message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Slime has a sequence of positive integers a_1, a_2, …, a_n.
In one operation Orac can choose an arbitrary subsegment [l … r] of this sequence and replace all values a_l, a_{l + 1}, …, a_r to the value of median of \\{a_l, a_{l + 1}, …, a_r\}.
In this problem, for the integer multiset s, the median of s is equal to the ⌊ (|s|+1)/(2)⌋-th smallest number in it. For example, the median of \{1,4,4,6,5\} is 4, and the median of \{1,7,5,8\} is 5.
Slime wants Orac to make a_1 = a_2 = … = a_n = k using these operations.
Orac thinks that it is impossible, and he does not want to waste his time, so he decided to ask you if it is possible to satisfy the Slime's requirement, he may ask you these questions several times.
Input
The first line of the input is a single integer t: the number of queries.
The first line of each query contains two integers n\ (1≤ n≤ 100 000) and k\ (1≤ k≤ 10^9), the second line contains n positive integers a_1,a_2,...,a_n\ (1≤ a_i≤ 10^9)
The total sum of n is at most 100 000.
Output
The output should contain t lines. The i-th line should be equal to 'yes' if it is possible to make all integers k in some number of operations or 'no', otherwise. You can print each letter in lowercase or uppercase.
Example
Input
5
5 3
1 5 2 6 1
1 6
6
3 2
1 2 3
4 3
3 1 2 3
10 3
1 2 3 4 5 6 7 8 9 10
Output
no
yes
yes
no
yes
Note
In the first query, Orac can't turn all elements into 3.
In the second query, a_1=6 is already satisfied.
In the third query, Orac can select the complete array and turn all elements into 2.
In the fourth query, Orac can't turn all elements into 3.
In the fifth query, Orac can select [1,6] at first and then select [2,10].
Submitted Solution:
```
ans = []
for _ in range(int(input())):
n, k = map(int, input().split())
a = list(map(int, input().split()))
if k not in a:
ans.append('no')
continue
if n == 1:
ans.append('yes' if a[0] == k else 'no')
continue
good = False
for i in range(n - 1):
if a[i] >= k and a[i + 1] >= k:
good = True
for i in range(1, n):
if a[i] >= k and a[i - 1] >= k:
good = True
for i in range(1, n - 1):
if a[i - 1] == k and a[i + 1] == k:
good = True
ans.append('yes' if good else 'no')
print('\n'.join(ans))
``` | instruction | 0 | 91,891 | 5 | 183,782 |
No | output | 1 | 91,891 | 5 | 183,783 |
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, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
import sys
z=sys.stdin.readline
o=[]
for _ in range(int(z())):
n=int(z());a=[*map(int,z().split())];s=sum(a)
if s%n:o.append(-1);continue
t=s//n;o.append(3*n-3)
for i in range(2,n+1):v=(i-a[i-1])%i;o+=[1,i,v,i,1,(a[i-1]+v)//i]
for i in range(2,n+1):o+=[1,i,t]
print(' '.join(map(str,o)))
``` | instruction | 0 | 91,916 | 5 | 183,832 |
Yes | output | 1 | 91,916 | 5 | 183,833 |
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, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
_str = str
BUFSIZE = 8192
def str(x=b''):
return x if type(x) is bytes else _str(x).encode()
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 inp():
return sys.stdin.readline()
def mpint():
return map(int, sys.stdin.readline().split(' '))
def itg():
return int(sys.stdin.readline())
# ############################## import
# ############################## main
from random import shuffle
def solve():
n = itg()
arr = [0] + list(mpint())
s = sum(arr)
if s % n:
print(-1)
return
s //= n
required = []
output = []
rng = list(range(2, n + 1))
shuffle(rng)
for i in rng: # collect arr[i] to arr[1]
x, r = divmod(arr[i], i)
i_r = i - r
if r and i_r < arr[1]: # add and collect
output.append((1, i, i_r))
arr[1] -= i_r
arr[i] += i_r
x += 1
if x:
output.append((i, 1, x))
arr[1] += x * i
arr[i] -= x * i
if arr[i]: # arr[i] < i
required.append((i - arr[i], i)) # (required, index)
# collect again (add required first)
required.sort()
for r, i in required:
if arr[1] >= r:
output.append((1, i, r))
output.append((i, 1, 1))
arr[1] += arr[i]
arr[i] = 0
else:
break
for j in range(2, n + 1):
x = s - arr[j]
if x < 0 or arr[1] < x:
break
if x:
output.append((1, j, x))
arr[1] -= x
arr[j] += x
del arr[0]
if arr != [s] * n:
print(-1)
else:
print(len(output))
for p in output:
print(*p)
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
if __name__ == '__main__':
# print("YES" if solve() else "NO")
# print("yes" if solve() else "no")
# solve()
for _ in range(itg()):
# print(*solve())
solve()
# Please check!
``` | instruction | 0 | 91,917 | 5 | 183,834 |
Yes | output | 1 | 91,917 | 5 | 183,835 |
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, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
import sys
input = sys.stdin.readline
for nt in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
s = sum(a)
if s%n!=0:
print (-1)
continue
m = s//n
inc, dec = [], []
num = a[0]
for i in range(1,n):
if a[i]>m:
inc.append([a[i],i+1])
else:
dec.append([a[i],i+1])
ans = []
# print(inc)
# print (dec)
while inc:
x = inc.pop()
diff = x[0]-m
count = (diff-1)//x[1] + 1
ans.append([x[1],1,count])
num += count*x[1]
x[0] -= count*x[1]
diff = m-x[0]
ans.append([1,x[1],diff])
num -= diff
# print (num)
while dec:
x = dec.pop()
diff = m-x[0]
ans.append([1,x[1],diff])
num -= diff
# print (num)
print (len(ans))
for i in ans:
print (*i)
``` | instruction | 0 | 91,919 | 5 | 183,838 |
No | output | 1 | 91,919 | 5 | 183,839 |
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, numbered from 1 to n. You can perform the following operation no more than 3n times:
1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9);
2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i.
After each operation, all elements of the array should be non-negative.
Can you find a sequence of no more than 3n operations after which all elements of the array are equal?
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 10^4.
Output
For each test case print the answer to it as follows:
* if there is no suitable sequence of operations, print -1;
* otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation.
If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.
Example
Input
3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3
Output
2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
ans=[]
if sum(a)%n!=0:
print(-1)
continue
b=sum(a)//n
for i in range(n-1,0,-1):
if a[i]%(i+1)==0:
ans.append([i+1,1,a[i]//(i+1)])
a[0]+=a[i]
a[i]=0
else:
x=i+1-a[i]%(i+1)
if a[x-1]>=x:
ans.append([x,i+1,1])
a[x-1]-=x
a[i]+=x
ans.append([i+1,1,a[i]//(i+1)])
a[0]+=a[i]
a[i]=0
else:
ans.append([1,i+1,x])
a[0]-=x
a[i]+=x
ans.append([i+1,1,a[i]//(i+1)])
a[0]+=a[i]
a[i]=0
for i in range(1,n):
ans.append([1,i+1,b])
l=len(ans)
print(l)
for i in range(l):
print(" ".join(map(str,ans[i])))
``` | instruction | 0 | 91,922 | 5 | 183,844 |
No | output | 1 | 91,922 | 5 | 183,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a square matrix n × n, consisting of non-negative integer numbers. You should find such a way on it that
* starts in the upper left cell of the matrix;
* each following cell is to the right or down from the current cell;
* the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
Input
The first line contains an integer number n (2 ≤ n ≤ 1000), n is the size of the matrix. Then follow n lines containing the matrix elements (non-negative integer numbers not exceeding 109).
