text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n positive integers. Find a non-empty subset of its elements such that their sum is even (i.e. divisible by 2) or determine that there is no such subset.
Both the given array and required subset may contain equal values.
Input
The first line contains a single integer t (1 ≤ t ≤ 100), number of test cases to solve. Descriptions of t test cases follow.
A description of each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 100), length of array a.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100), elements of a. The given array a can contain equal values (duplicates).
Output
For each test case output -1 if there is no such subset of elements. Otherwise output positive integer k, number of elements in the required subset. Then output k distinct integers (1 ≤ p_i ≤ n), indexes of the chosen elements. If there are multiple solutions output any of them.
Example
Input
3
3
1 4 3
1
15
2
3 5
Output
1
2
-1
2
1 2
Note
There are three test cases in the example.
In the first test case, you can choose the subset consisting of only the second element. Its sum is 4 and it is even.
In the second test case, there is only one non-empty subset of elements consisting of the first element, however sum in it is odd, so there is no solution.
In the third test case, the subset consisting of all array's elements has even sum.
Tags: brute force, dp, greedy, implementation
Correct Solution:
```
# coding: utf-8
# Your code here!
# coding: utf-8
# Your code here!
q=int(input())
for _ in range(q):
N=int(input())
A=list(map(int,input().split()))
even=[]
odd=[]
for i in range(len(A)):
if A[i]%2==0:
even.append(i+1)
break
else:
odd.append(i+1)
if even:
print(1)
print(even[0])
else:
if len(odd)>1:
print(2)
print(odd[0],odd[1])
else:
print(-1)
```
| 95,400 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers. Find a non-empty subset of its elements such that their sum is even (i.e. divisible by 2) or determine that there is no such subset.
Both the given array and required subset may contain equal values.
Input
The first line contains a single integer t (1 ≤ t ≤ 100), number of test cases to solve. Descriptions of t test cases follow.
A description of each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 100), length of array a.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100), elements of a. The given array a can contain equal values (duplicates).
Output
For each test case output -1 if there is no such subset of elements. Otherwise output positive integer k, number of elements in the required subset. Then output k distinct integers (1 ≤ p_i ≤ n), indexes of the chosen elements. If there are multiple solutions output any of them.
Example
Input
3
3
1 4 3
1
15
2
3 5
Output
1
2
-1
2
1 2
Note
There are three test cases in the example.
In the first test case, you can choose the subset consisting of only the second element. Its sum is 4 and it is even.
In the second test case, there is only one non-empty subset of elements consisting of the first element, however sum in it is odd, so there is no solution.
In the third test case, the subset consisting of all array's elements has even sum.
Submitted Solution:
```
t = int(input())
def solve(a):
n = len(a)
odds=[]
for j in range(n):
if a[j]%2==0:
print(1)
print(j+1)
return
else:
odds.append(j)
if len(odds)==2:
print(2)
print(odds[0]+1, odds[1]+1)
return
print(-1)
return
for i in range(t):
n = int(input())
a = list(map(int, input().split()))
solve(a)
```
Yes
| 95,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers. Find a non-empty subset of its elements such that their sum is even (i.e. divisible by 2) or determine that there is no such subset.
Both the given array and required subset may contain equal values.
Input
The first line contains a single integer t (1 ≤ t ≤ 100), number of test cases to solve. Descriptions of t test cases follow.
A description of each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 100), length of array a.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100), elements of a. The given array a can contain equal values (duplicates).
Output
For each test case output -1 if there is no such subset of elements. Otherwise output positive integer k, number of elements in the required subset. Then output k distinct integers (1 ≤ p_i ≤ n), indexes of the chosen elements. If there are multiple solutions output any of them.
Example
Input
3
3
1 4 3
1
15
2
3 5
Output
1
2
-1
2
1 2
Note
There are three test cases in the example.
In the first test case, you can choose the subset consisting of only the second element. Its sum is 4 and it is even.
In the second test case, there is only one non-empty subset of elements consisting of the first element, however sum in it is odd, so there is no solution.
In the third test case, the subset consisting of all array's elements has even sum.
Submitted Solution:
```
import sys, os.path
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
test=int(input())
for _ in range(test):
n=int(input())
arr=input().split()
flag=1
if n>1:
for i in range(n):
if int(arr[i])%2==0 and flag==1:
print(1)
print(i+1)
flag=0
if flag!=0:
print(2)
print(1,end=" ")
print(2)
elif n==1:
if int(arr[0])%2==0:
print(1)
print(1)
else:
print(-1)
```
Yes
| 95,402 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers. Find a non-empty subset of its elements such that their sum is even (i.e. divisible by 2) or determine that there is no such subset.
Both the given array and required subset may contain equal values.
Input
The first line contains a single integer t (1 ≤ t ≤ 100), number of test cases to solve. Descriptions of t test cases follow.
A description of each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 100), length of array a.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100), elements of a. The given array a can contain equal values (duplicates).
Output
For each test case output -1 if there is no such subset of elements. Otherwise output positive integer k, number of elements in the required subset. Then output k distinct integers (1 ≤ p_i ≤ n), indexes of the chosen elements. If there are multiple solutions output any of them.
Example
Input
3
3
1 4 3
1
15
2
3 5
Output
1
2
-1
2
1 2
Note
There are three test cases in the example.
In the first test case, you can choose the subset consisting of only the second element. Its sum is 4 and it is even.
In the second test case, there is only one non-empty subset of elements consisting of the first element, however sum in it is odd, so there is no solution.
In the third test case, the subset consisting of all array's elements has even sum.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
for i in range(n):
if a[i] % 2 == 0:
print(1)
print(i + 1)
break
if i > 0:
print(2)
print(1, 2)
break
else:
print(-1)
```
Yes
| 95,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers. Find a non-empty subset of its elements such that their sum is even (i.e. divisible by 2) or determine that there is no such subset.
Both the given array and required subset may contain equal values.
Input
The first line contains a single integer t (1 ≤ t ≤ 100), number of test cases to solve. Descriptions of t test cases follow.
A description of each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 100), length of array a.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100), elements of a. The given array a can contain equal values (duplicates).
Output
For each test case output -1 if there is no such subset of elements. Otherwise output positive integer k, number of elements in the required subset. Then output k distinct integers (1 ≤ p_i ≤ n), indexes of the chosen elements. If there are multiple solutions output any of them.
Example
Input
3
3
1 4 3
1
15
2
3 5
Output
1
2
-1
2
1 2
Note
There are three test cases in the example.
In the first test case, you can choose the subset consisting of only the second element. Its sum is 4 and it is even.
In the second test case, there is only one non-empty subset of elements consisting of the first element, however sum in it is odd, so there is no solution.
In the third test case, the subset consisting of all array's elements has even sum.
Submitted Solution:
```
from sys import stdin, stdout
input = stdin.readline
print = stdout.write
def main():
t = int(input())
for _ in [0] * t:
n = int(input())
odd = []
even = []
a = list(map(int, input().strip().split()))
for i in range(n):
if a[i] % 2:
odd.append(i)
if len(odd) == 2:
break
else:
even.append(i)
break
if not even and len(odd) == 2:
print(str(2) + '\n')
for i in odd:
print(str(i + 1) + ' ')
elif even:
print(str(1) + '\n')
print(str(even[0] + 1))
else:
print(str(-1))
print('\n')
main()
```
Yes
| 95,404 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers. Find a non-empty subset of its elements such that their sum is even (i.e. divisible by 2) or determine that there is no such subset.
Both the given array and required subset may contain equal values.
Input
The first line contains a single integer t (1 ≤ t ≤ 100), number of test cases to solve. Descriptions of t test cases follow.
A description of each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 100), length of array a.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100), elements of a. The given array a can contain equal values (duplicates).
Output
For each test case output -1 if there is no such subset of elements. Otherwise output positive integer k, number of elements in the required subset. Then output k distinct integers (1 ≤ p_i ≤ n), indexes of the chosen elements. If there are multiple solutions output any of them.
Example
Input
3
3
1 4 3
1
15
2
3 5
Output
1
2
-1
2
1 2
Note
There are three test cases in the example.
In the first test case, you can choose the subset consisting of only the second element. Its sum is 4 and it is even.
In the second test case, there is only one non-empty subset of elements consisting of the first element, however sum in it is odd, so there is no solution.
In the third test case, the subset consisting of all array's elements has even sum.
Submitted Solution:
```
from sys import stdin,stdout
t=int(stdin.readline().strip())
for _ in range(t):
n=int(stdin.readline().strip())
a=list(map(int,stdin.readline().strip().split()))
f=0
for i in range(n):
if a[i]%2==0:
ans=i+1
stdout.write("1"+"\n")
stdout.write(str(ans)+"\n")
f=1
if f==0:
if n==1:
stdout.write("-1"+"\n")
else:
stdout.write("2"+"\n"+str(1)+" "+str(2)+"\n")
```
No
| 95,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers. Find a non-empty subset of its elements such that their sum is even (i.e. divisible by 2) or determine that there is no such subset.
Both the given array and required subset may contain equal values.
Input
The first line contains a single integer t (1 ≤ t ≤ 100), number of test cases to solve. Descriptions of t test cases follow.
A description of each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 100), length of array a.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100), elements of a. The given array a can contain equal values (duplicates).
Output
For each test case output -1 if there is no such subset of elements. Otherwise output positive integer k, number of elements in the required subset. Then output k distinct integers (1 ≤ p_i ≤ n), indexes of the chosen elements. If there are multiple solutions output any of them.
Example
Input
3
3
1 4 3
1
15
2
3 5
Output
1
2
-1
2
1 2
Note
There are three test cases in the example.
In the first test case, you can choose the subset consisting of only the second element. Its sum is 4 and it is even.
In the second test case, there is only one non-empty subset of elements consisting of the first element, however sum in it is odd, so there is no solution.
In the third test case, the subset consisting of all array's elements has even sum.
Submitted Solution:
```
def rs(): return input().rstrip()
def ri(): return int(rs())
def ra(): return list(map(int, rs().split()))
def is_odd(a):
return (a % 2) == 1
def is_even(a):
return (a % 2) == 0
t = ri()
for _ in range(t):
n = ri()
arr = ra()
if len(arr) == 1:
if is_odd(arr[0]):
print(-1)
else:
print("1\n1")
elif len(arr) > 1:
f = arr[0]
s = arr[1]
if is_even(f):
print("1\n1")
elif is_even(s):
print("1\n2")
else:
print(1)
print(1, 2)
```
No
| 95,406 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers. Find a non-empty subset of its elements such that their sum is even (i.e. divisible by 2) or determine that there is no such subset.
Both the given array and required subset may contain equal values.
Input
The first line contains a single integer t (1 ≤ t ≤ 100), number of test cases to solve. Descriptions of t test cases follow.
A description of each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 100), length of array a.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100), elements of a. The given array a can contain equal values (duplicates).
Output
For each test case output -1 if there is no such subset of elements. Otherwise output positive integer k, number of elements in the required subset. Then output k distinct integers (1 ≤ p_i ≤ n), indexes of the chosen elements. If there are multiple solutions output any of them.
Example
Input
3
3
1 4 3
1
15
2
3 5
Output
1
2
-1
2
1 2
Note
There are three test cases in the example.
In the first test case, you can choose the subset consisting of only the second element. Its sum is 4 and it is even.
In the second test case, there is only one non-empty subset of elements consisting of the first element, however sum in it is odd, so there is no solution.
In the third test case, the subset consisting of all array's elements has even sum.
Submitted Solution:
```
def rs(): return input().rstrip()
def ri(): return int(rs())
def ra(): return list(map(int, rs().split()))
def is_odd(a):
return (a % 2) == 1
def is_even(a):
return (a % 2) == 0
t = ri()
for _ in range(t):
n = ri()
arr = ra()
if len(arr) == 1:
if is_odd(arr[0]):
print(-1)
else:
print("1\n1")
elif len(arr) > 1:
f = arr[0]
s = arr[1]
if is_even(f):
print("1\n1")
elif is_even(s):
print("1\n2")
else:
print("1\n1 2")
```
No
| 95,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n positive integers. Find a non-empty subset of its elements such that their sum is even (i.e. divisible by 2) or determine that there is no such subset.
Both the given array and required subset may contain equal values.
Input
The first line contains a single integer t (1 ≤ t ≤ 100), number of test cases to solve. Descriptions of t test cases follow.
A description of each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 100), length of array a.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100), elements of a. The given array a can contain equal values (duplicates).
Output
For each test case output -1 if there is no such subset of elements. Otherwise output positive integer k, number of elements in the required subset. Then output k distinct integers (1 ≤ p_i ≤ n), indexes of the chosen elements. If there are multiple solutions output any of them.
Example
Input
3
3
1 4 3
1
15
2
3 5
Output
1
2
-1
2
1 2
Note
There are three test cases in the example.
In the first test case, you can choose the subset consisting of only the second element. Its sum is 4 and it is even.
In the second test case, there is only one non-empty subset of elements consisting of the first element, however sum in it is odd, so there is no solution.
In the third test case, the subset consisting of all array's elements has even sum.
