')
```
No
| 87,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
from collections import deque
n, k = map(int, input().split())
d = set(int(x)-n for x in input().split())
q = deque()
q.append(0)
visited = {i : False for i in range(-1000, 1001)}
dist = {i : 0 for i in range(-1000, 1001)}
ans = -1
visited[0] = True
found = False
while q:
u = q.popleft()
for i in d:
if i + u == 0:
ans = dist[u] + 1
found = True
break
if i + u <= 1000 and i + u >= -1000 and not visited[i + u]:
visited[i + u] = True
dist[i + u] = dist[u] + 1
q.append(i + u)
if found:
break
print (ans)
```
| 87,180 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
if n < a[0] or n > a[k - 1]:
print(-1)
else:
b = []
b.append(a[0])
for i in a:
if i != b[len(b) - 1]:
b.append(i)
d = {}
for i in range(len(b)):
b[i] -= n
d[b[i]] = 1
ans = 1
while 0 not in d:
d1 = {}
for i in d:
for j in b:
if -1001 < i + j < 1001:
d1[i + j] = 1
d = d1
ans += 1
print(ans)
```
| 87,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
##
##
##
import sys
def line():
return sys.stdin.readline()
def numbers():
return list(map(int, line().split()))
def number():
return int(line())
adjlist = {}
n, k = 0, 0
mark = [False]*2010
edges = [False]*1010
# bfs for "ssph"
def bfs(s):
i = 0
frontier = [s]
while frontier:
if mark[s]:
break;
next_frontier = []
for u in frontier:
# check next state
for v, isState in enumerate(edges):
if isState:
# check new node
state = u + (n - 1000) - v
if state >= 0 and state <= 2000 and not mark[state]:
mark[state] = True
next_frontier.append(state)
frontier = next_frontier
i += 1
if mark[s]:
return i
else:
return -1
# main program
[n, k] = numbers()
concentrations = numbers()
# reading edges
for x in concentrations:
edges[x] = True
n = n + 1000
ans = bfs(1000)
print(ans)
# 1496438704903
# Made By Mostafa_Khaled
```
| 87,182 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
from collections import deque
n, k = map(int, input().split())
conc = set(int(x) - n for x in input().split())
q = deque()
q.append(0)
visited = {i : False for i in range(-1000, 1001)}
dist = {i : 0 for i in range(-1000, 1001)}
ans = -1
visited[0] = True
found = False
while q:
u = q.popleft()
for c in conc:
v = c + u
if v == 0:
ans=dist[u]+1
found=True
break
if v<=1000 and v>=-1000 and not visited[v]:
visited[v]=True
dist[v]=dist[u]+1
q.append(v)
if found:
break
print(ans)
```
| 87,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
##
##
##
import sys
def line():
return sys.stdin.readline()
def numbers():
return list(map(int, line().split()))
def number():
return int(line())
adjlist = {}
n, k = 0, 0
mark = [False]*2010
edges = [False]*1010
# bfs for "ssph"
def bfs(s):
i = 0
frontier = [s]
while frontier:
if mark[s]:
break;
next_frontier = []
for u in frontier:
# check next state
for v, isState in enumerate(edges):
if isState:
# check new node
state = u + (n - 1000) - v
if state >= 0 and state <= 2000 and not mark[state]:
mark[state] = True
next_frontier.append(state)
frontier = next_frontier
i += 1
if mark[s]:
return i
else:
return -1
# main program
[n, k] = numbers()
concentrations = numbers()
# reading edges
for x in concentrations:
edges[x] = True
n = n + 1000
ans = bfs(1000)
print(ans)
# 1496438704903
```
| 87,184 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Tags: dfs and similar, graphs, shortest paths
Correct Solution:
```
from collections import deque
MAX_A = 1000
def main():
n, k = map(int, input().split())
a = set(int(x) - n for x in input().split())
visited = [False] * (2 * MAX_A + 1)
visited[n] = True
Q = deque()
Q.append((n, 0))
result = None
while Q:
u, l = Q.popleft()
l += 1
for ai in a:
v = u + ai
if v == n:
result = l
break
if 0 <= v < len(visited) and not visited[v]:
visited[v] = True
Q.append((v, l))
if result is not None:
break
if result is None:
result = -1
print(result)
if __name__ == '__main__':
# import sys
# sys.stdin = open("E.txt")
main()
```
| 87,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Submitted Solution:
```
from collections import deque
Max = 1000
def MainBFS():
n, k = map(int, input().split())
conc = set(int(x) for x in input().split())
visited = [False] * (2 * Max + 1)
visited[0] = True
Q = deque()
Q.append((0, 0))
result = None
while Q:
u, l = Q.popleft()
l += 1
for c in conc:
v = u + c
hayResult = False
if v / l == n:
hayResult = True
if result is None:
result = l
elif result is not None and l < result:
result = l