Output
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
Examples
Input
3
1 2 3
4 5 6
7 8 9
Output
0
DDRR
Submitted Solution:
```
# def zero(num):
# ret = 0
# for char in str(num)[::-1]:
# if char != '0':
# break
# ret += 1
# return ret
# def count(num, m):
# ret = 0
# while num % m == 0:
# ret += 1
# num //= m
# return ret
# def compare(a, b):
# a2 = count(a, 2)
# a5 = count(a, 5)
# b2 = count(b, 2)
# b5 = count(b, 5)
# if a2 < b2 or a5 < b5:
# return 1
# else:
# return 0
n = int(input())
m = []
for i in range(n):
m.append(list(map(int, input().split())))
dp2 = [[0 for _ in range(n)] for __ in range(n)]
dp5 = [[0 for _ in range(n)] for __ in range(n)]
ans = 0
for i in range(n):
for j in range(n):
if (m[i][j] == 0):
ans = 1
path = 'R' * j + 'D' * i + 'R' * (n - 1 - j) + 'D' * (n - 1 - i)
m[i][j] = 10
for i in range(n):
for j in range(n):
x = m[i][j]
while x % 2 == 0 and x > 0:
dp2[i][j] += 1
x //= 2
while x % 5 == 0 and x > 0:
dp5[i][j] += 1
x //= 5
for i in range(1, n):
dp2[i][0] += dp2[i-1][0]
dp2[0][i] += dp2[0][i-1]
dp5[i][0] += dp5[i-1][0]
dp5[0][i] += dp5[0][i-1]
for i in range(1, n):
for j in range(1, n):
dp2[i][j] += min(dp2[i-1][j], dp2[i][j-1])
dp5[i][j] += min(dp5[i-1][j], dp5[i][j-1])
if dp2[-1][-1] < dp5[-1][-1]:
dp = dp2
else:
dp = dp5
if ans == 1 and dp[-1][-1]:
print(ans)
print(path)
else:
i, j = n - 1, n - 1
path = ''
while i or j:
if not i:
path += 'R'
j -= 1
elif not j:
path += 'D'
i -= 1
else:
if dp[i-1][j] < dp[i][j-1]:
path += 'D'
i -= 1
else:
path += 'R'
j -= 1
path = path[::-1]
print(dp[-1][-1])
print(path)
# dp = []
# for i in range(n):
# dp.append([])
# for j in range(n):
# dp[i].append(['', 1])
# for i in range(n):
# for j in range(n):
# if j == 0 and i == 0:
# dp[i][j][1] = m[i][j]
# elif j == 0:
# dp[i][j][1] = dp[i-1][j][1] * m[i][j]
# dp[i][j][0] = dp[i-1][j][0] + 'D'
# elif i == 0:
# dp[i][j][1] = dp[i][j-1][1] * m[i][j]
# dp[i][j][0] = dp[i][j-1][0] + 'R'
# else:
# down = dp[i-1][j][1] * m[i][j]
# right = dp[i][j-1][1] * m[i][j]
# # print(right, down, count(right, 2), count(
# # right, 5), count(down, 2), count(down, 5))
# if compare(right, down) and zero(right) <= zero(down):
# dp[i][j][1] = right
# dp[i][j][0] = dp[i][j-1][0] + 'R'
# else:
# dp[i][j][1] = down
# dp[i][j][0] = dp[i-1][j][0] + 'D'
# for i in range(n):
# print(dp[i])
# print(zero(dp[-1][-1][1]))
# print(dp[-1][-1][0])
``` | instruction | 0 | 92,061 | 5 | 184,122 |
Yes | output | 1 | 92,061 | 5 | 184,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a square matrix n × n, consisting of non-negative integer numbers. You should find such a way on it that
* starts in the upper left cell of the matrix;
* each following cell is to the right or down from the current cell;
* the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
Input
The first line contains an integer number n (2 ≤ n ≤ 1000), n is the size of the matrix. Then follow n lines containing the matrix elements (non-negative integer numbers not exceeding 109).
Output
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
Examples
Input
3
1 2 3
4 5 6
7 8 9
Output
0
DDRR
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from math import inf
def occ(x, a):
ans = 0
if x == 0:
return 0
while not x % a:
x //= a
ans += 1
return ans
def main():
n = int(input())
min_2 = [[inf for _ in range(n + 1)] for _ in range(n + 1)]
min_5 = [[inf for _ in range(n + 1)] for _ in range(n + 1)]
arr = [list(map(int, input().split())) for _ in range(n)]
min_2[0][1] = 0
min_5[0][1] = 0
fl = 0
mini = inf
for i in range(1, n + 1):
for j in range(1, n + 1):
min_2[i][j] = min(min_2[i - 1][j], min_2[i][j - 1]) + occ(arr[i - 1][j - 1], 2)
min_5[i][j] = min(min_5[i - 1][j], min_5[i][j - 1]) + occ(arr[i - 1][j - 1], 5)
mini = min(mini,min(min_2[i][j],min_5[i][j]))
if not arr[i-1][j-1] and mini:
fl,ind = 1,j-1
if fl:
print(1)
sys.stdout.write('R' * ind + 'D' * (n - 1) + 'R' * (n - ind - 1))
exit()
if min_2[-1][-1] < min_5[-1][-1]:
print(min_2[-1][-1])
flg = min_2
else:
print(min_5[-1][-1])
flg = min_5
ans = []
x, y = n, n
for _ in range(2 * n - 2):
if flg[x - 1][y] < flg[x][y - 1]:
ans.append('D')
x -= 1
else:
ans.append('R')
y -= 1
sys.stdout.write(''.join(ans[::-1]))
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | instruction | 0 | 92,062 | 5 | 184,124 |
No | output | 1 | 92,062 | 5 | 184,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bitwise exclusive OR (or bitwise addition modulo two) is a binary operation which is equivalent to applying logical exclusive OR to every pair of bits located on the same positions in binary notation of operands. In other words, a binary digit of the result is equal to 1 if and only if bits on the respective positions in the operands are different.
For example, if X = 10910 = 11011012, Y = 4110 = 1010012, then:
X xor Y = 6810 = 10001002.
Write a program, which takes two non-negative integers A and B as an input and finds two non-negative integers X and Y, which satisfy the following conditions:
* A = X + Y
* B = X xor Y, where xor is bitwise exclusive or.
* X is the smallest number among all numbers for which the first two conditions are true.