Submitted Solution:
```
t=int(input())
for test in range(t):
n=int(input())
a = list(map(int,input().split()))
if (n==1) and (a[0]%2)!=0:
print(-1)
elif (n==1) and (a[0]%2)==0:
print(1)
print(a[0])
else:
flag=False
for i in range(n):
if a[i]%2==0:
print(1)
print(i+1)
flag=True
ee=0
aa,b,x,y=0,1,0,0
count=0
if not(flag):
for i in range(n):
if a[i]%2==1:
aa=a[i]
ai = i+1
break
flag2=False
for i in range(n):
if a[i]%2==1 and a[i]!=aa:
b=a[i]
bi = i+1
flag2=True
break
if flag2:
print(2)
print(ai,bi)
else:
print(-1)
```
No
| 95,408 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Tags: dp, strings
Correct Solution:
```
import os
### START FAST IO ###
os_input = os.read(0, int(1e7)).split()
os_input_pos = -1
answer_list = []
def read_s():
global os_input_pos
os_input_pos += 1
return os_input[os_input_pos].decode()
def read_i():
return int(read_s())
def write_s(v):
answer_list.append(v)
def write_i(v):
write_s(str(v))
def print_ans():
os.write(1, "\n".join(answer_list).encode())
os.write(1, "\n".encode())
#### END FAST IO ####
T = read_i()
while T:
T -= 1
n = read_i()
s = read_s()
t = read_s()
if sorted(s) != sorted(t):
write_i(-1)
continue
s_count = [[0 for i in range(n+1)] for j in range(26)]
t_count = [[0 for i in range(n+1)] for j in range(26)]
for i in range(n):
for j in range(26):
s_count[j][i] = s_count[j][i-1]
t_count[j][i] = t_count[j][i-1]
s_count[ord(s[i]) - ord('a')][i] += 1
t_count[ord(t[i]) - ord('a')][i] += 1
dp = [[0 for i in range(n+1)] for j in range(n+1)]
for i in range(0, n):
for j in range(i, n):
dp[i][j] = dp[i-1][j] + 1
if s[i] == t[j]:
dp[i][j] = min(dp[i][j], dp[i-1][j-1])
c = ord(t[j]) - ord('a')
if s_count[c][i] < t_count[c][j]:
dp[i][j] = min(dp[i][j], dp[i][j-1])
write_i(dp[n-1][n-1])
print_ans()
```
| 95,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Tags: dp, strings
Correct Solution:
```
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
inf=10**9
for _ in range(II()):
n=II()
s = SI()
t = SI()
cnts=[[0]*26 for _ in range(n+1)]
cntt=[[0]*26 for _ in range(n+1)]
for i in range(n-1,-1,-1):
cnts[i][ord(s[i])-97]+=1
cntt[i][ord(t[i])-97]+=1
for j in range(26):
cnts[i][j]+=cnts[i+1][j]
cntt[i][j]+=cntt[i+1][j]
ng=False
for j in range(26):
if cnts[0][j]!=cntt[0][j]:
ng=True
break
if ng:
print(-1)
continue
dp=[[inf]*(n+1) for _ in range(n+1)]
for i in range(n+1):dp[i][0]=0
for i in range(n+1):
for j in range(i,n+1):
if i==n and j==n:break
pre=dp[i][j]
if pre==inf:continue
if i<j and i+1<=n:dp[i+1][j]=min(dp[i+1][j],pre+1)
if i+1<=n and j+1<=n and s[i]==t[j]:dp[i+1][j+1]=min(dp[i+1][j+1],pre)
if j+1<=n and cnts[i+1][ord(t[j])-97]>cntt[j+1][ord(t[j])-97]:
dp[i][j + 1] = min(dp[i][j + 1], pre)
#p2D(dp)
print(dp[n][n])
main()
```
| 95,410 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Tags: dp, strings
Correct Solution:
```
def num(c):return ord(c) - 97
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input());s1 = input().strip();s2 = input().strip();char1 = [0] * 26;char2 = [0] * 26
for c in s1:char1[num(c)] += 1
for c in s2:char2[num(c)] += 1
if char1 != char2:print(-1);continue
dp = [[(False, 0, 0) for j in range(n+1)] for i in range(n + 1)];dp[0][0] = [True, 0,[0]*26]
def upd(a, b, val, sett):
if not dp[a][b][0] or val > dp[a][b][1]:dp[a][b] = (True, val, sett)
for i in range(n):
for j in range(n):
valid, val, tab = dp[i][j]
if not valid:continue
top = s1[i];bot = s2[j]
if top == bot:
if not dp[i + 1][j + 1][0] or val + 1 > dp[i + 1][j + 1][1]:dp[i + 1][j + 1] = [True, val + 1, tab]
if tab[num(top)] > 0:
sett = tab[:];sett[num(top)] -= 1
if not dp[i + 1][j][0] or val > dp[i + 1][j][1]:dp[i + 1][j] = [True, val, sett]
sett = tab[:];sett[num(bot)] += 1
if not dp[i][j + 1][0] or val > dp[i][j + 1][1]:dp[i][j + 1] = [True, val, sett]
del dp[i][j][2]
poss = [dp[i][n][1] for i in range(n + 1)]
print(n - max(poss))
```
| 95,411 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Tags: dp, strings
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
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())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]+=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
#print(flag)
return flag
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
@bootstrap
def gdfs(r,p):
if len(g[r])==1 and p!=-1:
yield None
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
t=N()
for i in range(t):
n=N()
s=input()
k=input()
ssuf=[[] for i in range(n+1)]
ksuf=[[] for i in range(n+1)]
ssuf[-1]=[0]*26
ksuf[-1]=[0]*26
for i in range(n-1,-1,-1):
ssuf[i]=ssuf[i+1].copy()
ssuf[i][ord(s[i])-97]+=1
ksuf[i]=ksuf[i+1].copy()
ksuf[i][ord(k[i])-97]+=1
if ssuf[0]!=ksuf[0]:
print(-1)
continue
dp=[[inf]*(n+1) for i in range(n+1)]
for j in range(n+1):
dp[0][j]=0
for i in range(1,n+1):
for j in range(i,n+1):
if s[i-1]==k[j-1]:
dp[i][j]=dp[i-1][j-1]
else:
dp[i][j]=dp[i-1][j]+1
cur=ord(k[j-1])-97
if i<j and ssuf[i][cur]-ksuf[j][cur]>0:
dp[i][j]=min(dp[i][j],dp[i][j-1])
print(dp[n][n])
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
```
| 95,412 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Tags: dp, strings
Correct Solution:
```
def num(c):
return ord(c) - 97
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
s1 = input().strip()
s2 = input().strip()
char1 = [0] * 26
char2 = [0] * 26
for c in s1:
char1[num(c)] += 1
for c in s2:
char2[num(c)] += 1
if char1 != char2:
print(-1)
continue
dp = [[(False, 0, 0) for j in range(n+1)] for i in range(n + 1)]
dp[0][0] = [True, 0,[0]*26]
def upd(a, b, val, sett):
if not dp[a][b][0] or val > dp[a][b][1]:
dp[a][b] = (True, val, sett)
for i in range(n):
for j in range(n):
valid, val, tab = dp[i][j]
if not valid:
continue
top = s1[i]
bot = s2[j]
if top == bot:
#upd(i+1, j+1, val + 1, tab)
if not dp[i + 1][j + 1][0] or val + 1 > dp[i + 1][j + 1][1]:
dp[i + 1][j + 1] = [True, val + 1, tab]
if tab[num(top)] > 0:
sett = tab[:]
sett[num(top)] -= 1
#upd(i+1, j, val, sett)
if not dp[i + 1][j][0] or val > dp[i + 1][j][1]:
dp[i + 1][j] = [True, val, sett]
sett = tab[:]
sett[num(bot)] += 1
#upd(i, j + 1, val, sett)
if not dp[i][j + 1][0] or val > dp[i][j + 1][1]:
dp[i][j + 1] = [True, val, sett]
del dp[i][j][2]
poss = [dp[i][n][1] for i in range(n + 1)]
print(n - max(poss))
```
| 95,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Tags: dp, strings
Correct Solution:
```
def read_int():
return int(input())
def read_ints():
return map(int, input().split(' '))
t = read_int()
for case_num in range(t):
n = read_int()
cnt = [0 for i in range(26)]
ps = [[0 for j in range(n + 1)] for i in range(26)]
pt = [[0 for j in range(n + 1)] for i in range(26)]
s = input()
t = input()
for i in range(n):
ch = ord(s[i]) - ord('a')
cnt[ch] += 1
for j in range(26):
ps[j][i + 1] = ps[j][i] + (1 if ch == j else 0)
for i in range(n):
ch = ord(t[i]) - ord('a')
cnt[ch] -= 1
for j in range(26):
pt[j][i + 1] = pt[j][i] + (1 if ch == j else 0)
ok = True
for i in cnt:
if i != 0:
ok = False
break
if not ok:
print(-1)
else:
r = n
while r >= 1 and s[r - 1] == t[r - 1]:
r -= 1
inf = int(1e8)
dp = [[0 if i == 0 else inf for j in range(
r + 1)] for i in range(r + 1)]
for i in range(1, r + 1):
for j in range(i, r + 1):
if s[i - 1] == t[j - 1]:
dp[i][j] = min(dp[i][j], dp[i - 1][j - 1])
ch = ord(t[j - 1]) - ord('a')
if ps[ch][r] - ps[ch][i] > pt[ch][r] - pt[ch][j]:
dp[i][j] = min(dp[i][j], dp[i][j - 1])
dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1)
print(dp[r][r])
```
| 95,414 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Tags: dp, strings
Correct Solution:
```
INF = float('inf')
def solve():
N = int(input())
A = input().strip()
B = input().strip()
Acount = [[0] * 26 for _ in range(N + 1)]
Bcount = [[0] * 26 for _ in range(N + 1)]
for i in range(N - 1, -1, -1):
Acount[i][ord(A[i]) - ord('a')] += 1
Bcount[i][ord(B[i]) - ord('a')] += 1
for j in range(26):
Acount[i][j] += Acount[i + 1][j]
Bcount[i][j] += Bcount[i + 1][j]
for i in range(26):
if Acount[0][i] != Bcount[0][i]:
return -1
dp = [[INF] * (N + 1) for _ in range(N + 1)]
for j in range(N + 1):
dp[0][j] = 0
for i in range(1, N + 1):
s = ord(A[i - 1]) - ord('a')
for j in range(i, N + 1):
dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1)
t = ord(B[j - 1]) - ord('a')
if Acount[i][t] > Bcount[j][t]:
dp[i][j] = min(dp[i][j], dp[i][j - 1])
if s == t:
dp[i][j] = min(dp[i][j], dp[i - 1][j - 1])
# for row in dp:
# print(row)
return dp[N][N]
if __name__ == '__main__':
T = int(input())
for _ in range(T):
print(solve())
```
| 95,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Submitted Solution:
```
def rotated_compare(a, b):
d1, d2 = {}, {}
# getting all characters in a string and setting the count to number of occurrences
count = 1
for i in a:
if i not in d1.keys():
d1[i] = count
else:
d1[i] = count + 1
count = 1
for i in b:
if i not in d2.keys():
d2[i] = count
else:
d2[i] = count + 1
# comparing for different characters in a and b
lst_element = d1.keys()
flag = 0
for j in b:
if (j not in lst_element) or (j in lst_element and d1[j] != d2[j]):
flag = 1
break
if flag == 0:
if a == b:
return 0
len_a = len(a) - 1
for i in range(len(a)):
a = a[0:i] + a[len_a] + a[i:len_a]
if a == b:
return i+1
else:
return -1
def main():
num_of_tests = int(input())
lst_ans = []
for i in range(num_of_tests):
len = int(input())
str1 = input()
str2 = input()
lst_ans.append(rotated_compare(str1, str2))
for each in lst_ans:
print(each)
if __name__ == '__main__':
main()
```
No
| 95,416 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Submitted Solution:
```
def rotiation(a,b):
if (a=="abc" and b=="bca") or (a=="abb" and b=="bba"):
return 2
if len(a) != len(b):
return -1
d = list(a)
start=0
res=0
for word in a:
i=d.index(word)
if d==list(b):
break
x=b.find(word)
if x==-1 or b.count(word) != a.count(word):
return -1
if x==i:
continue
res +=1
start2,start=x,i
i +=1
x +=1
if (x>=len(d) or i>=len(d)):
d[start:i] = ""
# d[x - i:x + i - 1] = b[start2:x+1]
d[x :x + i - 1] = b[start2:x+1]
continue
while d[i]==b[x]:
i += 1
x += 1
if (x >= len(d) or i >= len(d)):
break
d[start:i] = ""
for tt in range(start2,x):
d.insert(tt,b[tt])
return res
n=int(input())
while n>0:
size=input()
a=input()
b=input()
res=rotiation(a,b)
print(res)
n -=1
```
No
| 95,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Submitted Solution:
```
def rotatestr(string, d):
n = len(string)
return string[n-d:] + string[0: n - d]
def rotating_substring(str1, str2, idx, count, memo):
n = len(str1)
if (idx, count) in memo:
return memo[(idx, count)]
if len(str2) == 0 or idx == n:
memo[(idx, count)] = count
return count
rotated = rotatestr(str2, 1)
unrotated = -1
withrotation = -1
if str1[idx] == rotated[0]:
withrotation = rotating_substring(str1, rotated, idx + 1, count + 1, memo)
if str1[idx] == str2[idx]:
unrotated = rotating_substring(str1, str2, idx + 1, count, memo)
result = -1
if unrotated == -1:
result = withrotation
elif withrotation == -1:
result = unrotated
else:
result = min(unrotated, withrotation)
memo[(idx, count)] = result
return result
def string_rotate():
t = int(input())
for i in range(t):
n = int(input())
str1 = input()
str2 = input()
memo = {}
ret1 = rotating_substring(str1, str2, 0, 0, memo)
memo = {}
ret2 = rotating_substring(str2, str1, 0, 0, memo)
if ret1 == -1:
print(ret2)
elif ret2 == -1:
print(ret1)
else:
result = min(ret1, ret2)
print(result)
string_rotate()
```
No
| 95,418 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.
You can perform the following operation on s any number of times to achieve it —
* Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string "abcde" becomes "adbce".
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Find the minimum number of operations required to convert s to t, or determine that it's impossible.
Input
The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings.
The second and the third lines contain strings s and t respectively.
The sum of n over all the test cases does not exceed 2000.
Output
For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.
Example
Input
6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc
Output
0
1
1
2
1
-1
Note
For the 1-st test case, since s and t are equal, you don't need to apply any operation.
For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.
For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.
For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.
For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.
For the 6-th test case, it is not possible to convert string s to t.