if v < len(visited) and not visited[v] and not hayResult:
visited[v] = True
Q.append((v, l))
if result is None:
result = -1
print(result)
if __name__ == '__main__':
MainBFS()
```
No
| 87,186 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Submitted Solution:
```
##
##
##
import sys
def line():
return sys.stdin.readline()
def numbers():
return list(map(int, line().split()))
def number():
return int(line())
adjlist = {}
n, k = 0, 0
mark = [False]*1010
edges = [False]*1010
# bfs for "ssph"
def bfs(s):
i = 0
frontier = [s]
while frontier:
if mark[s]:
break;
next_frontier = []
for u in frontier:
# check next state
for v, isState in enumerate(edges):
if isState:
# check new node
state = u + n - v
if state >= 0 and state <= n and not mark[state]:
mark[state] = True
next_frontier.append(state)
frontier = next_frontier
i += 1
if mark[s]:
return i
else:
return -1
# main program
[n, k] = numbers()
concentrations = numbers()
# reading edges
for x in concentrations:
edges[x] = True
ans = bfs(0)
print(ans)
# 1496437499797
```
No
| 87,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Submitted Solution:
```
n, k = map(int, input().split())
arr = list(map(int, input().split()))
possible = [1000000007]*1001
arr.sort()
for i in arr:
possible[i] = 1
for i in range(1, 1001):
for j in arr:
if(i-j < 1):
break
possible[i] = min(possible[i], 1+possible[i-j])
if(possible[n] == 1000000007):
print("-1")
for i in range(1, 1001):
if(i*n <= 1000 and i == possible[i*n]):
print(i)
exit(0)
print(-1)
```
No
| 87,188 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration <image>. Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration <image>. The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.
Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.
Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration <image>. Assume that the friends have unlimited amount of each Coke type.
Input
The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.
The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.
Output
Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration <image>, or -1 if it is impossible.
Examples
Input
400 4
100 300 450 500
Output
2
Input
50 2
100 25
Output
3
Note
In the first sample case, we can achieve concentration <image> using one liter of Coke of types <image> and <image>: <image>.
In the second case, we can achieve concentration <image> using two liters of <image> type and one liter of <image> type: <image>.
Submitted Solution:
```
from queue import Queue
ins = map(int, input().split())
n, k = next(ins), next(ins)
concentrations = list(map(int, input().split()))
q = Queue()
visited = {}
for i in range(len(concentrations)):
q.put((concentrations[i], 1))
visited[concentrations[i]] = True
done = False
while not q.empty():
top = q.get()
if top[0] == top[1] * n:
print(top[1])
done = True
break
else:
for val in concentrations:
new_val = val + top[0]
path = top[1] + 1
if new_val in visited or top[0] > 1000:
continue
visited[new_val] = True
q.put((new_val, path))
if not done:
print(-1)
```
No
| 87,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Tags: math, number theory
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
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")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
from math import factorial
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
from math import pow
# ------------------------------
def main():
for _ in range(N()):
a, b = RL()
mt = a*b
res = round(pow(mt, 1/3))
if res**3==mt and a%res==0 and b%res==0:
print('Yes')
else:
print('No')
if __name__ == "__main__":
main()
```
| 87,190 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Tags: math, number theory
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
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)
def input(): return sys.stdin.readline().rstrip("\r\n")
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def main():
for _ in range(int(input())):
# n=int(input())
# a=list(map(int, input().split()))
a,b=map(int, input().split())
k=a*b
f=0
# l=0
# r=k+1
# f=0
# while(l<=r):
# mid=(l+r)//2
# tr=mid*mid*mid
# if tr==k:
# f=mid
# break
# if tr>k:
# r=mid-1
# else:
# l=mid+1
root=round(k**(1/3))
if root*root*root==k:
f=root
if f:
if (a%f)==(b%f)==0:
print('YES')
else:
print('NO')
else:
print('NO')
return
if __name__=="__main__":
main()
```
| 87,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Tags: math, number theory