Input
The first line contains integer number A and the second line contains integer number B (0 ≤ A, B ≤ 264 - 1).
Output
The only output line should contain two integer non-negative numbers X and Y. Print the only number -1 if there is no answer.
Examples
Input
142
76
Output
33 109
Submitted Solution:
```
a = int(input())
b = int(input())
if (a+b)%2!=0:
print(-1)
else:
print((a-b)//2 , (a+b)//2)
``` | instruction | 0 | 92,275 | 5 | 184,550 |
Yes | output | 1 | 92,275 | 5 | 184,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bitwise exclusive OR (or bitwise addition modulo two) is a binary operation which is equivalent to applying logical exclusive OR to every pair of bits located on the same positions in binary notation of operands. In other words, a binary digit of the result is equal to 1 if and only if bits on the respective positions in the operands are different.
For example, if X = 10910 = 11011012, Y = 4110 = 1010012, then:
X xor Y = 6810 = 10001002.
Write a program, which takes two non-negative integers A and B as an input and finds two non-negative integers X and Y, which satisfy the following conditions:
* A = X + Y
* B = X xor Y, where xor is bitwise exclusive or.
* X is the smallest number among all numbers for which the first two conditions are true.
Input
The first line contains integer number A and the second line contains integer number B (0 ≤ A, B ≤ 264 - 1).
Output
The only output line should contain two integer non-negative numbers X and Y. Print the only number -1 if there is no answer.
Examples
Input
142
76
Output
33 109
Submitted Solution:
```
a = int(input())
b = int(input())
if (a - b) % 2 == 1:
print("-1")
exit(0)
x = (a - b) // 2
y = a - x
print(x, y)
``` | instruction | 0 | 92,276 | 5 | 184,552 |
Yes | output | 1 | 92,276 | 5 | 184,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bitwise exclusive OR (or bitwise addition modulo two) is a binary operation which is equivalent to applying logical exclusive OR to every pair of bits located on the same positions in binary notation of operands. In other words, a binary digit of the result is equal to 1 if and only if bits on the respective positions in the operands are different.
For example, if X = 10910 = 11011012, Y = 4110 = 1010012, then:
X xor Y = 6810 = 10001002.
Write a program, which takes two non-negative integers A and B as an input and finds two non-negative integers X and Y, which satisfy the following conditions:
* A = X + Y
* B = X xor Y, where xor is bitwise exclusive or.
* X is the smallest number among all numbers for which the first two conditions are true.
Input
The first line contains integer number A and the second line contains integer number B (0 ≤ A, B ≤ 264 - 1).
Output
The only output line should contain two integer non-negative numbers X and Y. Print the only number -1 if there is no answer.
Examples
Input
142
76
Output
33 109
Submitted Solution:
```
a = int(input())
b = int(input())
if (a - b) % 2 == 1:
print("-1")
exit(0)
x = (a - b) // 2
y = a - x
print(x, y)
# Made By Mostafa_Khaled
``` | instruction | 0 | 92,277 | 5 | 184,554 |
Yes | output | 1 | 92,277 | 5 | 184,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bitwise exclusive OR (or bitwise addition modulo two) is a binary operation which is equivalent to applying logical exclusive OR to every pair of bits located on the same positions in binary notation of operands. In other words, a binary digit of the result is equal to 1 if and only if bits on the respective positions in the operands are different.
For example, if X = 10910 = 11011012, Y = 4110 = 1010012, then:
X xor Y = 6810 = 10001002.
Write a program, which takes two non-negative integers A and B as an input and finds two non-negative integers X and Y, which satisfy the following conditions:
* A = X + Y
* B = X xor Y, where xor is bitwise exclusive or.
* X is the smallest number among all numbers for which the first two conditions are true.
Input
The first line contains integer number A and the second line contains integer number B (0 ≤ A, B ≤ 264 - 1).
Output
The only output line should contain two integer non-negative numbers X and Y. Print the only number -1 if there is no answer.
Examples
Input
142
76
Output
33 109
Submitted Solution:
```
x = int(input())
y = int(input())
if (y - x) % 2 == 0:
print((y - x) // 2, (x + y) // 2, sep=' ', end='\n')
else:
print(-1)
``` | instruction | 0 | 92,281 | 5 | 184,562 |
No | output | 1 | 92,281 | 5 | 184,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer N, find two permutations:
1. Permutation p of numbers from 1 to N such that pi ≠ i and pi & i = 0 for all i = 1, 2, ..., N.
2. Permutation q of numbers from 1 to N such that qi ≠ i and qi & i ≠ 0 for all i = 1, 2, ..., N.
& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The input consists of one line containing a single integer N (1 ≤ N ≤ 105).
Output
For each subtask, if the required permutation doesn't exist, output a single line containing the word "NO"; otherwise output the word "YES" in the first line and N elements of the permutation, separated by spaces, in the second line. If there are several possible permutations in a subtask, output any of them.
Examples
Input
3
Output
NO
NO
Input
6
Output
YES
6 5 4 3 2 1
YES
3 6 2 5 1 4
Submitted Solution:
```
from math import log2, floor
n=int(input())
# n=100
if n%2:
print('NO')
else:
print("YES")
k=floor(log2(n))+1
res = [0 for i in range(n)]
occupied = [0 for i in range(n)]
switch = 0
for i in range(n,0,-1):
val = 2**k-1-i
if val > n:
k-=1
val = 2**k-1-i
if occupied[val-1]:
while occupied[val-1]:
k-=1
val = 2**k-1-i
val = 2**k-1-i
occupied[val-1] = 1
res[i-1] = val
for i in res:
print(i,end=' ')
print()
if n in [1,2,3,4,5]:
print('NO')
elif int(log2(n)) == log2(n):
print('NO')
elif n == 6:
print('YES')
print('3 6 2 5 1 4')
else:
print('YES')
print('7 3 2 5 6 4 1',end=' ')
for i in range(8,n):
p = pow(2,floor(log2(i))+1)
if i < pow(2,floor(log2(n))):
print(p*3//2-1-i,end=' ')
else:
print(i+1,end=' ')
print(pow(2,floor(log2(n))))
``` | instruction | 0 | 92,333 | 5 | 184,666 |
No | output | 1 | 92,333 | 5 | 184,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer N, find two permutations:
1. Permutation p of numbers from 1 to N such that pi ≠ i and pi & i = 0 for all i = 1, 2, ..., N.
2. Permutation q of numbers from 1 to N such that qi ≠ i and qi & i ≠ 0 for all i = 1, 2, ..., N.
& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The input consists of one line containing a single integer N (1 ≤ N ≤ 105).
Output
For each subtask, if the required permutation doesn't exist, output a single line containing the word "NO"; otherwise output the word "YES" in the first line and N elements of the permutation, separated by spaces, in the second line. If there are several possible permutations in a subtask, output any of them.