Submitted Solution:
```
testcases=int(input())
for i in range(0,testcases):
length=int(input())
s=input().strip()
t=input().strip()
if(s==t):
print(0)
elif(s!=t and length>1):
for j in range(0,length):
s=s[length-1]+s[0:length-1]
if(s==t):
print(j+1)
else:
print(-1)
break
else:
print(-1)
```
No
| 95,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: brute force, dp, greedy
Correct Solution:
```
import math
######################################################
# ps template
def mi(): return map(int, input().split())
def ii(): return int(input())
def li(): return list(map(int, input().split()))
def si(): return input().split()
#######################################################
t = ii()
for _ in range(t):
n, k, l = mi()
a = li()
dp = [ 0 for i in range(n)]
ans = True
rev = False
if a[0]+k<=l:
dp[0] = 'F'
elif a[0]<=l:
dp[0] = l - a[0]
if dp[0] > 0:
rev = True
else:
rev = False
else:
dp[0] = -1
for i in range(1,n):
if dp[i-1]==-1 or a[i]>l:
ans = False
break
if a[i]+k<=l:
dp[i] = 'F'
elif dp[i-1]=='F':
dp[i] = l - a[i]
if dp[i]>0:
rev = True
else:
rev = False
elif rev:
if l-a[i]<=dp[i-1]-1:
dp[i] = l - a[i]
if dp[i]>0:
rev = True
else:
rev = False
else:
dp[i] = dp[i-1] - 1
if dp[i]>0:
rev = True
else:
rev = False
elif dp[i-1]+1+a[i]<=l:
dp[i] = dp[i-1]+1
if dp[i] == k:
rev= True
else:
ans = False
break
#print(dp)
if ans:
print("Yes")
else:
print('No')
```
| 95,420 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: brute force, dp, greedy
Correct Solution:
```
for _ in range(int(input())):
n,k,l=map(int,input().split())
depth=list(map(int,input().split()))
fg=1;mn=-k;mx=k
for i in depth:
if i>l:fg=0;break
mx=min(l-i,k)
if mx==k:mn=-k
else:mn=max(mn+1,-mx)
if mn>mx:fg=0;break
if fg:print("Yes")
else:print("No")
```
| 95,421 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: brute force, dp, greedy
Correct Solution:
```
import sys
readline = sys.stdin.readline
def solve():
N, K, L = map(int, readline().split())
D = list(map(int, readline().split()))
if any(d > L for d in D):
print('NO')
return
k = K + D[0] - L
for i, d in enumerate(D):
if d + K <= L:
if i == len(D) - 1:
break
k = K + D[i + 1] - L
else:
_k = k % (2 * K)
depth = d + (K - _k if _k <= K else _k - K)
if depth > L:
if k < K:
k = K + d - L
else:
print('NO')
return
k += 1
print('YES')
return
T = int(readline())
for i in range(T):
solve()
```
| 95,422 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: brute force, dp, greedy
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_right
from math import gcd,log
from collections import Counter
def main():
n,k,l=map(int,input().split())
arr=list(map(int,input().split()))
brr=list(range(0,k+1))+list(range(k-1,0,-1))
arr.insert(0,-1)
# print(arr)
# print(brr)
dp=[[0]*(2*k) for i in range(n+2)]
dp[0]=[1]*(2*k)
for i in range(1,n+1):
for j in range(2*k):
if arr[i]+brr[j]<=l:
dp[i][j]|=dp[i-1][(j-1)%(2*k)]
for j in range(2*k):
if arr[i]+brr[(j)%(2*k)]<=l:
dp[i][(j)%(2*k)]|=dp[i][(j-1)%(2*k)]
# print(dp)
if any(dp[n]):
print('Yes')
else:
print('No')
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")
# endregion
if __name__ == "__main__":
for _ in range(int(input())):
main()
```
| 95,423 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: brute force, dp, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n, k, l = map(int, input().split())
d = list(map(int, input().split()))
if max(d) > l:
print('No')
else:
p = [i for i in range(1, k)] + [k] + [i for i in range(k-1, 0, -1)]
groups = []
limit = l - k
g = []
for i in range(n):
if d[i] <= limit:
if g:
groups.append(g)
g = []
else:
g.append(d[i]-limit)
if i == n-1:
groups.append(g)
ans = True
for g in groups:
n_g = len(g)
if n_g <= len(p):
if any(all(p[j:(n_g+j)][i] >= g[i] for i in range(n_g)) for j in range(len(p)-n_g+1)):
pass
else:
ans = False
else:
ans = False
if ans:
print('Yes')
else:
print('No')
```
| 95,424 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: brute force, dp, greedy
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
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())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def binfun(x):
c=0
for w in arr:
c+=ceil(w/x)
return c
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [(ch,len(list(g))) for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
t=N()
for i in range(t):
n,k,l=RLL()
d=RLL()
d=[l-x for x in d]
if any(x<0 for x in d):
ans=False
else:
#print(d)
res=[]
cur=[]
for x in d:
if x>=k:
if cur:
res.append(cur)
cur=[]
else:
cur.append(x)
if cur:
res.append(cur)
flag=True
#print(res)
for a in res:
last=0
for x in a:
low=k-x
upp=k+x
cur=max(last+1,low)
if cur>upp:
flag=False
break
last=cur
if not flag:
break
ans=flag
print("YES" if ans else "NO")
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
```
| 95,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: brute force, dp, greedy
Correct Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
from functools import *
from fractions import Fraction as f
from copy import *
from bisect import *
from heapq import *
from math import *
from itertools import permutations
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
#sys.setrecursionlimit(10**6)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
def inc(d,c):
d[c]=d[c]+1 if c in d else 1
def li():
return [int(xx) for xx in input().split()]
def fli():
return [float(x) for x in input().split()]
def comp(a,b):
if(a>b):
return 2
return 2 if a==b else 0
def gi():
return [xx for x in input().split()]
def fi():
return int(input())
def pro(a):
return reduce(lambda a,b:a*b,a)
def swap(a,i,j):
a[i],a[j]=a[j],a[i]
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def graph(n,m):
for i in range(m):
x,y=mi()
a[x].append(y)
a[y].append(x)
def bo(i):
return ord(i)-ord('a')
t=fi()
def solve(i,time):
if i==n+1:
dp[i][time%(2*k)]=True
return True
if dp[i][time%(2*k)]!=-1:
return dp[i][time%(2*k)]
#print(i,d[i],p[time%(2*k)])
if d[i]+p[time%(2*k)]>l:
dp[i][time%(2*k)]=False
return False
for j in range(1,2*k+1):
if solve(i+1,time+j):
dp[i][time%(2*k)]=True
return True
if i!=0 and d[i]+p[(time+j)%(2*k)]>l:
dp[i][time%(2*k)]=False
return False
#print(i,time)
dp[i][time%(2*k)]=False
return False
while t>0:
t-=1
n,k,l=mi()
d=li()
p=[]
d.insert(0,0)
dp=[[-1 for i in range(2*k+1)] for j in range(n+2)]
for i in range(k+1):
p.append(i)
for i in range(k-1,-1,-1):
p.append(i)
print("Yes" if solve(0,0) else "No")
```
| 95,426 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: brute force, dp, greedy
Correct Solution:
```
# input = open('file.txt').readline
for _ in range( int(input()) ):
n , k , l = map( int , input().strip().split(" ") )
arr = list(map( int , input().strip().split(" ") ))
goods = []
bad = False
for i , a in enumerate( arr ):
if a + k <= l:
goods.append(i)
if a > l:
bad = True
break
if bad:
print('No')
continue
goods.append(n)
prev = -1
for g in goods:
st = prev
en = g
# print(st , en)
if st + 1 == en:
prev = g
continue
tk = k
while st < en-1 and tk > 0:
st += 1
tk -= 1
plc = arr[st] + tk
if plc > l:
tk -= ( plc - l )
if tk < 0:
bad = True
# print(st, en , tk , 'after')
if tk == 0:
while st < en-1:
st += 1
tk += 1
plc = arr[st] + tk
# print('inside',st,tk,plc)
if plc > l:
bad = True
break
if bad:
break
prev = g
if bad:
print('No')
else:
print('Yes')
```
| 95,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
from sys import stdin
def check(a,b):
global d,k,l
arr = [d[a+1:b]]
for _ in range(k):
arr.append(list(map(lambda x:x+1,arr[-1])))
for _ in range(k-1):
arr.append(list(map(lambda x:x-1,arr[-1])))
i=0
while i != len(arr):
j=fl=0
while j != len(arr[0]):
if arr[(i+j)%len(arr)][j] > l:
fl = 1
break
j += 1
if fl:
break
i += 1
if not fl:
return 1
return 0
for _ in range(int(stdin.readline())):
n,k,l = map(int,stdin.readline().split())
d = list(map(int,stdin.readline().split()))
safe = [-1]
for i in range(n):
if d[i]+k <= l:
safe.append(i)
safe.append(n)
ans = 'YES'
for i in range(1,len(safe)):
if not check(safe[i-1],safe[i]):
ans = 'NO'
break
print(ans)
```
Yes
| 95,428 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
import sys
input=sys.stdin.buffer.readline
def getAdditionalDepth(t):
t2=t%(2*k)
if t2>k:
t2=2*k-t2
return t2
t=int(input())
for _ in range(t):
n,k,l=[int(x) for x in input().split()]
d=[int(x) for x in input().split()]
ok=True
for i in range(n):
d[i]-=l #as long as d[i]<=0, Koa can stay at i
if d[i]>0:
print('No')
ok=False
break
if ok==False:
continue
# d.append(-float('inf'))
# print(d)##########
t=k #at peak d, decreasing
for i in range(-1,n-1):
if t>=2*k:
t-=2*k
if d[i+1]+k<=0:#set t back to peak
t=k
continue
nextDepth=d[i+1]+getAdditionalDepth(t+1)
# print('i:{} t:{} nextDepth:{}'.format(i,t,nextDepth))############
# print('i:{} t:{} nextDepth:{}'.format(i,t+nextDepth,d[i]+getAdditionalDepth(t+nextDepth)))############
if nextDepth<=0: #ok
t+=1
else: #wait for earliest opportunity to move on
if t<k: #must wait until past high-tide. Koa would drown
ok=False
break
else:
t+=nextDepth+1
if ok:
print('Yes')
else:
print('No')
```
Yes
| 95,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
for _ in range(int(input())):
n, k, l = map(int, input().split());dd = list(map(int, input().split()));t,curld,di = 0,0,False
for d in [-(2**30), *dd]:
if d > l:break
ld = l - d
if d + k <= l:curld,di = ld,False;continue
if di:
if ld >= curld + 1:curld += 1
else:break
else:
curld = min(curld - 1, ld)
if curld == 0:di = True
else:print('Yes');continue
print('No')
```
Yes
| 95,430 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
for _ in range(int(input())):
n,k,l=map(int,input().split())
di=list(map(int,input().split()))
safe=[0 for _ in range(n)]
p=[_ for _ in range(k+1)]
for _ in range(k-1,0,-1):p.append(_)
for i in range(n):
if di[i]>l:print("NO");break
if di[i]+k<=l:safe[i]=1
else:
i=0
safety=safe[0]
time=len(p)-(l-di[0])
while i<n-1:
height=di[i]+p[time%(2*k)]
if height>l:print("NO");break
if not safe[i]:
if di[i+1]+p[(time+1)%(2*k)]<=l:
i+=1
time+=1
else:
time+=1
else:
if safe[i]==1:
time=k
safe[i]=2
if di[i+1]+p[(time+1)%(2*k)]<=l:
i+=1
time+=1
else:time+=1
else:
print("YES")
```
Yes
| 95,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def riverCross(arr, n, k, l):
to = None
jump = 0
reverse = -1
ans = None
# increasingState = True
if n == 1:
if arr[0] > l:
ans = 'No'
return 'No'
else:
ans = 'Yes'
return 'Yes'
else:
for i in range(len(arr) - 1):
minimum = arr[i]
maximum = arr[i]+k
if maximum <= l:
nextMax = arr[i + 1] + k
if arr[i + 1] > l:
# print('No')
ans = 'No'
return 'No'
elif nextMax <= l:
jump += 1
to = arr[i + 1] + k
else:
to = nextMax
# while to > l:
# to = to + reverse
# if to == arr[i+1]:
# reverse = 1
# elif to == arr[i + 1] + k:
# reverse = -1
if to > l:
to = l
if to == arr[i+1]:
reverse = 1
elif to == arr[i + 1] + k:
reverse = -1
jump += 1
# reverse = -1
elif minimum > l:
# print('No')
ans = 'No'
return 'No'
# break
else:
nextMax = arr[i+1] + k
if arr[i+1] > l:
# print('No')
ans = 'No'
return 'No'
# break
elif nextMax <= l:
to = arr[i + 1] + k
if to == arr[i+1]:
reverse = 1
elif to == arr[i + 1] + k:
reverse = -1
jump += 1
else:
if to == None:
to = arr[i] + k - 1
# while to > l:
# to -= 1
if to > l:
to = l
# if to == arr[i+1]:
# reverse = 1
# elif to == arr[i + 1] + k:
# reverse = -1
if reverse == -1:
gapTime = (to - arr[i]) * 2 + 1
else:
gapTime = arr[i] + k - to
# if increasingState == True:
to = arr[i + 1] + k - maximum + to
if to == arr[i+1]:
reverse = 1
elif to == arr[i+1]+k:
reverse = -1
if reverse == -1:
to -= 1
else:
to += 1
# print(to, 'to', gapTime)
# while to > l:
# gapTime -= 1
# if gapTime == 0 and to > l:
# # print('No')
# ans = 'No'
# return 'No'
# # break
# to = to + reverse
# if to == arr[i + 1]:
# reverse = 1
if to > l:
if to - gapTime > l:
ans = 'No'
return 'No'
gapTime -= (to -l)
if gapTime <= 0:
ans = 'No'
return 'No'
to = l
if to == arr[i + 1]:
reverse = 1
elif to == arr[i + 1] + k:
reverse = -1
jump += 1
# reverse = -1
# print(to, jump)
# print(jump)
if to == arr[i+1]:
reverse = 1
elif to == arr[i+1]+ k:
reverse = -1
if jump == n - 1:
if to <= l:
return 'Yes'
else:
return 'No'
def solution():
values = list(map(lambda x: int(x), input().split()))
n = values[0]
k = values[1]
l = values[2]
arr = list(map(lambda x: int(x), input().split()))
print(riverCross(arr, n, k, l))
t = int(input());
for i in range(0, t):
solution()
```
No
| 95,432 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
import math
######################################################
# ps template
def mi(): return map(int, input().split())
def ii(): return int(input())
def li(): return list(map(int, input().split()))
def si(): return input().split()
#######################################################
t = ii()
for _ in range(t):
n, k, l = mi()
a = li()
dp = [ 0 for i in range(n)]
ans = True
rev = False
if a[0]+k<=l:
dp[0] = 'F'
elif a[0]<=l:
dp[0] = l - a[0]
if dp[0] > 0:
rev = True
else:
rev = False
for i in range(1,n):
if dp[i-1]==-1 or a[i]>l:
ans = False
break
if a[i]+k<=l:
dp[i] = 'F'
elif dp[i-1]=='F':
dp[i] = l - a[i]
if dp[i]>0:
rev = True
else:
rev = False
elif rev:
if l-a[i]<=dp[i-1]-1:
dp[i] = l - a[i]
if dp[i]>0:
rev = True
else:
rev = False
else:
dp[i] = dp[i-1] - 1
if dp[i]>0:
rev = True
else:
rev = False
elif dp[i-1]+1+a[i]<=l:
dp[i] = dp[i-1]+1
if dp[i] == k:
rev= True
else:
ans = False
break
#print(dp)
if ans:
print("Yes")
else:
print('No')
```
No
| 95,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
import sys
import math
from collections import defaultdict
import heapq
t=int(sys.stdin.readline())
for _ in range(t):
n,k,l=map(int,sys.stdin.readline().split())
d=list(map(int,sys.stdin.readline().split()))
p=[]
# for i in range(k+1):
# p.append(i)
# for i in range(k-1,0,-1):
# p.append(i)
# z=True
# for i in range(2*k):
# st=i
# m=True
# for j in range(n):
# st+=1
# curd=d[j]+p[st%(2*k)]
# if curd>l:
# print(curd,'curd',l,'l',j,'j')
# m=False
# break
# if m:
# z=False
# break
z=True
dic=defaultdict(list)
arr=[]
for i in range(n):
arr.append(l-d[i])
if arr[i]<0:
z=False
if not z:
print("NO")
continue
z=True
last=arr[0]
cnt=1
arr1=[]
for i in range(n):
cnt=0
for j in range(i,n):
if arr[j]<=arr[i]:
cnt+=1
else:
break
arr1.append(cnt)
for i in range(n):
if arr[i]>=k:
continue
else:
if arr1[i]>arr[i]+1:
z=False
break
# print(arr1,'arr1')
# print(arr,'arr')
# for i in range(1,n):
# if arr[i]==last:
# cnt+=1
# else:
# dic[last].append(cnt)
# cnt=1
# last=arr[i]
# dic[last].append(cnt)
# print(dic,'dic')
# for x in dic:
# if x>=k:
# continue
# for y in dic[x]:
# if y>x+1:
# z=False
# break
# if not z:
# break
if not z:
print("NO")
else:
print("YES")
```
No
| 95,434 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
def dfs (G, node, depths, l, V):
V.add (node)
pos = node[0]
if pos == len(depths): return
wave = node[1]
depth = depths[pos] + wave
if depth > l:
return
for v in G[node]:
if v not in V:
dfs (G, v, depths, l, V)
def main():
t = int(input())
for _ in range (t):
n,k,l = map(int, input().split())
D = [0] + list(map(int, input().split()))
G = dict()
for pos in range (0, n+1):
for k in range (k+1):
G[(pos, k, 1)] = list()
G[(pos, k, -1)] = list()
start = (0,0,1)
dest = (n+1,0,0)
G[start] = list()
G[dest] = list()
for node in G.keys():
if node != dest:
pos = node[0]
wave = node[1]
sign = node[2]
if sign == 1:
if wave == k:
newSign = -1
newWave = k-1
else:
newSign = 1
newWave = wave+1
else:
if wave == 0:
newSign = 1
newWave = 1
else:
newSign = -1
newWave = wave-1
G[node].append ((pos, newWave, newSign))
if pos<n:
G[node].append ((pos+1, newWave, newSign))
else:
G[node].append ((pos+1, 0, 0))
# print (G)
visited = set()
dfs (G, start, D, l, visited)
if dest in visited:
print ("Yes")
else:
print ("No")
# 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 = lambda s: self.buffer.write(s.encode()) if self.writable else None
def read(self):
if self.buffer.tell():
return self.buffer.read().decode("ascii")
return os.read(self._fd, os.fstat(self._fd).st_size).decode("ascii")
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().decode("ascii")
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
def print(*args, sep=" ", end="\n", file=sys.stdout, flush=False):
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(end)
if flush:
file.flush()
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
sys.setrecursionlimit(int(1e5))
if __name__ == "__main__":
main()
```
No
| 95,435 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are lower. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 100; 1 ≤ k ≤ 100; 1 ≤ l ≤ 100) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 100) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 100.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
pr=stdout.write
import heapq
raw_input = stdin.readline
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return (map(int,stdin.read().split()))
range = xrange # not for python 3.0+
for t in range(ni()):
n,k,l=li()
arr=li()
p=[ i for i in range(k+1)]
p = p+[i for i in range(k-1,0,-1)]
f=0
for st in range(2*k):
f1=0
prev=st
t1=0
pos=0
while pos<n-1:
if not arr[pos]+p[(prev+t1)%(2*k)]<=l:
f1=1
break
if t1>2*k:
f1=1
break
if arr[pos+1]+p[(prev+t1+1)%(2*k)]<=l:
prev=prev+t1+1
t1=0
pos+=1
continue
t1+=1
if not f1:
f=1
break
if f:
pr('Yes\n')
else:
pr('No\n')
```
No
| 95,436 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Tags: implementation, strings
Correct Solution:
```
gans = []
for _ in range(int(input())):
n, k = map(int, input().split())
u = list(input())
if u == ['?'] * n:
gans.append('YES')
continue
cnt1 = cnt0 = 0
for j in range(k):
ok1 = False
ok0 = False
for i in range(j, n, k):
if u[i] == '0':
ok0 = True
elif u[i] == '1':
ok1 = True
if ok1 and ok0:
gans.append('NO')
break
if ok1:
cnt1 += 1
elif ok0:
cnt0 += 1
else:
if abs(cnt1 - cnt0) <= k - cnt1 - cnt0:
gans.append('YES')
else:
gans.append('NO')
print('\n'.join(gans))
```
| 95,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Tags: implementation, strings
Correct Solution:
```
# cook your dish here
def checker(A,k):
tot=0
o,z=0,0
for i in range(k):
# tot+=A[i]
if A[i]==1:
o+=1
elif A[i]==-1:
z+=1
# if tot!=0:
# return False
if z>k//2 or o>k//2:
return False
for j in range(k,len(A)):
if A[j-k]==1:
o-=1
elif A[j-k]==-1:
z-=1
if A[j]==1:
o+=1
elif A[j]==-1:
z+=1
if z>k//2 or o>k//2:
return False
# tot-=A[j-k]
# tot+=A[j]
# if tot!=0:
# return False
return True
def func(A,k):
if k==len(A):
o,z=0,0
for i in A:
if i=='0':
z+=1
elif i=='1':
o+=1
if o>k//2 or z>k//2:
return 'NO'
return 'YES'
res=[0]*len(A)
one,zero,ques=[],[],[]
for i in range(k):
new=set()
ind=[]
for j in range(i,len(A),k):
new.add(A[j])
ind.append(j)
if '0' in new and '1' in new:
return "NO"
elif '0' in new:
zero.extend(ind)
elif '1' in new:
one.extend(ind)
else:
ques.extend(ind)
# print(one, zero, ques)
for i in one:
res[i]=1
for i in zero:
res[i]=-1
# res2=res.copy()
for i in ques:
res[i]=0
# res2[i]=0
# print(res, res2)
# print(res,res2)
if checker(res,k):
return "YES"
return "NO"
t=int(input())
for i in range(t):
# n=int(input())
n,k=list(map(int, input().split()))
A=input()
# for i in A[::-1]:
# print(i,end=' ')
print(func(A,k))
```
| 95,438 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
n,k = map(int,input().split())
s = list(str(input()))
tt = 0
c = 0
a = 0
aa = 0
bb = 0
for i in range(n):
if i+k<n and s[i+k] == "?":
if s[i]!="?":
s[i+k] = s[i]
if i-k>=0 and s[i-k] == "?":
s[i-k] = s[i]
if s[i] == "?":
if i+k<n and s[i+k]!="?":
s[i] = s[i+k]
elif i-k>=0 and s[i-k] != "?":
s[i] = s[i-k]
for i in range(n):
if s[i] == "0":
c+=1
if i+k<n and s[i+k] == "1":
tt = 1
break
if c+a == k and c!=a:
tt = 1
break
if c+a == k and c == a:
c = 0
a = 0
elif s[i] == "1":
bb+=1
a+=1
if i+k<n and s[i+k] == "0":
tt = 1
break
if c+a == k and c!=a:
tt = 1
break
if i>=k-1:
if c>k//2 or a>k//2:
tt = 1
break
else:
if s[aa] == "0":
c-=1
elif s[aa] == "1":
a-=1
aa+=1
if tt == 1:
print("NO")
else:
print("YES")
```
| 95,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Tags: implementation, strings
Correct Solution:
```
#from math import *
from bisect import *
from collections import *
from decimal import *
def inp():
return int(input())
def st():
return input().rstrip('\n')
def lis():
return list(map(int,input().split()))
def ma():
return map(int,input().split())
t=inp()
while(t):
t-=1
n,k=ma()
s=st()
s=list(s)
fl=0
for i in range(k,n):
if(s[i]!='?'):
if(s[i%k]=='?'):
s[i%k]=s[i]
elif(s[i%k]!=s[i]):
fl=1
break
o,z=0,0
for i in range(k):
if(s[i]=='1'):
o+=1
elif(s[i]=='0'):
z+=1
if(o> k//2 or z > k//2):
fl=1
s='YES'
if(fl):
s='NO'
print(s)
```
| 95,440 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Tags: implementation, strings
Correct Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineering College
'''
from os import path
import sys
from heapq import heappush,heappop
from functools import cmp_to_key as ctk
from collections import deque,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input().rstrip()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
mod=1000000007
# mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('a')
file = 1
def solve():
n,k = mi()
s = list(si())
for i in range(n):
if s[i] != '?' and s[i%k] == '?':
s[i%k] = s[i]
for i in range(n):
if s[i] != '?' and s[i%k] != '?' and s[i] != s[i%k]:
print('NO')
return
k1 = k//2
cnt0,cnt1 = 0 ,0
for i in range(k):
if s[i]=='1':
cnt1+=1
if s[i]=='0':
cnt0+=1
if cnt1>k1 or cnt0>k1:
print('NO')
return
print('YES')
if __name__ =="__main__":
if(file):
if path.exists('input.txt'):
sys.stdin=open('input.txt', 'r')
sys.stdout=open('output.txt','w')
else:
input=sys.stdin.readline
for _ in range(ii()):
solve()
```
| 95,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Tags: implementation, strings
Correct Solution:
```
"""
Satwik_Tiwari ;) .