Correct Solution:
```
import os, sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
for _ in range(int(input())):
a, b = map(int, input().split())
q = a * b
t = round(q ** (1 / 3))
if t * t * t == q and a % t == b % t == 0:
print('YES')
else:
print('NO')
```
| 87,192 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Tags: math, number theory
Correct Solution:
```
import os
import sys
from math import ceil, pow
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def main():
for _ in range(int(input())):
a, b = map(int, input().split())
if (a * b == (ceil(pow(a * b, 1 / 3)) ** 3)):
temp = ceil(pow(a * b, 1 / 3))
if ((a % temp) == (b % temp) == 0):
print("Yes")
continue
print("No")
if __name__ == "__main__":
main()
```
| 87,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Tags: math, number theory
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------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
avl=AvlTree()
#-----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default='z', func=lambda a, b: min(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: 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 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------------------------------------
for i in range(int(input())):
a,b=map(int,input().split())
c=a*b
l=int(c**(1./3)+0.5)
if l**3==a*b and a%l==0 and b%l==0:
print("YES")
else:
print("NO")
```
| 87,194 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Tags: math, number theory
Correct Solution:
```
import sys
n = int(input())
ans = []
arr = sys.stdin.read().split()
d = {}
for i in range(1,1001):
d[i**3] = i
for i in range(n):
a, b = int(arr[i<<1]), int(arr[i<<1|1])
if a == b:
if a in d:
ans.append('Yes')
else:
ans.append('No')
continue
if a > b: a, b = b, a
x = d.get(a*a//b,-1)
if x == -1:
ans.append('No')
continue
if a % (x*x):
ans.append('No')
continue
y = a //(x*x)
if x * x * y == a and x * y * y == b: ans.append('Yes')
else: ans.append('No')
print('\n'.join(ans))
```
| 87,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Tags: math, number theory
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
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-----------------------------------------------------
for t in range (int(input())):
a,b=map(int,input().split())
p=a*b
#print(p)
c=int(round(p**(1./3)))
#print (c)
if c**3==p and a%c==0 and b%c==0:
print("Yes")
else:
print("No")
```
| 87,196 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Tags: math, number theory
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
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)
def input(): return sys.stdin.readline().rstrip("\r\n")
n = int(input())
output = []
for i in range(1, n + 1):
a, b = map(int, input().split())
p = a * b
root = round(p ** (1 / 3))
ok = root * root * root == p and a % root == b % root == 0
output.append('Yes' if ok else 'No')
print('\n'.join(output))
```
| 87,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def PrimeFactors(n):
factors=[]
while(n%2==0):
factors.append(2)
n//=2
for i in range(3,int(sqrt(n))+1,2):
while(n%i==0):
factors.append(i)
n//=i
if(n>1):
factors.append(n)
return factors
for _ in range(Int()):
a,b=value()
root=ceil((a*b)**(1/3))
# print(root)
ok="No"
if(root**3==a*b and a%root==0 and b%root==0):
ok="Yes"
print(ok)
```
Yes
| 87,198 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input
In the first string, the number of games n (1 ≤ n ≤ 350000) is given.
Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109) – the results of Slastyona and Pushok, correspondingly.
Output
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Example
Input
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000
Output
Yes
Yes
Yes
No
No
Yes
Note
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
Submitted Solution:
```
import sys
input = sys.stdin.readline
print = sys.stdout.write
cbrt = {i**3:i for i in range(1001)}
n = int(input())
all_res = []
for _ in range(n):
a, b = map(int, input().split())
if a == b:
all_res.append('Yes' if a in cbrt else 'No')
continue
if a > b:
a, b = b, a
r = cbrt.get(a * a // b, 0)
if r == 0 or a % (r * r) > 0:
all_res.append('No')
continue
y = a //(r * r)
if r * r * y == a and r * y * y == b:
all_res.append('Yes')
else:
all_res.append('No')
print('\n'.join(all_res))
```
Yes
| 87,199 |
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