Examples
Input
3
Output
NO
NO
Input
6
Output
YES
6 5 4 3 2 1
YES
3 6 2 5 1 4
Submitted Solution:
```
def test(x, i):
i = list(i)
ok = True
for j in range(x):
if (i[j] == j+1 or (i[j]&(j+1) != 0)):
ok = False
if ok:
print(i)
def comp(n):
return 2**len(bin(n)[2:])-1-n
n = int(input())
nn = n
if (n%2 == 0):
x = []
while (n != 0):
#add n to comp(n) to the front of x
for i in range(comp(n), n+1):
x.append(i)
n = comp(n)-1
x.reverse()
print("YES")
print(' '.join([str(i) for i in x]))
else:
print("NO")
print("NO")
pow2 = [2**i for i in range(20)]
def make(n):
if n <= 5: return []
if n == 6: return [3, 6, 1, 5, 4, 2]
if n == 7: return [3, 6, 1, 5, 4, 7, 2]
if n in pow2:
return []
shift = 2**(len(bin(n)[2:])-1)
array = [i for i in range(shift, n+1)]
array = array[1:] + [array[0]]
return make(shift-1) + array
n = nn
k = make(n)
if k == []:
print("NO")
print("NO")
else:
print("YES")
print(' '.join([str(i) for i in k]))
``` | instruction | 0 | 92,334 | 5 | 184,668 |
No | output | 1 | 92,334 | 5 | 184,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer N, find two permutations:
1. Permutation p of numbers from 1 to N such that pi ≠ i and pi & i = 0 for all i = 1, 2, ..., N.
2. Permutation q of numbers from 1 to N such that qi ≠ i and qi & i ≠ 0 for all i = 1, 2, ..., N.
& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The input consists of one line containing a single integer N (1 ≤ N ≤ 105).
Output
For each subtask, if the required permutation doesn't exist, output a single line containing the word "NO"; otherwise output the word "YES" in the first line and N elements of the permutation, separated by spaces, in the second line. If there are several possible permutations in a subtask, output any of them.
Examples
Input
3
Output
NO
NO
Input
6
Output
YES
6 5 4 3 2 1
YES
3 6 2 5 1 4
Submitted Solution:
```
from math import log2, floor
n=int(input())
# n=3
if n%2:
print('NO')
else:
print("YES")
k=floor(log2(n))+1
res = [0 for i in range(n)]
occupied = [0 for i in range(n)]
switch = 0
for i in range(n,0,-1):
val = 2**k-1-i
if val > n:
k-=1
val = 2**k-1-i
if occupied[val-1]:
while occupied[val-1]:
k-=1
val = 2**k-1-i
val = 2**k-1-i
occupied[val-1] = 1
res[i-1] = val
for i in res:
print(i,end=' ')
print()
if n in [1,2,3,4,5]:
print('NO')
elif int(log2(n)) == log2(n):
print('NO')
elif n == 6:
print('YES')
print('3 6 2 5 1 4')
else:
print('YES')
print('7 3 2 5 6 4 1',end=' ')
for i in range(8,n+1):
p = pow(2,floor(log2(n)))
if i < p:
print(p*3//2-1-i,end=' ')
else:
print(n+p-i,end=' ')
print()
``` | instruction | 0 | 92,335 | 5 | 184,670 |
No | output | 1 | 92,335 | 5 | 184,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given an integer N, find two permutations:
1. Permutation p of numbers from 1 to N such that pi ≠ i and pi & i = 0 for all i = 1, 2, ..., N.
2. Permutation q of numbers from 1 to N such that qi ≠ i and qi & i ≠ 0 for all i = 1, 2, ..., N.
& is the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
Input
The input consists of one line containing a single integer N (1 ≤ N ≤ 105).
Output
For each subtask, if the required permutation doesn't exist, output a single line containing the word "NO"; otherwise output the word "YES" in the first line and N elements of the permutation, separated by spaces, in the second line. If there are several possible permutations in a subtask, output any of them.
Examples
Input
3
Output
NO
NO
Input
6
Output
YES
6 5 4 3 2 1
YES
3 6 2 5 1 4
Submitted Solution:
```
from math import log2, floor
n=int(input())
# n=100
if n%2:
print('NO')
else:
print("YES")
k=floor(log2(n))+1
res = [0 for i in range(n)]
occupied = [0 for i in range(n)]
switch = 0
for i in range(n,0,-1):
val = 2**k-1-i
if val > n:
k-=1
val = 2**k-1-i
if occupied[val-1]:
while occupied[val-1]:
k-=1
val = 2**k-1-i
val = 2**k-1-i
occupied[val-1] = 1
res[i-1] = val
for i in res:
print(i,end=' ')
print()
if n in [1,2,3,4,5]:
print('NO')
elif int(log2(n)) == log2(n):
print('NO')
elif n == 6:
print('YES')
print('3 6 2 5 1 4')
else:
print('YES')
print('7 3 2 5 6 4 1',end=' ')
for i in range(8,n+1):
p = pow(2,floor(log2(i))+1)
if i < pow(2,floor(log2(n))):
print(p*3//2-1-i,end=' ')
else:
print(n+p//2-i,end=' ')
print()
``` | instruction | 0 | 92,336 | 5 | 184,672 |
No | output | 1 | 92,336 | 5 | 184,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4
Submitted Solution:
```
N = int(input())
A = []
for i in range(N):
A.append(int(input()))
count = 1
tmp = A[0]
for i in range(1,N):
if tmp >= A[i]:
count = count+1
tmp = A[i]
print(count)
``` | instruction | 0 | 92,446 | 5 | 184,892 |
No | output | 1 | 92,446 | 5 | 184,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4
Submitted Solution:
```
N = int(input())
A = [int(input()) for _ in range(N)]
B = [A[0]]
for i in range(1,N):
if A[i]<=B[-1]:
B.append(A[i])
elif A[i]>B[0]:
B[0] = A[i]
else:
l = 0
r = len(B)-1
while l+1<r:
c = (l+r)//2
if B[c]>=A[i]:
l = c
elif B[c]<A[i]:
r = c
if A[l]<A[i]:
A[l] = A[i]
else:
A[r] = A[i]
print(len(B))
``` | instruction | 0 | 92,448 | 5 | 184,896 |
No | output | 1 | 92,448 | 5 | 184,897 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There is an arithmetic progression A with the number of terms N, the first term a, and the tolerance d. Since M statements that update the sequence are given in the following format, find the value of the K item when the sequence A is updated M times in the given order.
* The i-th statement is represented by the three integers xi, yi, zi (1 ≤ i ≤ M).
* If xi is 0, swap the value of yi item and the value of zi item.
* If xi is 1, rewrite the value of yi item to the value of zi item.
Constraints
* 1 ≤ N ≤ 108
* 1 ≤ a ≤ 5
* 1 ≤ d ≤ 5
* 1 ≤ M ≤ 10
* 0 ≤ xi ≤ 1 (1 ≤ i ≤ M)
* 1 ≤ yi ≤ N (1 ≤ i ≤ M)
* 1 ≤ zi ≤ N (1 ≤ i ≤ M)
* yi ≠ zi (1 ≤ i ≤ M)
* 1 ≤ K ≤ N
Input
N
a d
M
x1 y1 z1
x2 y2 z2
...
xM yM zM
K
The first line is given one integer N. On the second line, two integers a and d are given, separated by blanks. On the third line, one integer M is given. Of the M lines from the 4th line, the ith line is given three integers xi, yi, and zi representing the i-th statement, separated by blanks. The last line is given one integer K.