6th Sept , 2020 - Sunday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned
#==============================================================================================
#fast I/O region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#===============================================================================================
# code here ;))
def solve(case):
n,k = sep()
s = list(inp())
f = True
for kk in range(k):
arr= []
if(not f):
break
for i in range(kk,n,k):
arr.append(s[i])
# print(arr)
if(arr.count('?') == len(arr)):
continue
else:
if(arr.count('0')>0 and arr.count('1')>0):
f = False
break
else:
if(arr.count('0')>0):
chng = '0'
else:
chng = '1'
for i in range(kk,n,k):
s[i] = chng
# print(s)
if(f):
sum = 0
cnt = 0
for i in range(k):
if(s[i] == '?'):
cnt+=1
else:
sum+=int(s[i])
# print(sum,cnt)
if(sum>(k//2) or (sum+cnt<(k//2))):
print('NO')
else:
print('YES')
else:
print('NO')
# testcase(1)
testcase(int(inp()))
```
| 95,442 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Tags: implementation, strings
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 print_list(l):
print(' '.join(map(str,l)))
# import sys
# sys.setrecursionlimit(5010)
# import heapq as hq
# from collections import deque as dq
# from math import ceil,floor,sqrt,pow
# import bisect as bs
from collections import Counter
from collections import defaultdict as dc
for _ in range(N()):
n,k = RL()
s = input()
dic = dc(set)
for i in range(n):
dic[i%k].add(s[i])
flag = True
one,zero = 0,0
for i in range(k):
if '1' in dic[i] and '0' in dic[i]:
flag = False
break
elif '1' in dic[i]:
dic[i] = '1'
one+=1
elif '0' in dic[i]:
dic[i] = '0'
zero+=1
if not flag:
print('NO')
else:
if one>(k>>1) or zero>(k>>1):
print('NO')
else:
print('YES')
```
| 95,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Tags: implementation, strings
Correct Solution:
```
def cktrue(n,k,s1):
G={'1':0,'0':0,'?':0}
for j in range(k):
i=s1[j]
G[i]=G[i]+1
if (G['0']>k//2 or G['1']>k//2):
return 0
for i in range(k,n):
G[s1[i-k]]=G[s1[i-k]]-1
G[s1[i]]=G[s1[i]]+1
if (G['0']>k//2 or G['1']>k//2):
return 0
return 1
def fna(n,k,s):
flag=0
s1=[]
nz,no,nq=0,0,0
for j in range(k):
i=s[j]
s1.append(i)
if (i=='1'):
no=no+1
elif(i=='0'):
nz=nz+1
else:
nq=nq+1
if (no>k//2 or nz>k//2):
return 0
for i in range(k,n):
if (s[i]=='?'):
s1.append(s1[i-k])
elif(s1[i-k]=='?'):
s1[i-k]=s[i]
s1.append(s[i])
else:
if (s[i]!=s1[i-k]):
return 0
s1.append(s1[i-k])
return cktrue(n,k,s1)
T=int(input())
for _ in range(T):
n,k=map(int,input().split())
s=list(input())
flag=(fna(n,k,s))
if(flag==0):
print("NO")
else:
print("YES")
```
| 95,444 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
for t in range(int(input())):
n,m=map(int,input().split())
a=input()
a='0'+a
flag=0
visit=["?" for i in range(m)]
for i in range(1,n+1):
if(visit[i%m]=='?'):
visit[i%m]=a[i]
elif(visit[i%m]=='0'):
if(a[i]=='1'):
flag=-1
break
elif(visit[i%m]=='1'):
if(a[i]=='0'):
flag=-1
break
add,count=0,0
if(flag==-1):
print("NO")
continue
for i in range(m):
if(visit[i]=='1'):
add+=1
elif(visit[i]=="0"):
count+=1
if(add<=m//2 and count<=m//2):
print("YES")
else:
print("NO")
```
Yes
| 95,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
import sys
import math
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
t = II()
for q in range(t):
n,k = MI()
s = list(SI())
boo = True
su = 0
for i in range(n):
if i-k>=0 and s[i]!=s[i-k]:
if s[i] == "?":
s[i] = s[i-k]
elif s[i-k]!="?":
boo = False
break
if i+k<n and s[i]!=s[i+k]:
if s[i+k] == "?":
s[i+k] = s[i]
elif s[i]!="?":
boo = False
break
o = s[:k].count("1")
z = s[:k].count("0")
if k//2-o<0 or k//2-z<0:
boo = False
for i in range(k,n):
if boo == False:
break
if s[i-k] == "1":
o-=1
elif s[i-k] == "0":
z-=1
if s[i] == "1":
o+=1
elif s[i] == "0":
z+=1
if k//2-o<0 or k//2-z<0:
boo = False
print("YES" if boo else "NO")
#7 4
#1?0??1?
```
Yes
| 95,446 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
# -*- coding: utf-8 -*-
# import bisect
# import heapq
# import math
# import random
# from collections import Counter, defaultdict, deque
# from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
# from fractions import Fraction
# from functools import lru_cache, reduce
# from itertools import combinations, combinations_with_replacement, product, permutations, accumulate
# from operator import add, mul, sub, itemgetter, attrgetter
import sys
# sys.setrecursionlimit(10**6)
# readline = sys.stdin.buffer.readline
readline = sys.stdin.readline
INF = 2**62-1
def read_int():
return int(readline())
def read_int_n():
return list(map(int, readline().split()))
def read_float():
return float(readline())
def read_float_n():
return list(map(float, readline().split()))
def read_str():
return readline().strip()
def read_str_n():
return readline().strip().split()
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.perf_counter()
ret = f(*args, **kwargs)
e = time.perf_counter()
error_print(e - s, 'sec')
return ret
return wrap
from collections import Counter, defaultdict
from itertools import product
def ref(N, K, S):
S = [c for c in S]
n = Counter(S)['?']
iq = [i for i in range(N) if S[i] == '?']
for j in product('01', repeat=n):
for k, v in zip(iq, j):
S[k] = v
ok = True
cc = Counter(S[:K])
if cc['0'] > K//2 or cc['1'] > K//2:
ok = False
for i in range(N-K):
a = S[i]
b = S[i+K]
if a != b:
ok = False
if ok:
return 'YES'
return 'NO'
# @mt
def slv(N, K, S):
sk = defaultdict(Counter)
for i in range(N):
sk[i%K][S[i]] += 1
c = Counter()
for i in range(K):
if '0' in sk[i] and '1' in sk[i]:
return 'NO'
if '0' in sk[i]:
c['0'] += 1
elif '1' in sk[i]:
c['1'] += 1
if c['0'] > K//2 or c['1'] > K//2:
return 'NO'
return 'YES'
def main():
for _ in range(read_int()):
N, K = read_int_n()
S = read_str()
print(slv(N, K, S))
# import random
# for _ in range(1000):
# N = 10
# K = random.randint(1, 3) * 2
# S = random.choices('01?', k=N)
# a = ref(N, K, S)
# b = slv(N, K, S)
# if a != b:
# print(N, K)
# print(S)
# print(a, b)
# break
if __name__ == '__main__':
main()
```
Yes
| 95,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
# @uthor : Kaleab Asfaw
from sys import stdin, stdout
# Fast IO
def input():
a = stdin.readline()
if a[-1] == "\n": a = a[:-1]
return a
def print(*argv, end="\n", sep=" "):
n = len(argv)
for i in range(n):
if i == n-1: stdout.write(str(argv[i]))
else: stdout.write(str(argv[i]) + sep)
stdout.write(end)
# Others
mod = 10**9+7
def lcm(x, y): return (x * y) / (gcd(x, y))
def comb(lst, x): return list(c(lst, x))
def fact(x, mod=mod):
ans = 1
for i in range(1, x+1): ans = (ans * i) % mod
return ans
def arr2D(n, m, default=0):
lst = []
for i in range(n): temp = [default] * m; lst.append(temp)
return lst
def sortDictV(x): return {k: v for k, v in sorted(x.items(), key = lambda item : item[1])}
def smaller(lst, x): return bisect_left(lst, x) -1
def smallerEq(lst, x): return bisect_right(lst, x) -1
def solve(n, k, lst):
val = {}
for i in range(n):
if lst[i] != "?":
if val.get(i%k) == None: val[i%k] = lst[i]
elif lst[i] != val[i%k]: return "NO"
# print(lst)
# print(val)
for i in range(n):
if val.get(i):
if lst[i] == "?":
lst[i] = val[i%k]
elif lst[i] != val[i%k]: return "aNO"
lstS = "".join(lst[:k])
q = lstS.count("?")
o = lstS.count("1")
z = lstS.count("0")
if abs(o-z) > q: return "NO"
for i in range(1, n-k):
if lst[i-1] == "?": q -= 1
elif lst[i-1] == "1": o -= 1
else: z -= 1
if lst[i+k-1] == "?": q += 1
elif lst[i+k-1] == "1": o += 1
else: z += 1
if abs(o-z) > q: return "cNO"
return "YES"
for _ in range(int(input())): # Multicase
n, k = list(map(int, input().split()))
lst = list(input())
print(solve(n, k, lst))
```
Yes
| 95,448 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n,k=map(int,input().split())
s=input()
s=list(s)
s_=s[0:k]
count1=0
count0=0
countQ=0
for i in range(k):
if s[i]=='1':
count1+=1
elif s[i]=='0':
count0+=1
else:
countQ+=1
if count1<=(k//2) and count0<=(k//2):
i=0
while count1!=int(k/2):
if s_[i]=='?':
s_[i]='1'
count1+=1
i+=1
while count0!=int(k/2):
if s_[i]=='?':
s_[i]='0'
count0+=1
i+=1
else:
print("NO")
continue
f=0
for i in range(k,n):
if s[i]==s_[0]:
s_.remove(s_[0])
s_.append(s[i])
continue
elif s[i]=='?' or s_[0]=='?':
if s[i]=='?':
s[i]=s_[0]
s_.remove(s_[0])
s_.append(s[i])
else:
f=1
print("NO")
break
if f==0:
print("YES")
```
No
| 95,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
from sys import stdin
###############################################################
def iinput(): return int(stdin.readline())
def minput(): return map(int, stdin.readline().split())
def linput(): return list(map(int, stdin.readline().split()))
###############################################################
from math import ceil
t = iinput()
while t:
t -= 1
n, k = minput()
s = list(input())
ans = 'YES'
if n == k:
if s.count('1') > n//2 or s.count('0') > n//2:
ans = 'NO'
elif all(e == '?' for e in s):
ans = 'YES'
else:
zeros = s[:k].count('0')
ones = s[:k].count('1')
for i in range(1, n-k+1):
if zeros > k//2 or ones > k//2:
ans = 'NO'
break
if s[i-1] == '0': zeros -= 1
elif s[i-1] == '1': ones -= 1
if s[i+k-1] == '0': zeros += 1
elif s[i+k-1] == '1': ones += 1
if zeros > k // 2 or ones > k // 2:
ans = 'NO'
if ans == 'YES':
# if s.count('?') < ceil(n/2):
for i in range(k):
if s[i] != '?':
for j in range(i+k, n, k):
if s[j] == '?':
s[j] = s[i]
elif s[j] != s[i]:
ans = 'NO'
break
for j in range(i-k, -1, -k):
if s[j] == '?':
s[j] = s[i]
elif s[j] != s[i]:
ans = 'NO'
break
if ans == 'NO':
break
# print(s)
zeros = s[:k].count('0')
ones = s[:k].count('1')
for i in range(1, n - k + 1):
if zeros > k // 2 or ones > k // 2:
ans = 'NO'
break
if s[i - 1] == '0':
zeros -= 1
elif s[i - 1] == '1':
ones -= 1
if s[i + k - 1] == '0':
zeros += 1
elif s[i + k - 1] == '1':
ones += 1
if zeros > k // 2 or ones > k // 2:
ans = 'NO'
print(ans)
```
No
| 95,450 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
flag, zero, one = True, 0, 0
for i in range(k):
letter = None
for j in range(i,n,k):
if s[i] != '?':
if letter and s[j] != letter:
flag = False
break
letter = s[j]
if letter:
zero += 1 if letter == '0' else 0
one += 1 if letter == '1' else 0
if max(zero, one) > k // 2:
flag = False
print("YES" if flag else "NO")
```
No
| 95,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
from sys import stdin
from collections import Counter
N = int(stdin.readline())
for case in range(N):
length, k = map(int, stdin.readline().split()) # 1 count and 0 count equal k/2
idx = 0
build = ""
string = str(stdin.readline())
flag = True
count = Counter(string)
onecount = count.get("1", 0)
zerocount = count.get("0", 0)
for i in range(k):
if string[i] == "?":
if i+k <= length-1:
if string[i+k] != "?":
build += string[i+k]
if string[i+k] == "1":
onecount += 1
else:
zerocount += 1
else:
if onecount >= zerocount:
build += "0"
zerocount += 1
else:
build += "1"
onecount += 1
elif onecount >= zerocount:
build += "0"
zerocount += 1
else:
build += "1"
onecount += 1
else:
build += string[i]
if onecount != k//2:
print("NO")
continue
for i in range(k, length):
if string[i] == "?":
build += build[i-k]
else:
build += string[i]
if build[i] != build[i-k]:
print("NO")
flag = False
break
if flag:
print("YES")
```
No
| 95,452 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Tags: brute force, constructive algorithms, flows, greedy, math
Correct Solution:
```
import sys
n=int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
mx=min(a[0],b[1])+min(a[1],b[2])+min(a[2],b[0])
if(a[0]>b[0]+b[2]):
mn=a[0]-b[0]-b[2]
elif(a[1]>b[0]+b[1]):
mn=a[1]-b[0]-b[1]
elif(a[2]>b[1]+b[2]):
mn=a[2]-b[1]-b[2]
else:
mn=0
print(mn,mx)
```
| 95,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Tags: brute force, constructive algorithms, flows, greedy, math
Correct Solution:
```
n = int(input())
a1,a3,a2 = map(int,input().split())
b1,b3,b2 = map(int,input().split())
mx = min(a1,b3) + min(a2,b1) + min(a3,b2)
mi = max(0,a1+b3-n , a2+b1-n , a3+b2-n)
print(mi,mx)
```
| 95,454 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Tags: brute force, constructive algorithms, flows, greedy, math
Correct Solution:
```
#1426E
from itertools import permutations
import sys
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
maxa=min(a[0],b[1])+min(a[1],b[2])+min(a[2],b[0])
combi=((0,0),(1,1),(2,2),(0,2),(1,0),(2,1))
mina=sys.maxsize
for i in permutations((combi)):
a1=a[:];b1=b[:]
for j in i:
x=min(a1[j[0]],b1[j[1]])
a1[j[0]]-=x
b1[j[1]]-=x
mina=min(mina,min(a1[0],b1[1])+min(a1[1],b1[2])+min(a1[2],b1[0]))
print("%d %d" %(mina, maxa))
```
| 95,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Tags: brute force, constructive algorithms, flows, greedy, math
Correct Solution:
```
n = int(input())
r,s,p = map(int,input().split())
a,b,c = map(int,input().split())
maxi = min(r,b)+min(s,c)+min(a,p)
mini = max(0,r+b-n)+max(0,s+c-n)+max(0,p+a-n)
print(mini,maxi)
```
| 95,456 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Tags: brute force, constructive algorithms, flows, greedy, math
Correct Solution:
```
def maxV(first, second):
return str(min(first[0], second[1]) + min(first[1], second[2]) + min(first[2], second[0]))
def minV(first, second):
vitPedra = first[0] - (second[0] + second[2])
vitTesoura = first[1] - (second[0] + second[1])
vitPapel = first[2] - (second[1] + second[2])
return str(max(vitPedra, 0) + max(vitTesoura, 0) + max(vitPapel, 0))
n = int(input())
alice = [int(k) for k in input().split()]
bob = [int(k) for k in input().split()]
print(minV(alice, bob), maxV(alice, bob))
```
| 95,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Tags: brute force, constructive algorithms, flows, greedy, math
Correct Solution:
```
import sys
def main():
#n = iinput()
#k = iinput()
#m = iinput()
n = int(sys.stdin.readline().strip())
#n, k = rinput()
#n, m = rinput()
#m, k = rinput()
#n, k, m = rinput()
#n, m, k = rinput()
#k, n, m = rinput()
#k, m, n = rinput()
#m, k, n = rinput()
#m, n, k = rinput()
#n, t = map(int, sys.stdin.readline().split())
#q = list(map(int, sys.stdin.readline().split()))
#q = linput()
a,a1,a2= map(int, sys.stdin.readline().split())
b,b1,b2= map(int, sys.stdin.readline().split())
res = min(a, b1) + min(a1, b2) + min(a2, b)
if b >= a1 + a:
ans = a2 - (n - b)
elif b1 >= a1 + a2:
ans = a - (n - b1)
elif b2 >= a2 + a:
ans = a1 - (n - b2)
elif (b == a and b1 == a1 and b2 == a2):
ans = 0
elif a > b and a1 > b1:
b2 -= a2
a -= b2
a1 -= b1
if a <= 0:
ans = a1 - b
else:
ans = a + a1 - b
elif a2 > b2 and a1 > b1:
b -= a
a1 -= b
a2 -= b2
if a1 <= 0:
ans = a2 - b1
else:
ans = a2 + a1 - b1
elif a > b and a2 > b2:
b1 -= a1
a2 -= b1
a -= b
if a2 <= 0:
ans = a - b
else:
ans = a2 + a - b
elif a > b:
a -= b
b1 -= a1
a2 -= b1
a1 = 0
ans = a + a2 - b2
elif a1 > b1:
a1 -= b1
b2 -= a2
a -= b2
a2 = 0
ans = a + a1 - b
elif a2 > b2:
a2 -= b2
b -= a
a1 -= b
a = 0
ans = a2 + a1 - b1
print(max(0, ans), res)
for i in range(1):
main()
```
| 95,458 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Tags: brute force, constructive algorithms, flows, greedy, math
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
mini=0
maxi=min(a[0],b[1])+min(a[1],b[2])+min(a[2],b[0])
if a[0]>(b[0]+b[2]):
mini=a[0]-b[2]-b[0]
elif a[1]>(b[1]+b[0]):
mini=a[1]-b[1]-b[0]
elif a[2]>(b[2]+b[1]):
mini=a[2]-b[2]-b[1]
else:
mini=0
print(mini,maxi)
```
| 95,459 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Tags: brute force, constructive algorithms, flows, greedy, math
Correct Solution:
```
import itertools
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
M = min(a[0], b[1]) + min(a[1], b[2]) + min(a[2], b[0])
order = []
order.append((0, 0))
order.append((0, 2))
order.append((1, 1))
order.append((1, 0))
order.append((2, 2))
order.append((2, 1))
orders = itertools.permutations(order)
# ors = [ordr for ordr in orders]
# print(len(ors))
m = float("inf")
for ordr in orders:
ac = a[:]
bc = b[:]
# print(ac, bc)
for i in range(6):
cnt = min(ac[ordr[i][0]], bc[ordr[i][1]])
ac[ordr[i][0]] -= cnt
bc[ordr[i][1]] -= cnt
cand = min(ac[0], bc[1]) + min(ac[1], bc[2]) + min(ac[2], bc[0])
m = min(m, cand)
print(m, M)
```
| 95,460 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Submitted Solution:
```
import sys
import math
from collections import defaultdict,Counter
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdout=open("CP1/output.txt",'w')
# sys.stdin=open("CP1/input.txt",'r')
# mod=pow(10,9)+7
n=int(input())
a1,a2,a3=map(int,input().split())
b1,b2,b3=map(int,input().split())
ma=min(a1,b2)+min(a2,b3)+min(a3,b1)
s1=min(b1,a2)
s2=min(b2,a3)
s3=min(b3,a1)
mi=s1+s2+s3
mi=min(b1,a2)+min(b2,a3)+min(b3,a1)
mi+=min(a1,b1-s1)+min(a2,b2-s2)+min(a3,b3-s3)
# print(n-mi)
# ans=str(n-mi)+' '+str(ma)
print(n-mi,ma)
```
Yes
| 95,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Submitted 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(RL())
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)
def toord(c): return ord(c)-ord('a')
def lcm(a, b): return a*b//gcd(a, b)
mod = 998244353
INF = float('inf')
from math import factorial, sqrt, ceil, floor, gcd
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
# ------------------------------
# f = open('./input.txt')
# sys.stdin = f
def main():
n = N()
arra = RLL()
arrb = RLL()
ma = min(arrb[0], arra[2]) + min(arrb[1], arra[0]) + min(arrb[2], arra[1])
mi = n - (min(arra[0], arrb[0] + arrb[2]) + min(arra[1], arrb[1] + arrb[0]) + min(arra[2], arrb[2] + arrb[1]))
print(mi, ma)
if __name__ == "__main__":
main()
```
Yes
| 95,462 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Submitted Solution:
```
import sys,math
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
n = inp()
a = inpl()
b = inpl()
sua = sum(a)
sub = sum(b)
mi_res = 0
if a[0] > b[0]+b[2]: mi_res = a[0]-b[0]-b[2]
if a[1] > b[0]+b[1]: mi_res = a[1]-b[0]-b[1]
if a[2] > b[2]+b[1]: mi_res = a[2]-b[2]-b[1]
ma_res = min(a[0],b[1]) + min(a[1],b[2]) + min(a[2],b[0])
print(mi_res,ma_res)
```
Yes
| 95,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Submitted Solution:
```
# Coder : Hakesh D #
import sys
#input=sys.stdin.readline
from collections import deque
from math import ceil,sqrt,gcd,factorial
from bisect import bisect_right,bisect_left
mod = 1000000007
INF = 10**18
NINF = -INF
def INT():return int(input())
def MAP():return map(int,input().split())
def LIST():return list(map(int,input().split()))
def modi(x):return pow(x,mod-2,mod)
def lcm(x,y):return (x*y)//gcd(x,y)
def write(l):
for i in l:
print(i,end=' ')
print()
########################################################################################
n = int(input())
r,s,p=MAP()
rr,ss,pp=MAP()
maxa = min(r,ss) + min(s,pp) + min(p,rr)
mini = min(r,rr+pp) + min(s,ss + rr) + min(p,pp + ss)
print(n - mini,maxa)
```
Yes
| 95,464 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Submitted Solution:
```
n = int(input())
a1, a2, a3 = map(int, input().split())
b1, b2, b3 = map(int, input().split())
x1, x2, x3 = a1, a2, a3
y1, y2, y3 = b1, b2, b3
d1, d2 = 0, 0
for i in range(n):
if a1 != 0 and b2 != 0:
d2 += 1
a1 -= 1
b2 -= 1
if a2 != 0 and b3 != 0:
d2 += 1
a2 -= 1
b3 -= 1
if a3 != 0 and b1 != 0:
d2 += 1
a3 -= 1
b1 -= 1
a1, a2, a3 = x1, x2, x3
b1, b2, b3 = y1, y2, y3
del x1, x2, x3, y1, y2, y3
for i in range(n):
if max(a1,a2,a3) == a1:
while a1 != 0 and b3 != 0:
a1 -= 1
b3 -= 1
elif max(a1,a2,a3) == a2:
while a2 != 0 and b1 != 0:
a2 -= 1
b1 -= 1
elif max(a1,a2,a3) == a3:
while a3 != 0 and b2 != 0:
a3 -= 1
b2 -= 1
for i in range(n):
if max(a1,a2,a3) == a1:
while a1 != 0 and b1 != 0:
a1 -= 1
b1 -= 1
elif max(a1,a2,a3) == a2:
while a2 != 0 and b2 != 0:
a2 -= 1
b2 -= 1
elif max(a1,a2,a3) == a3:
while a3 != 0 and b3 != 0:
a3 -= 1
b3 -= 1
for i in range(n):
if a1 != 0 and b2 != 0:
d1 += 1
a1 -= 1
b2 -= 1
if a2 != 0 and b3 != 0:
d1 += 1
a2 -= 1
b3 -= 1
if a3 != 0 and b1 != 0:
d1 += 1
a3 -= 1
b1 -= 1
print(d1,d2)
```
No
| 95,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Submitted Solution:
```
def main():
n = int(input())
r2, s2, p2 = map(int, input().split())
r1, s1, p1 = map(int, input().split())
ma = min(r1, p2)+min(s1, r2)+min(p1, s2)
if s2>=r1+s1:
mi = s2-r1-s1
else:
mi = n
k1 = max(0, s2-s1)
k2 = min(r1, s2)
c1 = r1-r2
c2 = p2-s1+s2
if k1<=c1:
mi = c1-min(k2, c1)
if k2>=c2:
mi = min(mi, c2-max(k1, c2))
if c2>c1+1 and (c2>k1 and c1<k2) or (k1>c1 and k2<c2):
mi = 0
print(mi, ma)
main()
```
No
| 95,466 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Submitted Solution:
```
def find_max_winning_rounds(p1,p2):
result=0
for i in range(3):
max_winnable=min(p1[i],p2[(i+1)%3])
result+=max_winnable
return result
def find_min_winning_rounds(p1,p2):
for i in range(3):
find_bigger_item_to_subtract(p2, i, p1)
result = 0
for i in range(3):
result += p1[i]
return result
def find_bigger_item_to_subtract(p2, i, p1):
if p2[(i+2)%3]>=p2[i]:
p1[i],p2[(i+2)%3]=sub(p1[i],p2[(i+2)%3])
if p1[i]>0:
p1[i],p2[i]=sub(p1[i],p2[i])
else:
p1[i],p2[i]=sub(p1[i],p2[i])
if p1[i]>0:
p1[i],p2[(i+2)%3]=sub(p1[i],p2[(i+2)%3])
def sub(a,b):
if a >= b:
return a-b,0
else:
return 0,b-a
rounds=int(input())
player1_strategy=list(map(int,input().split()))
player2_strategy=list(map(int,input().split()))
max_win=find_max_winning_rounds(player1_strategy,player2_strategy)
min_win=find_min_winning_rounds(player1_strategy,player2_strategy)
print(min_win,max_win)
```
No
| 95,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have decided to play the game "Rock, Paper, Scissors".
The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:
* if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
* if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
* if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.
Alice and Bob decided to play exactly n rounds of the game described above. Alice decided to show rock a_1 times, show scissors a_2 times and show paper a_3 times. Bob decided to show rock b_1 times, show scissors b_2 times and show paper b_3 times. Though, both Alice and Bob did not choose the sequence in which they show things. It is guaranteed that a_1 + a_2 + a_3 = n and b_1 + b_2 + b_3 = n.
Your task is to find two numbers:
1. the minimum number of round Alice can win;
2. the maximum number of rounds Alice can win.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^{9}) — the number of rounds.
The second line of the input contains three integers a_1, a_2, a_3 (0 ≤ a_i ≤ n) — the number of times Alice will show rock, scissors and paper, respectively. It is guaranteed that a_1 + a_2 + a_3 = n.
The third line of the input contains three integers b_1, b_2, b_3 (0 ≤ b_j ≤ n) — the number of times Bob will show rock, scissors and paper, respectively. It is guaranteed that b_1 + b_2 + b_3 = n.
Output
Print two integers: the minimum and the maximum number of rounds Alice can win.
Examples
Input
2
0 1 1
1 1 0
Output
0 1
Input
15
5 5 5
5 5 5
Output
0 15
Input
3
0 0 3
3 0 0
Output
3 3
Input
686
479 178 29
11 145 530
Output
22 334
Input
319
10 53 256
182 103 34
Output
119 226
Note
In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.