Output
Output the K item when the sequence A is updated M times in the order given by the input.
Examples
Input
5
2 3
3
0 2 5
0 3 5
1 2 4
2
Output
11
Input
6
5 5
4
0 1 2
1 1 2
0 4 1
1 1 6
3
Output
15 | instruction | 0 | 92,560 | 5 | 185,120 |
"Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
N = int(input())
a, d = map(int, input().split())
M = int(input())
data = {}
def get_i(data, i):
if i in data:
return data[i]
else:
return a + d * i
def data_swap(y, z, data):
tmp = get_i(data, y)
data[y] = get_i(data, z)
data[z] = tmp
def data_write(y, z, data):
data[y] = get_i(data, z)
for m in range(M):
# for i in range(5):
# print(get_i(data, i))
# print("---")
x, y, z = map(int, input().split())
y -= 1
z -= 1
if x == 0:
data_swap(y, z, data)
else:
data_write(y, z, data)
K = int(input())
print(get_i(data, K-1))
``` | output | 1 | 92,560 | 5 | 185,121 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There is an arithmetic progression A with the number of terms N, the first term a, and the tolerance d. Since M statements that update the sequence are given in the following format, find the value of the K item when the sequence A is updated M times in the given order.
* The i-th statement is represented by the three integers xi, yi, zi (1 ≤ i ≤ M).
* If xi is 0, swap the value of yi item and the value of zi item.
* If xi is 1, rewrite the value of yi item to the value of zi item.
Constraints
* 1 ≤ N ≤ 108
* 1 ≤ a ≤ 5
* 1 ≤ d ≤ 5
* 1 ≤ M ≤ 10
* 0 ≤ xi ≤ 1 (1 ≤ i ≤ M)
* 1 ≤ yi ≤ N (1 ≤ i ≤ M)
* 1 ≤ zi ≤ N (1 ≤ i ≤ M)
* yi ≠ zi (1 ≤ i ≤ M)
* 1 ≤ K ≤ N
Input
N
a d
M
x1 y1 z1
x2 y2 z2
...
xM yM zM
K
The first line is given one integer N. On the second line, two integers a and d are given, separated by blanks. On the third line, one integer M is given. Of the M lines from the 4th line, the ith line is given three integers xi, yi, and zi representing the i-th statement, separated by blanks. The last line is given one integer K.
Output
Output the K item when the sequence A is updated M times in the order given by the input.
Examples
Input
5
2 3
3
0 2 5
0 3 5
1 2 4
2
Output
11
Input
6
5 5
4
0 1 2
1 1 2
0 4 1
1 1 6
3
Output
15 | instruction | 0 | 92,561 | 5 | 185,122 |
"Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
#B
n = int(input())
a,d = map(int,input().split())
A = []
m = int(input())
x = [0] * m
y = [0] * m
z = [0] * m
NumOfTerms = []
for i in range(m):
x[i],y[i],z[i] = map(int,input().split())
NumOfTerms.append(y[i])
NumOfTerms.append(z[i])
num = set(NumOfTerms)
for i in num:
A.append([i, a+d*(i-1)])
for i in range(len(x)):
if x[i] == 0:
for j in range(len(A)):
if A[j][0] == y[i]:
yj = j
if A[j][0] == z[i]:
zj = j
tmp = A[yj][1]
A[yj][1] = A[zj][1]
A[zj][1] = tmp
else:
for j in range(len(A)):
if A[j][0] == y[i]:
yj = j
if A[j][0] == z[i]:
zj = j
A[yj][1] = A[zj][1]
k = int(input())
for j in range(len(num)):
if A[j][0] == k:
print(A[j][1])
exit()
print(a+d*(k-1))
``` | output | 1 | 92,561 | 5 | 185,123 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There is an arithmetic progression A with the number of terms N, the first term a, and the tolerance d. Since M statements that update the sequence are given in the following format, find the value of the K item when the sequence A is updated M times in the given order.
* The i-th statement is represented by the three integers xi, yi, zi (1 ≤ i ≤ M).
* If xi is 0, swap the value of yi item and the value of zi item.
* If xi is 1, rewrite the value of yi item to the value of zi item.
Constraints
* 1 ≤ N ≤ 108
* 1 ≤ a ≤ 5
* 1 ≤ d ≤ 5
* 1 ≤ M ≤ 10
* 0 ≤ xi ≤ 1 (1 ≤ i ≤ M)
* 1 ≤ yi ≤ N (1 ≤ i ≤ M)
* 1 ≤ zi ≤ N (1 ≤ i ≤ M)
* yi ≠ zi (1 ≤ i ≤ M)
* 1 ≤ K ≤ N
Input
N
a d
M
x1 y1 z1
x2 y2 z2
...
xM yM zM
K
The first line is given one integer N. On the second line, two integers a and d are given, separated by blanks. On the third line, one integer M is given. Of the M lines from the 4th line, the ith line is given three integers xi, yi, and zi representing the i-th statement, separated by blanks. The last line is given one integer K.
Output
Output the K item when the sequence A is updated M times in the order given by the input.
Examples
Input
5
2 3
3
0 2 5
0 3 5
1 2 4
2
Output
11
Input
6
5 5
4
0 1 2
1 1 2
0 4 1
1 1 6
3
Output
15 | instruction | 0 | 92,562 | 5 | 185,124 |
"Correct Solution:
```
# coding:utf-8
import sys
from collections import Counter, defaultdict
INF = float('inf')
MOD = 10 ** 9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x) - 1 for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())
def SI(): return input()
n = II()
a, b = LI()
m = II()
D = defaultdict(int)
for i in range(m):
x, y, z = LI()
y -= 1
z -= 1
tmp_y = D[y] if y in D.keys() else a + b * y
tmp_z = D[z] if z in D.keys() else a + b * z
if x == 0:
D[z] = tmp_y
D[y] = tmp_z
k = II()
k -= 1
print(D[k] if k in D.keys() else a + b * k)
``` | output | 1 | 92,562 | 5 | 185,125 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There is an arithmetic progression A with the number of terms N, the first term a, and the tolerance d. Since M statements that update the sequence are given in the following format, find the value of the K item when the sequence A is updated M times in the given order.
* The i-th statement is represented by the three integers xi, yi, zi (1 ≤ i ≤ M).
* If xi is 0, swap the value of yi item and the value of zi item.
* If xi is 1, rewrite the value of yi item to the value of zi item.
Constraints
* 1 ≤ N ≤ 108
* 1 ≤ a ≤ 5
* 1 ≤ d ≤ 5
* 1 ≤ M ≤ 10
* 0 ≤ xi ≤ 1 (1 ≤ i ≤ M)
* 1 ≤ yi ≤ N (1 ≤ i ≤ M)
* 1 ≤ zi ≤ N (1 ≤ i ≤ M)
* yi ≠ zi (1 ≤ i ≤ M)
* 1 ≤ K ≤ N
Input
N
a d
M
x1 y1 z1
x2 y2 z2
...
xM yM zM
K
The first line is given one integer N. On the second line, two integers a and d are given, separated by blanks. On the third line, one integer M is given. Of the M lines from the 4th line, the ith line is given three integers xi, yi, and zi representing the i-th statement, separated by blanks. The last line is given one integer K.
Output
Output the K item when the sequence A is updated M times in the order given by the input.
Examples
Input
5
2 3
3
0 2 5
0 3 5
1 2 4
2
Output
11
Input
6
5 5
4
0 1 2
1 1 2
0 4 1
1 1 6
3
Output
15 | instruction | 0 | 92,563 | 5 | 185,126 |
"Correct Solution:
```
input()
a, d = map(int, input().split())
x = [tuple(map(int, input().split())) for _ in range(int(input()))]
k = int(input())
for i in x[::-1]:
if i[0] == 1:
if i[1] == k:
k = i[2]
else:
if i[1] == k:
k = i[2]
elif i[2] == k:
k = i[1]
print(a + d * (k - 1))
``` | output | 1 | 92,563 | 5 | 185,127 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There is an arithmetic progression A with the number of terms N, the first term a, and the tolerance d. Since M statements that update the sequence are given in the following format, find the value of the K item when the sequence A is updated M times in the given order.
* The i-th statement is represented by the three integers xi, yi, zi (1 ≤ i ≤ M).
* If xi is 0, swap the value of yi item and the value of zi item.
* If xi is 1, rewrite the value of yi item to the value of zi item.
Constraints
* 1 ≤ N ≤ 108
* 1 ≤ a ≤ 5
* 1 ≤ d ≤ 5
* 1 ≤ M ≤ 10
* 0 ≤ xi ≤ 1 (1 ≤ i ≤ M)
* 1 ≤ yi ≤ N (1 ≤ i ≤ M)
* 1 ≤ zi ≤ N (1 ≤ i ≤ M)
* yi ≠ zi (1 ≤ i ≤ M)
* 1 ≤ K ≤ N
Input
N
a d
M
x1 y1 z1
x2 y2 z2
...
xM yM zM
K
The first line is given one integer N. On the second line, two integers a and d are given, separated by blanks. On the third line, one integer M is given. Of the M lines from the 4th line, the ith line is given three integers xi, yi, and zi representing the i-th statement, separated by blanks. The last line is given one integer K.
Output
Output the K item when the sequence A is updated M times in the order given by the input.
Examples
Input
5
2 3
3
0 2 5
0 3 5
1 2 4
2
Output
11
Input
6
5 5
4
0 1 2
1 1 2
0 4 1
1 1 6
3
Output
15 | instruction | 0 | 92,564 | 5 | 185,128 |
"Correct Solution:
```
# AOJ 1563: Array Update
# Python3 2018.7.13 bal4u
def get(id):
global a, d
if id in dic: ans = dic[id]
else: ans = a + (id-1)*d;
return ans
dic = {}
N = int(input())
a, d = map(int, input().split())
M = int(input())
for i in range(M):
x, y, z = map(int, input().split())
if x == 0:
vy, vz = get(y), get(z)
dic[y] = vz
dic[z] = vy
else:
vz = get(z)
dic[y] = vz
K = int(input())
print(get(K))
``` | output | 1 | 92,564 | 5 | 185,129 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There is an arithmetic progression A with the number of terms N, the first term a, and the tolerance d. Since M statements that update the sequence are given in the following format, find the value of the K item when the sequence A is updated M times in the given order.
* The i-th statement is represented by the three integers xi, yi, zi (1 ≤ i ≤ M).
* If xi is 0, swap the value of yi item and the value of zi item.
* If xi is 1, rewrite the value of yi item to the value of zi item.
Constraints
* 1 ≤ N ≤ 108
* 1 ≤ a ≤ 5
* 1 ≤ d ≤ 5
* 1 ≤ M ≤ 10
* 0 ≤ xi ≤ 1 (1 ≤ i ≤ M)
* 1 ≤ yi ≤ N (1 ≤ i ≤ M)
* 1 ≤ zi ≤ N (1 ≤ i ≤ M)
* yi ≠ zi (1 ≤ i ≤ M)
* 1 ≤ K ≤ N
Input
N
a d
M
x1 y1 z1
x2 y2 z2
...
xM yM zM
K
The first line is given one integer N. On the second line, two integers a and d are given, separated by blanks. On the third line, one integer M is given. Of the M lines from the 4th line, the ith line is given three integers xi, yi, and zi representing the i-th statement, separated by blanks. The last line is given one integer K.
Output
Output the K item when the sequence A is updated M times in the order given by the input.
Examples
Input
5
2 3
3
0 2 5
0 3 5
1 2 4
2
Output
11
Input
6
5 5
4
0 1 2
1 1 2
0 4 1
1 1 6
3
Output
15 | instruction | 0 | 92,565 | 5 | 185,130 |
"Correct Solution:
```
n = int(input())
a, d = map(int, input().split())
mem = {}
def get_mem(num):
if num in mem:return mem[num]
else:return num
m = int(input())
for _ in range(m):
x, y, z = map(int, input().split())
if x == 0:
ny, nz = get_mem(y), get_mem(z)
mem[y] = nz
mem[z] = ny
if x == 1:
nz = get_mem(z)
mem[y] = nz
k = get_mem(int(input()))
print(a + d * (k - 1))
``` | output | 1 | 92,565 | 5 | 185,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
There is an arithmetic progression A with the number of terms N, the first term a, and the tolerance d. Since M statements that update the sequence are given in the following format, find the value of the K item when the sequence A is updated M times in the given order.
* The i-th statement is represented by the three integers xi, yi, zi (1 ≤ i ≤ M).
* If xi is 0, swap the value of yi item and the value of zi item.
* If xi is 1, rewrite the value of yi item to the value of zi item.
Constraints
* 1 ≤ N ≤ 108
* 1 ≤ a ≤ 5
* 1 ≤ d ≤ 5
* 1 ≤ M ≤ 10
* 0 ≤ xi ≤ 1 (1 ≤ i ≤ M)
* 1 ≤ yi ≤ N (1 ≤ i ≤ M)
* 1 ≤ zi ≤ N (1 ≤ i ≤ M)
* yi ≠ zi (1 ≤ i ≤ M)
* 1 ≤ K ≤ N
Input
N
a d
M
x1 y1 z1
x2 y2 z2
...
xM yM zM
K
The first line is given one integer N. On the second line, two integers a and d are given, separated by blanks. On the third line, one integer M is given. Of the M lines from the 4th line, the ith line is given three integers xi, yi, and zi representing the i-th statement, separated by blanks. The last line is given one integer K.
Output
Output the K item when the sequence A is updated M times in the order given by the input.