In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.
In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.
Submitted Solution:
```
n=int(input())
a1,a2,a3=map(int,input().split())
b1,b2,b3=map(int,input().split())
maxx=0
maxx=maxx+min(a1,b2)
maxx=maxx+min(a2,b3)
maxx=maxx+min(a3,b1)
minn=0
if (a1-b3)>0:
minn=minn+(a1-b3)
if (a2-b1)>0:
minn=minn+(a2-b1)
if (a3-b2)>0:
minn=minn+(a3-b2)
print(minn,maxx)
```
No
| 95,468 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Tags: data structures, dfs and similar, graphs, shortest paths
Correct Solution:
```
from heapq import heappop, heappush
nodes, edges, capitalIndex = map(int, input().split())
adj = [dict() for i in range(nodes + 1)]
visited = [False] * (nodes + 1)
costList = [float("inf")] * (nodes + 1)
costList[capitalIndex] = 0
for i in range(edges):
start, end, dist = map(int, input().split())
adj[start][end] = min(adj[start].get(end, float("inf")), dist)
adj[end][start] = adj[start][end]
distanceToCapital = int(input())
# dijkstra
pq = []
counter = 0
heappush(pq, (0, counter, capitalIndex))
counter += 1
while pq:
cost, count, index = heappop(pq)
if not visited[index]:
visited[index] = True
for nextIndex, nextCost in adj[index].items():
totalCost = cost + nextCost
if not visited[nextIndex] and totalCost < costList[nextIndex]:
costList[nextIndex] = totalCost
heappush(pq, (totalCost, counter, nextIndex))
counter += 1
# is there a silo on the city?
silos = costList.count(distanceToCapital)
# is there an edge containing a silo?
for start in range(len(adj)):
costStart = costList[start]
for end, costEdge in adj[start].items():
totalCost = costStart + costEdge
# make sure capitalDistance is between road
if(costStart < distanceToCapital < totalCost):
roadTraveled = distanceToCapital - costStart
roadLeft = totalCost - distanceToCapital
# make sure path from end to road isn't shorter so this is indeed shortest path
if distanceToCapital <= costList[end] + roadLeft:
silos += 1
# make sure we aren't double counting silos in the middle
if distanceToCapital == costList[end] + roadLeft:
silos -= start > end
print(silos)
```
| 95,469 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Tags: data structures, dfs and similar, graphs, shortest paths
Correct Solution:
```
n, m, s = map(int, input().split())
p = [[] for i in range(n + 1)]
for i in range(m):
u, v, w = map(int, input().split())
p[u].append((v, w))
p[v].append((u, w))
l = int(input())
t = [l + 1] * (n + 1)
t[s], q = 0, {s}
while q:
u = q.pop()
r = t[u]
for v, w in p[u]:
if r + w < t[v]:
q.add(v)
t[v] = r + w
s, r = 0, 2 * l
for u in range(1, n + 1):
d = t[u]
if d < l:
for v, w in p[u]:
if d + w > l and (t[v] + d + w > r or (u < v and t[v] + d + w == r)): s += 1
elif d == l:
s += 1
print(s)
# Made By Mostafa_Khaled
```
| 95,470 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Tags: data structures, dfs and similar, graphs, shortest paths
Correct Solution:
```
from heapq import heappush, heappop
def iin():
return (map(int, input().split()))
nodes, edges, start = iin()
start -= 1
graph = [[] for x in range(nodes)]
edge_list = []
shortpath = [float('inf') for x in range(nodes)]
for x in range(edges):
left, right, cost = iin()
left -= 1; right -= 1
graph[left].append((right, cost))
graph[right].append((left, cost))
edge_list.append((left, right, cost))
l, = iin()
pq = [(0, start)]
while pq:
path_cost, node = heappop(pq)
if shortpath[node] < path_cost: continue
shortpath[node] = path_cost
for adj, cost in graph[node]:
if path_cost + cost < shortpath[adj]:
heappush(pq, (path_cost + cost, adj))
silos = 0
for p in shortpath:
silos += 0 if p != l else 1
for left, right, cost in edge_list:
if shortpath[left] < l and shortpath[left] + cost > l:
silos += 1 if shortpath[right] + (cost - (l-shortpath[left])) >= shortpath[left] + (l-shortpath[left]) else 0
if shortpath[right] < l and shortpath[right] + cost > l:
silos += 1 if shortpath[left] + (cost - (l-shortpath[right])) > shortpath[right] + (l-shortpath[right]) else 0
print(silos)
```
| 95,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Tags: data structures, dfs and similar, graphs, shortest paths
Correct Solution:
```
from heapq import heappush, heappop
def i_in():
return (map(int, input().split()))
nodes, edges, start = i_in()
start -= 1
graph = [[] for x in range(nodes)]
edge_list = []
shortest_path = [float('inf') for x in range(nodes)]
for x in range(edges):
left, right, cost = i_in()
left -= 1; right -= 1
graph[left].append((right, cost))
graph[right].append((left, cost))
edge_list.append((left, right, cost))
l, = i_in()
pq = [(0, start)]
while pq:
path_cost, node = heappop(pq)
if shortest_path[node] < path_cost: continue
shortest_path[node] = path_cost
for adj, cost in graph[node]:
if cost + path_cost < shortest_path[adj]:
shortest_path[adj] = cost + path_cost
heappush(pq, (shortest_path[adj], adj))
silos = 0
for p in shortest_path:
silos += 1 if p == l else 0
for left, right, cost in edge_list:
if shortest_path[left] < l and shortest_path[left] + cost > l:
silo_dist = l - shortest_path[left]
silos += 1 if shortest_path[right] + (cost - silo_dist) >= shortest_path[left] + silo_dist else 0
if shortest_path[right] < l and shortest_path[right] + cost > l:
silo_dist = l - shortest_path[right]
silos += 1 if shortest_path[left] + (cost - silo_dist) > shortest_path[right] + silo_dist else 0
print(silos)
```
| 95,472 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Tags: data structures, dfs and similar, graphs, shortest paths
Correct Solution:
```
import sys
n,m,s=map(int,sys.stdin.readline().split())
p=[[] for i in range(n+1)]
for i in range(m):
u, v, w = map(int, input().split())
p[u].append((v, w))
p[v].append((u, w))
l = int(input())
t = [l + 1] * (n + 1)
t[s], q = 0, {s}
while q:
u = q.pop()
r = t[u]
for v, w in p[u]:
if r + w < t[v]:
q.add(v)
t[v] = r + w
s, r = 0, 2 * l
for u in range(1,n+1):
d=t[u]
if d < l:
for v, w in p[u]:
if d + w > l and (t[v] + d + w > r or (u < v and t[v] + d + w == r)): s += 1
elif d == l:
s += 1
print(s)
```
| 95,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Tags: data structures, dfs and similar, graphs, shortest paths
Correct Solution:
```
import sys
n,m,s=map(int,sys.stdin.readline().split())
p=[[] for i in range(n+1)]
for i in range(m):
u, v, w = map(int, input().split())
p[u].append((v, w))
p[v].append((u, w))
l = int(sys.stdin.readline())
t = [l + 1] * (n + 1)
t[s], q = 0, {s}
while q:
u = q.pop()
r = t[u]
for v, w in p[u]:
if r + w < t[v]:
q.add(v)
t[v] = r + w
s, r = 0, 2 * l
for u in range(1,n+1):
d=t[u]
if d < l:
for v, w in p[u]:
if d + w > l and (t[v] + d + w > r or (u < v and t[v] + d + w == r)): s += 1
elif d == l:
s += 1
print(s)
```
| 95,474 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Tags: data structures, dfs and similar, graphs, shortest paths
Correct Solution:
```
from sys import stdin
from heapq import heappush, heappop
INF = 2 * 10 ** 9
def dijkstra(graph, n, s):
d = [INF] * n
d[s] = 0
# class VertexIndex(int):
# def __cmp__(self, other):
# return cmp(d[self], d[other])
q = []
heappush(q, (0, s))
while q:
tmp_d, v = heappop(q)
dv = d[v]
if tmp_d != dv:
continue
for to, w in graph[v]:
if dv + w < d[to]:
d[to] = dv + w
heappush(q, (d[to], to))
return d
def main():
next_line = stdin.readline
n, m, s = map(int, next_line().split())
s -= 1
graph = [[] for _ in range(n)]
def init(i):
line = next_line().split()
v = int(line[0]) - 1
u = int(line[1]) - 1
w = int(line[2])
graph[v].append([u, w])
graph[u].append([v, w])
return v, u, w
edges = list(map(init, range(m)))
# for i in range(m):
# line = next_line().split()
# v = int(line[0]) - 1
# u = int(line[1]) - 1
# w = int(line[2])
# graph[v].append([u, w])
# graph[u].append([v, w])
# edges[i] = (v, u, w)
l = int(next_line())
ds = dijkstra(graph, n, s)
def calc(edge):
v, u, w = edge
dv = ds[v]
du = ds[u]
res = 0
if dv < l < dv + w and du + w - (l - dv) > l:
res += 1
if du < l < du + w and dv + w - (l - du) > l:
res += 1
if dv < l and du < l and dv + du + w == 2 * l:
res += 1
return res
res = sum(d == l for d in ds) + sum(map(calc, edges))
print(res)
if __name__ == '__main__':
main()
```
| 95,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Tags: data structures, dfs and similar, graphs, shortest paths
Correct Solution:
```
from heapq import heappush, heappop
def i_in():
return (map(int, input().split()))
nodes, edges, start = i_in()
start -= 1
graph = [[] for x in range(nodes)]
edge_list = []
shortest_path = [float('inf') for x in range(nodes)]
for x in range(edges):
left, right, cost = i_in()
left -= 1; right -= 1
graph[left].append((right, cost))
graph[right].append((left, cost))
edge_list.append((left, right, cost))
l, = i_in()
pq = [(0, start)]
while pq:
path_cost, node = heappop(pq)
if shortest_path[node] < path_cost: continue
shortest_path[node] = path_cost
for adj, cost in graph[node]:
if cost + path_cost < shortest_path[adj]:
shortest_path[adj] = cost + path_cost
heappush(pq, (shortest_path[adj], adj))
silos = 0
for p in shortest_path:
silos += 1 if p == l else 0
for left, right, cost in edge_list:
if shortest_path[left] < l and shortest_path[left] + cost > l:
silo_dist = l - shortest_path[left]
silos += 1 if shortest_path[right] + (cost - silo_dist) >= l else 0
if shortest_path[right] < l and shortest_path[right] + cost > l:
silo_dist = l - shortest_path[right]
silos += 1 if shortest_path[left] + (cost - silo_dist) > l else 0
print(silos)
```
| 95,476 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Submitted Solution:
```
from sys import stdin
from heapq import heappush, heappop
INF = 2 * 10 ** 9
def dijkstra(graph, n, s):
d = [INF] * n
d[s] = 0
# class VertexIndex(int):
# def __cmp__(self, other):
# return cmp(d[self], d[other])
q = []
heappush(q, (0, s))
while q:
tmp_d, v = heappop(q)
dv = d[v]
if tmp_d != dv:
continue
for to, w in graph[v]:
if dv + w < d[to]:
d[to] = dv + w
heappush(q, (d[to], to))
return d
def main():
next_line = stdin.readline
n, m, s = map(int, next_line().split())
s -= 1
edges = [None] * m
graph = [[] for _ in range(n)]
for i in range(m):
line = next_line().split()
v = int(line[0]) - 1
u = int(line[1]) - 1
w = int(line[2])
graph[v].append([u, w])
graph[u].append([v, w])
edges[i] = (v, u, w)
l = int(next_line())
ds = dijkstra(graph, n, s)
# res = 0
# for v, u, w in edges:
# dv = ds[v]
# du = ds[u]
# if dv < l < dv + w and du + w - (l - dv) > l:
# res += 1
# if du < l < du + w and dv + w - (l - du) > l:
# res += 1
# if dv < l and du < l and dv + du + w == 2 * l:
# res += 1
def is_satisfies(edge):
v, u, w = edge
dv = ds[v]
du = ds[u]
res = 0
if dv < l < dv + w and du + w - (l - dv) > l:
res += 1
if du < l < du + w and dv + w - (l - du) > l:
res += 1
if dv < l and du < l and dv + du + w == 2 * l:
res += 1
return res
res = sum(d == l for d in ds) + sum(map(is_satisfies, edges))
print(res)
if __name__ == '__main__':
main()
```
Yes
| 95,477 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Submitted Solution:
```
from heapq import heappush, heappop
def get_graph_from_input(m):
G = {}
for _ in range(m):
u, v, w = input().split()
w = int(w)
if not u in G:
G[u] = {}
if not v in G:
G[v] = {}
G[u][v] = w
G[v][u] = w
return G
def get_num_missile_silos(G, l, s):
min_distance = {}
road_nodes = {}
q = [(0, s)]
while(q):
dist, node = heappop(q)
if not node in min_distance:
min_distance[node] = dist
else:
continue
if not '.' in node:
for neighbor in G[node]:
w = G[node][neighbor]
new_dist = dist + w
if dist < l and new_dist > l:
dist_to_silo = l - dist
min_node, max_node = min(neighbor, node), max(neighbor, node)
if min_node != node:
dist_to_silo = w - dist_to_silo
road_silo = (min_node, max_node)
if not road_silo in min_distance:
min_distance[road_silo] = set([])
min_distance[road_silo].add(dist_to_silo)
if new_dist <= l and not neighbor in min_distance:
heappush(q, (dist + w, neighbor))
silos = 0
for node in min_distance:
if type(node) is tuple:
u, v = node
if not u in min_distance or not v in min_distance or ((min_distance[u] + min_distance[v] + G[u][v]) >= 2 * l):
silos += len(min_distance[node])
else:
if min_distance[node] == l:
silos += 1
return silos
if __name__ == "__main__":
n, m, s = map(int, input().split())
G = get_graph_from_input(m)
l = int(input())
print(get_num_missile_silos(G, l, str(s)))
```
Yes
| 95,478 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Submitted Solution:
```
from heapq import *
from sys import stdin
if __name__ == '__main__':
V, E, start = map(int, stdin.readline().rstrip().split())
start -= 1
adj_list = []
for v in range(0, V):
adj_list.append({})
shortest_dist = []
for e in range(0, E):
v1, v2, length = map(int, stdin.readline().rstrip().split())
adj_list[v1 - 1][v2 - 1] = length
adj_list[v2 - 1][v1 - 1] = length
goal_dist = int(stdin.readline().rstrip())
heapq = []
shortest_dist = [100000000] * V
shortest_dist[start] = 0
seen = [False] * V
heappush(heapq, (0, start))
while heapq:
(dist, v1) = heappop(heapq)
if not seen[v1]:
seen[v1] = True
neighbors = adj_list[v1]
to_change = []
for v2, length in neighbors.items():
if not seen[v2]:
new_dist = length + dist
if dist < goal_dist < new_dist:
length1 = goal_dist - dist
to_change.append((V, v2, length1))
length2 = new_dist - goal_dist
neighbors2 = adj_list[v2]
neighbors2[V] = length2
del neighbors2[v1]
adj_list.append({v1: length1, v2: length2})
seen.append(False)
shortest_dist.append(100000000)
v2 = V
new_dist = goal_dist
V += 1
if shortest_dist[v2] > new_dist:
shortest_dist[v2] = new_dist
heappush(heapq, (new_dist, v2))
for (v_add, v_remove, length) in to_change:
del neighbors[v_remove]
neighbors[v_add] = length
# print(shortest_dist)
count = 0
for sd in shortest_dist:
if sd == goal_dist:
count += 1
print(count)
# print(adj_list)
```
Yes
| 95,479 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Submitted Solution:
```
from sys import stdin
from heapq import heappush, heappop
INF = 2 * 10 ** 9
def dijkstra(graph, n, s):
d = [INF] * n
d[s] = 0
# class VertexIndex(int):
# def __cmp__(self, other):
# return cmp(d[self], d[other])
q = []
heappush(q, (0, s))
while q:
tmp_d, v = heappop(q)
dv = d[v]
if tmp_d != dv:
continue
graph_v = graph[v]
for to, w in graph_v:
if dv + w < d[to]:
d[to] = dv + w
heappush(q, (d[to], to))
return d
def main():
next_line = stdin.readline
n, m, s = map(int, next_line().split())
s -= 1
graph = [[] for _ in range(n)]
def init(i):
line = next_line().split()
v = int(line[0]) - 1
u = int(line[1]) - 1
w = int(line[2])
graph[v].append([u, w])
graph[u].append([v, w])
return v, u, w
edges = list(map(init, range(m)))
# for i in range(m):
# line = next_line().split()
# v = int(line[0]) - 1
# u = int(line[1]) - 1
# w = int(line[2])
# graph[v].append([u, w])
# graph[u].append([v, w])
# edges[i] = (v, u, w)
l = int(next_line())
ds = dijkstra(graph, n, s)
def calc(edge):
v, u, w = edge
dv = ds[v]
du = ds[u]
res = 0
if dv < l < dv + w and du + w - (l - dv) > l:
res += 1
if du < l < du + w and dv + w - (l - du) > l:
res += 1
if dv < l and du < l and dv + du + w == 2 * l:
res += 1
return res
res = sum(d == l for d in ds) + sum(map(calc, edges))
print(res)
if __name__ == '__main__':
main()
```
Yes
| 95,480 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Submitted Solution:
```
from sys import stdin
from heapq import heappush, heappop
INF = 2 * 10 ** 9
def dijkstra(graph, n, s):
d = [INF] * n
d[s] = 0
class VertexIndex(int):
def __lt__(self, other):
return d[self] < d[other]
q = []
heappush(q, VertexIndex(s))
used = [False] * n
while q:
v = heappop(q)
dv = d[v]
if used[v]:
continue
used[v] = True
graph_v = graph[v]
for to, w in graph_v:
if dv + w < d[to]:
d[to] = dv + w
heappush(q, VertexIndex(to))
return d
def main():
next_line = stdin.readline
n, m, s = map(int, next_line().split())
s -= 1
graph = [[] for _ in range(n)]
def init(i):
line = next_line().split()
v = int(line[0]) - 1
u = int(line[1]) - 1
w = int(line[2])
graph[v].append([u, w])
graph[u].append([v, w])
return v, u, w
edges = list(map(init, range(m)))
l = int(next_line())
ds = dijkstra(graph, n, s)
def calc(edge):
v, u, w = edge
dv = ds[v]
du = ds[u]
res = 0
if dv < l < dv + w and du + w - (l - dv) > l:
res += 1
if du < l < du + w and dv + w - (l - du) > l:
res += 1
if dv < l and du < l and dv + du + w == 2 * l:
res += 1
return res
res = sum(d == l for d in ds) + sum(map(calc, edges))
print(res)
if __name__ == '__main__':
main()
```
No
| 95,481 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Submitted Solution:
```
import sys
import heapq
from sys import stdin, stdout
n, m, inizio = [int(x) for x in (stdin.readline().split())]
inizio = inizio - 1
grafo = []
dist = []
for i in range(n):
grafo.append([])
dist.append(sys.maxsize)
for i in range(m):
da, a, peso = [int(x) for x in (stdin.readline().split())]
da = da - 1
a = a - 1
grafo[da].append([peso,a, False])
grafo[a].append([peso,da, False])
for i in range(n):
grafo[i].sort()
dist[inizio] = 0
h = []
count = 1
heapq.heappush(h, (0, inizio, -1, -1))
l = int(stdin.readline())
tot = 0
while count != 0:
gianni = heapq.heappop(h)
count = count - 1
if gianni[2]!=-1 and grafo[gianni[3]][gianni[2]][2] == False:
grafo[gianni[3]][gianni[2]][2] = True
if gianni[0] >= l:
tot = tot + 1
if gianni[0] == dist[gianni[1]]:
for i in range(len(grafo[gianni[1]])):
adj = grafo[gianni[1]][i]
if adj[0]+gianni[0] < dist[adj[1]]:
dist[adj[1]] = adj[0]+gianni[0]
heapq.heappush(h, (dist[adj[1]], adj[1], i, gianni[1]))
count = count + 1
stdout.write(str(tot))
```
No
| 95,482 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Submitted Solution:
```
from heapq import heappush, heappop
def get_graph_from_input(m):
G = {}
for _ in range(m):
u, v, w = input().split()
w = int(w)
if not u in G:
G[u] = {}
if not v in G:
G[v] = {}
G[u][v] = w
G[v][u] = w
return G
def get_num_missile_silos(G, l, s):
min_distance = {}
q = [(0, s)]
while(q):
dist, node = heappop(q)
if not node in min_distance:
min_distance[node] = dist
else:
continue
if not '.' in node:
for neighbor in G[node]:
w = G[node][neighbor]
new_dist = dist + w
if dist < l and new_dist > l:
dist_to_silo = l - dist
min_node, max_node = min(neighbor, node), max(neighbor, node)
if min_node != node:
dist_to_silo = w - dist_to_silo
road_silo = f"{min_node}.{max_node}.{dist_to_silo}"
if not road_silo in min_distance:
min_distance[road_silo] = l
if new_dist <= l and not neighbor in min_distance:
heappush(q, (dist + w, neighbor))
return sum(min_distance[k] == l for k in min_distance)
if __name__ == "__main__":
n, m, s = map(int, input().split())
G = get_graph_from_input(m)
l = int(input())
print(get_num_missile_silos(G, l, str(s)))
```
No
| 95,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, <image>, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
* between any two cities no more than one road exists;
* each road connects two different cities;
* from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Examples
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2
Output
3
Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4
Output
3
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
Submitted Solution:
```
from heapq import heappush, heappop
def get_graph_from_input(m):
G = {}
for _ in range(m):
u, v, w = input().split()
w = int(w)
if not u in G:
G[u] = {}
if not v in G:
G[v] = {}
G[u][v] = w
G[v][u] = w
return G
def get_num_missile_silos(G, l, s):
min_distance = {}
q = [(0, s)]
while(q):
dist, node = heappop(q)
if not node in min_distance:
min_distance[node] = dist
if not '.' in node:
for neighbor in G[node]:
w = G[node][neighbor]
new_dist = dist + w
if dist < l and new_dist > l:
dist_to_silo = l - dist
if min(neighbor, node) != node:
dist_to_silo = w - dist_to_silo
road_silo = f"{min(int(neighbor), int(node))}.{max(int(neighbor), int(node))}.{dist_to_silo}"
if not road_silo in min_distance:
min_distance[road_silo] = l
if new_dist <= l and not neighbor in min_distance:
heappush(q, (dist + w, neighbor))
return sum(min_distance[k] == l for k in min_distance)
if __name__ == "__main__":
n, m, s = map(int, input().split())
G = get_graph_from_input(m)
l = int(input())
print(get_num_missile_silos(G, l, str(s)))
```
No
| 95,484 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Tags: data structures, dp, implementation, strings
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
for _ in range(int(input())):
n,m=map(int,input().split())
op=input()
pre=[[0,0,0]]
suf=[[0,0,0]]
out=0
for i in op:
if i=='-':
out-=1
else:
out+=1
pre.append([out,min(pre[-1][1],out),max(pre[-1][2],out)])
out=0
for i in reversed(op):
if i=='+':
out-=1
else:
out+=1
suf.append([out,min(suf[-1][1],out),max(suf[-1][2],out)])
for i in range(m):
l,r=map(int,input().split())
l-=1
minpre=pre[l][1]
maxpre=pre[l][2]
minsuf=-suf[n-r][0]+suf[n-r][1]+pre[l][0] #potential minimum candidate
maxsuf=-suf[n-r][0]+suf[n-r][2]+pre[l][0] #potential maximum candidate
print(max(maxpre,maxsuf)-min(minpre,minsuf)+1)
```
| 95,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Tags: data structures, dp, implementation, strings
Correct Solution:
```
import sys
input=sys.stdin.readline
for z in range(int(input())):
n,m=map(int,input().split())
s=input()
a=[0]
for i in range(n):
if s[i]=="+":
a.append(a[-1]+1)
else:
a.append(a[-1]-1)
nminl=[0]
nmaxl=[0]
nminr=[a[-1]]*(n+1)
nmaxr=[a[-1]]*(n+1)
for i in range(1,n+1):
nminl.append(min(nminl[-1],a[i]))
nmaxl.append(max(nmaxl[-1],a[i]))
for i in range(n-1,-1,-1):
nminr[i]=min(nminr[i+1],a[i])
nmaxr[i]=max(nmaxr[i+1],a[i])
for i in range(m):
l,r=map(int,input().split())
num2=a[l-1]-a[r]
if r!=n:
ni=min(nminl[l-1],num2+nminr[min(n,r+1)])
na=max(nmaxl[l-1],num2+nmaxr[min(n,r+1)])
else:
ni=nminl[l-1]
na=nmaxl[l-1]
print(na-ni+1)
```
| 95,486 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Tags: data structures, dp, implementation, strings
Correct Solution:
```
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
from collections import Counter, OrderedDict
from itertools import permutations as perm
from fractions import Fraction
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
ans = []
t, = gil()
for _ in range(t):
n, m = gil()
a = []
for v in g():
if v == "+":
a.append(1)
else:
a.append(-1)
l, r = a[:], a[:]
for i in range(1, n):
l[i] += l[i-1]
lstack = []
mi, ma = 0, 0
for i in range(n):
mi = min(mi, l[i])
ma = max(ma, l[i])
lstack.append((mi, ma))
r[-1] *= -1
for i in reversed(range(n-1)):
r[i] *= -1
r[i] += r[i+1]
rstack = [None for _ in range(n)]
mi, ma = 0, 0
# if a[i] == 1:
# mi, ma = -1, -1
# else:
# mi, ma = 1, 1
# stack[-1] = (mi, ma)
for i in reversed(range(n)):
mi = min(mi, r[i])
ma = max(ma, r[i])
rstack[i] = (mi-r[i], ma-r[i])
# print(l)
# print(lstack)
# print()
# print(r)
# print(rstack)
# exit()
# print()
for _ in range(m):
li, ri = gil()
li -= 1
x = 0
var = 0
mi, ma = 0, 0
if li:
x += l[li-1]
mi = min(mi, lstack[li-1][0])
ma = max(ma, lstack[li-1][1])
if ri < n:
mi = min(mi, x+rstack[ri][0])
ma = max(ma, x+rstack[ri][1])
ans.append(str(ma-mi+1))
# print(mi, ma, ma-mi+1)
# print(ma-mi+1)
print("\n".join(ans))
```
| 95,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Tags: data structures, dp, implementation, strings
Correct Solution:
```
from itertools import accumulate
from sys import stdin, stdout
readline = stdin.readline
write = stdout.write
def read_ints():
return map(int, readline().split())
def read_string():
return readline()[:-1]
def write_int(x):
write(str(x) + '\n')
def find_prefix_ranges(op_seq):
x = 0
min_x = max_x = x
result = [(min_x, max_x)]
for op in op_seq:
x += op
if x < min_x:
min_x = x
if x > max_x:
max_x = x
result.append((min_x, max_x))
return result
def find_suffix_ranges(op_seq):
min_x = max_x = 0
result = [(min_x, max_x)]
for op in op_seq[::-1]:
min_x += op
max_x += op
if op < min_x:
min_x = op
if op > max_x:
max_x = op
result.append((min_x, max_x))
return result[::-1]
t_n, = read_ints()
for i_t in range(t_n):
n, q_n = read_ints()
s = read_string()
op_seq = [+1 if char == '+' else -1 for char in s]
prefix_sums = [0] + list(accumulate(op_seq))
prefix_ranges = find_prefix_ranges(op_seq)
suffix_ranges = find_suffix_ranges(op_seq)
for i_q in range(q_n):
l, r = read_ints()
range_before = prefix_ranges[l-1]
range_after = suffix_ranges[r]
delta_for_after = prefix_sums[l-1]
min_x, max_x = range_after
min_x += delta_for_after
max_x += delta_for_after
range_after = (min_x, max_x)
a, b = range_before, range_after
if a > b:
a, b = b, a
assert a <= b
if a[1] < b[0]:
result = a[1] - a[0] + 1 + b[1] - b[0] + 1
else:
final_range = (a[0], max(a[1], b[1]))
result = final_range[1] - final_range[0] + 1
write_int(result)
```
| 95,488 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Tags: data structures, dp, implementation, strings
Correct Solution:
```
from sys import stdin
t=int(stdin.readline())
for _ in range(t):
n,m=map(int,stdin.readline().split())
s=stdin.readline()[:-1]
maxt=0
mint=0
maxil=[0]*(n+1)
minil=[0]*(n+1)
val=[0]*(n+1)
x=0
for i in range(n):
if s[i]=='+':
x+=1
else:
x-=1
val[i+1]=x
maxt=max(maxt,x)
mint=min(mint,x)
maxil[i+1]=maxt
minil[i+1]=mint
maxt=0
mint=0
maxir=[0]*(n+1)
minir=[0]*(n+1)
var=[0]*(n+1)
x=0
for i in range(n-1,-1,-1):
if s[i]=='+':
x-=1
else:
x+=1
var[i]=x
maxt=max(maxt,x)
mint=min(mint,x)
maxir[i]=maxt
minir[i]=mint
for i in range(m):
l,r=map(int,stdin.readline().split())
lrange=min(minil[l-1],val[l-1]-(var[r]-minir[r]))
rrange=max(maxil[l-1],val[l-1]+(maxir[r]-var[r]))
print(rrange-lrange+1)
```
| 95,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Tags: data structures, dp, implementation, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n, m = list(map(int, input().split()))
s = input()
sums = [0] * (n + 1)
pref_mins = [0] * (n + 1)
pref_maxs = [0] * (n + 1)
for i in range(n):
sums[i + 1] = sums[i] + (1 if s[i] == '+' else -1)
pref_mins[i + 1] = min(pref_mins[i], sums[i + 1])
pref_maxs[i + 1] = max(pref_maxs[i], sums[i + 1])
suf_mins = [0] * (n + 1)
suf_maxs = [0] * (n + 1)
suf_maxs[n] = sums[n]
suf_mins[n] = sums[n]
for i in reversed(range(n)):
suf_maxs[i] = max(sums[i], suf_maxs[i + 1])
suf_mins[i] = min(sums[i], suf_mins[i + 1])
for i in range(m):
l, r = list(map(int, input().split()))
sum = sums[r] - sums[l - 1]
ans = max(pref_maxs[l - 1], suf_maxs[r] - sum) - min(pref_mins[l - 1], suf_mins[r] - sum) + 1
print(ans)
```
| 95,490 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Tags: data structures, dp, implementation, strings
Correct Solution:
```
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")
import typing
def _ceil_pow2(n: int) -> int:
x = 0
while (1 << x) < n:
x += 1
return x
def _bsf(n: int) -> int:
x = 0
while n % 2 == 0:
x += 1
n //= 2
return x
class SegTree:
def __init__(self,
op: typing.Callable[[typing.Any, typing.Any], typing.Any],
e: typing.Any,
v: typing.Union[int, typing.List[typing.Any]]) -> None:
self._op = op
self._e = e
if isinstance(v, int):
v = [e] * v
self._n = len(v)
self._log = _ceil_pow2(self._n)
self._size = 1 << self._log
self._d = [e] * (2 * self._size)
for i in range(self._n):
self._d[self._size + i] = v[i]
for i in range(self._size - 1, 0, -1):
self._update(i)
def set(self, p: int, x: typing.Any) -> None:
assert 0 <= p < self._n
p += self._size
self._d[p] = x
for i in range(1, self._log + 1):
self._update(p >> i)
def get(self, p: int) -> typing.Any:
assert 0 <= p < self._n
return self._d[p + self._size]
def prod(self, left: int, right: int) -> typing.Any:
assert 0 <= left <= right <= self._n
sml = self._e
smr = self._e
left += self._size
right += self._size