Examples
Input
5
2 3
3
0 2 5
0 3 5
1 2 4
2
Output
11
Input
6
5 5
4
0 1 2
1 1 2
0 4 1
1 1 6
3
Output
15
Submitted Solution:
```
import sys
def init_array(N, a, d):
return [a+d*n for n in range(N)]
def main():
N = int(input())
ins = input().split()
a = int(ins[0])
b = int(ins[1])
M = int(input())
arrays = init_array(N, a, b)
for i in range(M):
ops = [int(x) for x in input().split()]
ops[1] -= 1
ops[2] -= 1
if ops[0] == 0:
tmp = arrays[ops[1]]
arrays[ops[1]] = arrays[ops[2]]
arrays[ops[2]] = tmp
else:
arrays[ops[1]] = arrays[ops[2]]
K = int(input())
print(arrays[K-1])
return 0
if __name__ == "__main__":
sys.exit(main())
``` | instruction | 0 | 92,566 | 5 | 185,132 |
No | output | 1 | 92,566 | 5 | 185,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
There is an arithmetic progression A with the number of terms N, the first term a, and the tolerance d. Since M statements that update the sequence are given in the following format, find the value of the K item when the sequence A is updated M times in the given order.
* The i-th statement is represented by the three integers xi, yi, zi (1 ≤ i ≤ M).
* If xi is 0, swap the value of yi item and the value of zi item.
* If xi is 1, rewrite the value of yi item to the value of zi item.
Constraints
* 1 ≤ N ≤ 108
* 1 ≤ a ≤ 5
* 1 ≤ d ≤ 5
* 1 ≤ M ≤ 10
* 0 ≤ xi ≤ 1 (1 ≤ i ≤ M)
* 1 ≤ yi ≤ N (1 ≤ i ≤ M)
* 1 ≤ zi ≤ N (1 ≤ i ≤ M)
* yi ≠ zi (1 ≤ i ≤ M)
* 1 ≤ K ≤ N
Input
N
a d
M
x1 y1 z1
x2 y2 z2
...
xM yM zM
K
The first line is given one integer N. On the second line, two integers a and d are given, separated by blanks. On the third line, one integer M is given. Of the M lines from the 4th line, the ith line is given three integers xi, yi, and zi representing the i-th statement, separated by blanks. The last line is given one integer K.
Output
Output the K item when the sequence A is updated M times in the order given by the input.
Examples
Input
5
2 3
3
0 2 5
0 3 5
1 2 4
2
Output
11
Input
6
5 5
4
0 1 2
1 1 2
0 4 1
1 1 6
3
Output
15
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
#B
n = int(input())
a,d = map(int,input().split())
A = []
for i in range(0, n):
A.append(a + d*i)
m = int(input())
for i in range(m):
x,y,z = map(int,input().split())
if x == 0:
tmp = A[y-1]
A[y-1] = A[z-1]
A[z-1] = tmp
else:
A[y-1] = A[z-1]
k = int(input())
print(A[k-1])
``` | instruction | 0 | 92,567 | 5 | 185,134 |
No | output | 1 | 92,567 | 5 | 185,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
There is an arithmetic progression A with the number of terms N, the first term a, and the tolerance d. Since M statements that update the sequence are given in the following format, find the value of the K item when the sequence A is updated M times in the given order.
* The i-th statement is represented by the three integers xi, yi, zi (1 ≤ i ≤ M).
* If xi is 0, swap the value of yi item and the value of zi item.
* If xi is 1, rewrite the value of yi item to the value of zi item.
Constraints
* 1 ≤ N ≤ 108
* 1 ≤ a ≤ 5
* 1 ≤ d ≤ 5
* 1 ≤ M ≤ 10
* 0 ≤ xi ≤ 1 (1 ≤ i ≤ M)
* 1 ≤ yi ≤ N (1 ≤ i ≤ M)
* 1 ≤ zi ≤ N (1 ≤ i ≤ M)
* yi ≠ zi (1 ≤ i ≤ M)
* 1 ≤ K ≤ N
Input
N
a d
M
x1 y1 z1
x2 y2 z2
...
xM yM zM
K
The first line is given one integer N. On the second line, two integers a and d are given, separated by blanks. On the third line, one integer M is given. Of the M lines from the 4th line, the ith line is given three integers xi, yi, and zi representing the i-th statement, separated by blanks. The last line is given one integer K.
Output
Output the K item when the sequence A is updated M times in the order given by the input.
Examples
Input
5
2 3
3
0 2 5
0 3 5
1 2 4
2
Output
11
Input
6
5 5
4
0 1 2
1 1 2
0 4 1
1 1 6
3
Output
15
Submitted Solution:
```
n = int(input())
a, d = map(int, input().split())
s = [0] + [a + d * i for i in range(n)]
for _ in range(int(input())):
x, y, z = map(int, input().split())
if x == 0:s[y], s[z] = s[z], s[y]
else:s[y] = s[z]
print(s[int(input())])
``` | instruction | 0 | 92,568 | 5 | 185,136 |
No | output | 1 | 92,568 | 5 | 185,137 |
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
mmemewwemeww
Output
Cat | instruction | 0 | 92,575 | 5 | 185,150 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
s = S()
l = len(s)
d = collections.defaultdict(int)
m = {}
m[''] = True
def f(s):
if s in m:
return m[s]
if s[0] != 'm' or s[-1] != 'w':
m[s] = False
return False
l = len(s)
for i in range(1,l-1):
if s[i] != 'e':
continue
if f(s[1:i]) and f(s[i+1:-1]):
m[s] = True
return True
m[s] = False
return False
if f(s):
return 'Cat'
return 'Rabbit'
print(main())
``` | output | 1 | 92,575 | 5 | 185,151 |
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
mmemewwemeww
Output
Cat | instruction | 0 | 92,576 | 5 | 185,152 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**5)
def LI(): return [int(x) for x in input().split()]
def LF(): return [float(x) for x in input().split()]
def LI_(): return [-1*int(x) for x in input().split()]
def II(): return int(input())
def IF(): return float(input())
def LM(func,n): return [[func(x) for x in input().split()]for i in range(n)]
mod = 1000000007
inf = float('INF')
def isCat(S):
#print(S)
if len(S) == 0:
return True
if S[0] != "m" or S[-1] != "w":
return False
c = 0
i=0
for i in range(1,len(S)-1):
if S[i] == 'm':
c += 1
elif S[i] == 'w':
c -= 1
if c ==0:
break
if S[1] == 'e':
i = 0
return S[i+1] == 'e' and isCat(S[1:i+1]) and isCat(S[i+2:-1])
if(isCat(input())):
print("Cat")
else:
print("Rabbit")
``` | output | 1 | 92,576 | 5 | 185,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S ⊆ \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j ∈ [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is ∑_{i ∈ S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 ≤ n ≤ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 ≤ b_i ≤ 10^5).