while left < right:
if left & 1:
sml = self._op(sml, self._d[left])
left += 1
if right & 1:
right -= 1
smr = self._op(self._d[right], smr)
left >>= 1
right >>= 1
return self._op(sml, smr)
def all_prod(self) -> typing.Any:
return self._d[1]
def max_right(self, left: int,
f: typing.Callable[[typing.Any], bool]) -> int:
assert 0 <= left <= self._n
assert f(self._e)
if left == self._n:
return self._n
left += self._size
sm = self._e
first = True
while first or (left & -left) != left:
first = False
while left % 2 == 0:
left >>= 1
if not f(self._op(sm, self._d[left])):
while left < self._size:
left *= 2
if f(self._op(sm, self._d[left])):
sm = self._op(sm, self._d[left])
left += 1
return left - self._size
sm = self._op(sm, self._d[left])
left += 1
return self._n
def min_left(self, right: int,
f: typing.Callable[[typing.Any], bool]) -> int:
assert 0 <= right <= self._n
assert f(self._e)
if right == 0:
return 0
right += self._size
sm = self._e
first = True
while first or (right & -right) != right:
first = False
right -= 1
while right > 1 and right % 2:
right >>= 1
if not f(self._op(self._d[right], sm)):
while right < self._size:
right = 2 * right + 1
if f(self._op(self._d[right], sm)):
sm = self._op(self._d[right], sm)
right -= 1
return right + 1 - self._size
sm = self._op(self._d[right], sm)
return 0
def _update(self, k: int) -> None:
self._d[k] = self._op(self._d[2 * k], self._d[2 * k + 1])
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
t=int(input())
for _ in range(t):
n,m=map(int,input().split())
s=input()
a=[0]
curr=0
for i in range(n):
if s[i]=='+':
curr+=1
else:
curr-=1
a.append(curr)
segtreemax=SegTree(max,-10**9,a)
segtreemin=SegTree(min,10**9,a)
for i in range(m):
l,r=map(int,input().split())
max1=segtreemax.prod(0,l)
min1=segtreemin.prod(0,l)
if r<n:
max1=max(max1,a[l-1]+segtreemax.prod(r+1,n+1)-a[r])
min1=min(min1,a[l-1]+segtreemin.prod(r+1,n+1)-a[r])
print(max1-min1+1)
```
| 95,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Tags: data structures, dp, implementation, strings
Correct Solution:
```
from collections import defaultdict,deque
import sys
import bisect
import math
input=sys.stdin.readline
mod=1000000007
def check(base, higher, lower):
return higher - base, lower - base
for t in range(int(input())):
n, m = map(int, input().split())
st = input()
store = [0 for i in range(n + 2)]
add = 0
for i in range(n):
if (st[i] == "-"):
add -= 1
else:
add += 1
store[i + 1] = add
add = 0
f_mini_store = [0] * (n + 2)
f_maxi_store = [0] * (n + 2)
b_mini_store = [0] * (n + 2)
b_maxi_store = [0] * (n + 2)
maxi = -9999999999
mini = 99999999999
for i in range(1, n + 1):
maxi = max(maxi, store[i])
mini = min(mini, store[i])
f_mini_store[i] = mini
f_maxi_store[i] = maxi
maxi, mini = -99999999999, 99999999999
for i in range(n, 0, -1):
maxi = max(maxi, store[i])
mini = min(mini, store[i])
b_mini_store[i] = mini
b_maxi_store[i] = maxi
set1 = [0, 0]
set2 = [0, 0]
count = 0
for i in range(m):
set1, set2 = [0, 0], [0, 0]
count = 0
l, r = map(int, input().split())
add = 0
if (l > 1):
set2[0], set2[1] = f_maxi_store[l - 1], f_mini_store[l - 1]
add += 3
if (r < n):
set1[0], set1[1] = check(store[r], b_maxi_store[r + 1], b_mini_store[r + 1])
add += 2
set1[0] = store[l - 1] + set1[0]
set1[1] = store[l - 1] + set1[1]
upper = max(set1[0], set2[0])
down = min(set1[1], set2[1])
if (add == 0):
print(1)
elif (add == 3):
if (set2[0] >= 0 and set2[1] <= 0):
print(set2[0] - set2[1] + 1)
else:
print(set2[0] - set2[1] + 2)
elif (add == 2):
if (set1[0] >= 0 and set1[1] <= 0):
print(set1[0] - set1[1] + 1)
else:
print(set1[0] - set1[1] + 2)
else:
if (upper >= 0 and down <= 0):
print(upper - down + 1)
else:
print(upper - down + 2)
```
| 95,492 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Tags: data structures, dp, implementation, strings
Correct Solution:
```
""" Python 3 compatibility tools. """
from __future__ import division, print_function
import itertools
import sys
if sys.version_info[0] < 3:
input = raw_input
range = xrange
filter = itertools.ifilter
map = itertools.imap
zip = itertools.izip
import os
from atexit import register
from io import BytesIO
sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
sys.stdout = BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))
input = lambda: sys.stdin.readline().rstrip('\r\n')
cases = int(input())
for _ in range(cases):
n, q = map(int, input().split())
instructions = input()
minA = [0] * (n + 1)
maxA = [0] * (n + 1)
prefix = [0]
current = 0
for i, s in enumerate(instructions):
if s == "+":
current += 1
else:
current -= 1
index = i + 1
minA[index] = min(minA[index - 1], current)
maxA[index] = max(current, maxA[index - 1])
prefix.append(current)
current = 0
minD = [prefix[-1]] * (n + 1)
maxD = [prefix[-1]] * (n + 1)
for index in range(n - 1, 0, -1):
minD[index] = min(minD[index + 1], prefix[index])
maxD[index] = max(prefix[index], maxD[index + 1])
for _ in range(q):
start, end = map(int, input().split())
rangeStart = [minA[start - 1], maxA[start - 1]]
ignored = prefix[end] - prefix[start - 1]
rangeEnd = [minD[end] - ignored, maxD[end] - ignored]
maxComb = max(rangeEnd[1], rangeStart[1])
minComb = min(rangeEnd[0], rangeStart[0])
print(maxComb - minComb + 1)
```
| 95,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Submitted Solution:
```
import sys
t = int(sys.stdin.readline().strip())
for _ in range(t):
n, m = map(int, sys.stdin.readline().split())
s = sys.stdin.readline().strip()
psa = [0 for k in range(n+1)]
for j in range(1, n+1):
if s[j-1] == "+":
psa[j] = psa[j-1]+1
else:
psa[j] = psa[j-1]-1
sufMax = [0 for k in range(n+1)]
sufMin = [0 for k in range(n+1)]
preMax = [0 for k in range(n+1)]
preMin = [0 for k in range(n+1)]
sufMax[-1] = psa[-1]
sufMin[-1] = psa[-1]
preMax[1] = psa[1]
preMin[1] = psa[1]
for i in range(1, n+1):
preMax[i] = max(psa[i], preMax[i-1])
preMin[i] = min([psa[i], preMin[i-1]])
for i in range(n-1, 0, -1):
sufMax[i] = max(psa[i], sufMax[i+1])
sufMin[i] = min(psa[i], sufMin[i+1])
for j in range(m):
l, r = map(int, sys.stdin.readline().split())
bestMax = preMax[l-1]
bestMin = preMin[l-1]
if r == n:
print(bestMax - bestMin + 1)
else:
temp = psa[l-1]-psa[r]
bestMax = max(bestMax, sufMax[r+1]+temp)
bestMin = min(bestMin, sufMin[r+1]+temp)
print(bestMax-bestMin+1)
```
Yes
| 95,494 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Submitted Solution:
```
'''
#python-io
'''
import sys
# need this
input = sys.stdin.readline
MAXN = 2 * 100_000 + 5
cnt, top, bot = [0]*MAXN, [[0]*MAXN for i in range(2)], [[0]*MAXN for i in range(2)]
def program(n, m, s):
for i in range(n):
if s[i] == '+': cnt[i+1] = cnt[i] + 1
else: cnt[i+1] = cnt[i] - 1
for i in range(1, n+1):
top[0][i] = bot[0][i] = top[1][i] = bot[1][i] = cnt[i]
for i in range(1, n+1):
top[0][i] = max(top[0][i], top[0][i-1])
bot[0][i] = min(bot[0][i], bot[0][i-1])
for i in range(n-1, -1, -1):
top[1][i] = max(top[1][i], top[1][i+1])
bot[1][i] = min(bot[1][i], bot[1][i+1])
for _ in range(m):
l, r = map(int, input().strip().split(' '))
diff = cnt[l-1] - cnt[r]
if r < n:
t = max(top[0][l-1], top[1][r+1] + diff)
b = min(bot[0][l-1], bot[1][r+1] + diff)
else:
t, b = top[0][l-1], bot[0][l-1]
# may not be necessary
sys.stdout.write(str(t-b+1) + '\n')
def main():
for t in range(int(input().strip())):
n, m = map(int, input().strip().split(' '))
s = input().strip()
program(n, m, s)
if __name__ == "__main__":
main()
```
Yes
| 95,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
for _ in range(int(input())):
n, m = map(int,input().split())
a, x = [0], 0
for i in input().rstrip():
x += (i == "+") - (i == "-")
a.append(x)
maxl, minl = [a[0]], [a[0]]
for i in range(1, n + 1, 1):
maxl.append(max(maxl[-1], a[i]))
minl.append(min(minl[-1], a[i]))
maxr, minr = [a[n]], [a[n]]
for i in range(n - 1, -1, -1):
minr.append(min(minr[-1], a[i]))
maxr.append(max(maxr[-1], a[i]))
maxr.reverse()
minr.reverse()
b=[]
for i in range(m):
l, r = map(int, input().split())
if r == n:
b.append(maxl[l - 1] - minl[l - 1] + 1)
else:
z = a[r] - a[l - 1]
mi_l, ma_l, mi_r, ma_r = minl[l - 1], maxl[l - 1], minr[r + 1] - z, maxr[r + 1] - z
if min(ma_l, ma_r) < max(mi_l, mi_r):
b.append(ma_r - mi_r + ma_l - mi_l + 2)
else:
b.append(max(ma_l, ma_r) - min(mi_r, mi_l) + 1)
print(*b,sep="\n")
# 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")
if __name__ == "__main__":
main()
```
Yes
| 95,496 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Submitted Solution:
```
#region Header
#!/usr/bin/env python3
# from typing import *
import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq
def input():
return sys.stdin.readline()[:-1]
# sys.setrecursionlimit(1000000)
#endregion
# _INPUT = """2
# 8 4
# -+--+--+
# 1 8
# 2 8
# 2 5
# 1 1
# 4 10
# +-++
# 1 1
# 1 2
# 2 2
# 1 3
# 2 3
# 3 3
# 1 4
# 2 4
# 3 4
# 4 4
# """
# sys.stdin = io.StringIO(_INPUT)
class BIT:
"""
Binary Indexed Tree (Fenwick Tree), 1-indexed
"""
def __init__(self, n):
"""
Parameters
----------
n : int
要素数。index は 0..n になる。
"""
self.size = n
self.data = [0] * (n+1)
# self.depth = n.bit_length()
def add(self, i, x):
while i <= self.size:
self.data[i] += x
i += i & -i
def get_sum(self, i):
s = 0
while i > 0:
s += self.data[i]
i -= i & -i
return s
def get_rsum(self, l, r):
"""
[l, r) の sum
"""
return self.get_sum(r) - self.get_sum(l-1)
class BIT_Max:
"""
Binary Indexed Tree (Fenwick Tree), 1-indexed
"""
def __init__(self, n):
"""
Parameters
----------
n : int
要素数。index は 0..n になる。
"""
self.size = n
self.data = [0] * (n+1)
# self.depth = n.bit_length()
def update(self, i, x):
while i <= self.size:
self.data[i] = max(x, self.data[i])
i += i & -i
def get_max(self, i):
# 1からiまで
s = 0
while i > 0:
s = max(s, self.data[i])
i -= i & -i
return s
class BIT_Min:
"""
Binary Indexed Tree (Fenwick Tree), 1-indexed
"""
def __init__(self, n):
"""
Parameters
----------
n : int
要素数。index は 0..n になる。
"""
self.size = n
self.data = [0] * (n+1)
# self.depth = n.bit_length()
def update(self, i, x):
while i <= self.size:
self.data[i] = min(x, self.data[i])
i += i & -i
def get_min(self, i):
# 1からiまで
s = 0
while i > 0:
s = min(s, self.data[i])
i -= i & -i
return s
def solve(N, M, S, Q):
bit_sum = BIT(N)
bit_max = BIT_Max(N)
bit_min = BIT_Min(N)
v = 0
for i in range(N):
if S[i] == '+':
v += 1
bit_sum.add(i+1, 1)
else:
v -= 1
bit_sum.add(i+1, -1)
bit_max.update(i+1, v)
bit_min.update(i+1, v)
last = v
bit_sum_r = BIT(N)
bit_max_r = BIT_Max(N)
bit_min_r = BIT_Min(N)
v = 0
for i in range(N):
if S[N-1-i] == '-':
v += 1
bit_sum_r.add(i+1, 1)
else:
v -= 1
bit_sum_r.add(i+1, -1)
bit_max_r.update(i+1, v)
bit_min_r.update(i+1, v)
for (l, r) in Q:
if r == N-1:
if l == 0:
print(1)
else:
max1 = bit_max.get_max(l)
min1 = bit_min.get_min(l)
print(max1 - min1 + 1)
else:
# [0, l) and [r, N+1)
max1 = bit_max.get_max(l)
min1 = bit_min.get_min(l)
a = bit_sum.get_sum(r+1) - bit_sum.get_sum(l)
max2 = bit_max_r.get_max(N-r-1) + last - a
min2 = bit_min_r.get_min(N-r-1) + last - a
print(max(max1, max2) - min(min1, min2) + 1)
def main():
T0 = int(input())
for _ in range(T0):
N, M = map(int, input().split())
S = input()
Q = []
for _ in range(M):
l, r = map(int, input().split())
Q.append((l-1, r-1))
solve(N, M, S, Q)
if __name__ == '__main__':
main()
```
Yes
| 95,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Submitted Solution:
```
import sys
input=sys.stdin.readline
t = int(input())
while t:
ans = []
t -= 1
n, q = map(int, input().split())
s = input()
left = [[0, 0, 0]]
right = [[0, 0]]
now = plus = minus = 0
for i in s:
if i == "-":
now -= 1
if now < minus:
minus -= 1
elif i == "+":
now += 1
if now > plus:
plus += 1
left.append([plus, minus, now])
now = plus = minus = 0
a = b = 0
for i in s[::-1]:
if i == "-":
a -= 1
b -= 1
a = max(0, a)
elif i == "+":
a += 1
b += 1
b = min(b, 0)
right.append([a, b])
right = right[::-1]
print(left)
print(right)
for i in range(q):
l, r = map(int, input().split())
a, b, now = left[l - 1]
c, d = right[r]
print(a, b, now, c, d)
a = max(a, now + c)
b = min(b, now + d)
ans.append(a-b+1)
print(ans)
```
No
| 95,498 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types:
* increase x by 1;
* decrease x by 1.
You are given m queries of the following format:
* query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the description of t testcases follows.
The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.
The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.
Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.
The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5.
Output
For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.
Example
Input
2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4
Output
1
2
4
4
3
3
4
2
3
2
1
2
2
2
Note
The instructions that remain for each query of the first testcase are:
1. empty program — x was only equal to 0;
2. "-" — x had values 0 and -1;
3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them;
4. "+--+--+" — the distinct values are 1, 0, -1, -2.
Submitted Solution:
```
import sys
#py import math
input = sys.stdin.readline
# x,y,z=map(int,input().split())
# a = list(map(int,input().split()))
# string = input().rstrip()
tests = int(input())
for test in range(tests):
n,m =map(int,input().split())
program = input().rstrip()
value = [0]
currentValue = 0
for step in program:
if step == "-":
currentValue += -1
else:
currentValue += 1
value.append(currentValue)
print(value)
for query in range(m):
l,r =map(int,input().split())
maxLeft = max(value[0:l])
minLeft = min(value[0:l])
if r == n:
print(maxLeft-minLeft+1)
else:
shift = value[l-1]-value[r]
maxValue = max(maxLeft , shift+max(value[r:]))
minValue = min(minLeft , shift+min(value[r:]))
print(maxValue-minValue+1)
```
No
| 95,499 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.