Output
Print one integer — the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Submitted Solution:
```
print('quypnpnpnp')
``` | instruction | 0 | 92,861 | 5 | 185,722 |
No | output | 1 | 92,861 | 5 | 185,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S ⊆ \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j ∈ [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is ∑_{i ∈ S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 ≤ n ≤ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 ≤ b_i ≤ 10^5).
Output
Print one integer — the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Submitted Solution:
```
n = int(input())
count = 0
for x in range (1,n-2):
for y in range (1,n-2):
if ((x % y) == 0):
count += 1
print (count % 998244353)
``` | instruction | 0 | 92,862 | 5 | 185,724 |
No | output | 1 | 92,862 | 5 | 185,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Note that the memory limit is unusual.
You are given an integer n and two sequences a_1, a_2, ..., a_n and b_1, b_2, ..., b_n.
Let's call a set of integers S such that S ⊆ \{1, 2, 3, ..., n\} strange, if, for every element i of S, the following condition is met: for every j ∈ [1, i - 1], if a_j divides a_i, then j is also included in S. An empty set is always strange.
The cost of the set S is ∑_{i ∈ S} b_i. You have to calculate the maximum possible cost of a strange set.
Input
The first line contains one integer n (1 ≤ n ≤ 3000).
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100).
The third line contains n integers b_1, b_2, ..., b_n (-10^5 ≤ b_i ≤ 10^5).
Output
Print one integer — the maximum cost of a strange set.
Examples
Input
9
4 7 3 4 5 6 7 8 13
-2 3 -19 5 -6 7 -8 9 1
Output
16
Input
2
42 42
-37 13
Output
0
Input
2
42 42
13 -37
Output
13
Note
The strange set with the maximum cost in the first example is \{1, 2, 4, 8, 9\}.
The strange set with the maximum cost in the second example is empty.
Submitted Solution:
```
# Author: yumtam
# Created at: 2021-03-02 20:19
def main():
n = int(input())
A = [int(t) for t in input().split()]
cost = [int(t) for t in input().split()]
g = Flow(n+2)
s, t = n, n+1
last = [-1] * 101
for i, x in enumerate(A):
for d in range(1, 101):
if x % d == 0 and last[d] >= 0:
g.add_edge(i, last[d], 10**5)
last[x] = i
if cost[i] >= 0:
g.add_edge(s, i, cost[i])
else:
g.add_edge(i, t, -cost[i])
min_cut = g.calc(s, t)
ans = sum(max(c, 0) for c in cost) - min_cut
print(ans)
class Flow:
def __init__(self, n):
self.n = n
self.g = [dict() for _ in range(n)]
def add_edge(self, u, v, w):
self.g[u][v] = w
self.g[v][u] = 0
def bfs(self, s, t):
q = [s]
vis = [0] * self.n
vis[s] = 1
prev = [-1] * self.n
for ver in q:
for nei, w in self.g[ver].items():
if not vis[nei] and w > 0:
vis[nei] = 1
prev[nei] = ver
q.append(nei)
if not vis[t]:
return 0
flow = float('inf')
ver = t
while ver != s:
p = prev[ver]
flow = min(flow, self.g[p][ver])
ver = p
ver = t
while ver != s:
p = prev[ver]
self.g[p][ver] -= flow
self.g[ver][p] += flow
ver = p
return flow
def calc(self, s, t):
res = 0
while True:
flow = self.bfs(s, t)
res += flow
if not flow:
return res
main()
``` | instruction | 0 | 92,863 | 5 | 185,726 |
No | output | 1 | 92,863 | 5 | 185,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
def Solve():
x,y,m=map(int,input().split())
if(x>=m or y>=m):
return 0
if(x<=0 and y<=0):
return -1
ans=0
if(y<=0):
ans=abs(y)//x+1
y+=ans*x
elif(x<=0):
ans=abs(x)//y+1
x+=ans*y
while(x<m and y<m):
if(x<y):
x+=y
else:
y+=x
ans+=1
return ans
print(Solve())
``` | instruction | 0 | 92,983 | 5 | 185,966 |
Yes | output | 1 | 92,983 | 5 | 185,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
x, y, m = [int(number) for number in input().split()]
x, y = min(x, y), max(x, y)
if y >= m:
print(0)
quit()
if y <= 0:
print(-1)
quit()
answer = 0
if x < 0:
answer += (-x + y - 1) // y
x += answer * y
x, y = min(x, y), max(x, y)
if y >= m:
print(answer)
quit()
while y < m:
x, y = y, x + y
answer += 1
print(answer)
``` | instruction | 0 | 92,984 | 5 | 185,968 |
Yes | output | 1 | 92,984 | 5 | 185,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
def doit():
x, y, m = [int(k) for k in input().strip().split()]
if x < y:
x, y = y, x
if x >= m:
print(0)
return
if x<=0 and y<=0:
print(-1)
return
k = 0
if y < 0:
k = (-y+x-1)//x
y += k*x
assert(y >= 0)
if x < y:
x, y = y, x
while x < m:
k += 1
x, y = x+y, x
if x < y:
x, y = y, x
print(k)
doit()
``` | instruction | 0 | 92,985 | 5 | 185,970 |
Yes | output | 1 | 92,985 | 5 | 185,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
x, y, m = map(int, input().split())
if x < y: x, y = y, x
if x >= m: print(0)
elif x > 0:
s = 0
if y < 0:
s = (-y) // x + 1
y += x * s
while x < m:
x, y = x + y, x
s += 1
print(s)
elif x < m: print(-1)
else: print(0)
``` | instruction | 0 | 92,986 | 5 | 185,972 |
Yes | output | 1 | 92,986 | 5 | 185,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
nums = input().split(' ')
x = int(nums[0])
y = int(nums[1])
m = int(nums[2])
def ispr(x, y, m):
def ispr_recur(x, y, m, n):
if x <= 0 and y <= 0 and (m > x or m > y):
return -1
if (x >= m or y >= m):
return n
elif x >= y:
n +=1
return ispr_recur(x, x + y, m, n)
elif x < y:
n +=1
return ispr_recur(x + y, y, m, n)
return ispr_recur(x, y, m, 0)
print(ispr(x, y, m))
``` | instruction | 0 | 92,987 | 5 | 185,974 |
No | output | 1 | 92,987 | 5 | 185,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
# -*- coding: utf-8 -*-
def main():
x, y, m = map(int, input().split())
b1 = min(x, y)
b2 = max(x, y)
count = 0
while b2 < m:
b = b1 + b2
if (b <= b1):
print(-1)
return
b1 = b2
b2 = b
count += 1
print(count)
return
if __name__ == '__main__':
main()
``` | instruction | 0 | 92,988 | 5 | 185,976 |
No | output | 1 | 92,988 | 5 | 185,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
#_________________ Mukul Mohan Varshney _______________#
#Template
import sys
import os
import math
import copy
from math import gcd
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations, combinations
#define function
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Ncr(n,r,p): return ((fact[n])*((ifact[r]*ifact[n-r])%p))%p
def Most_frequent(list): return max(set(list), key = list.count)
def Mat2x2(n): return [List() for _ in range(n)]
def btod(n):
return int(n,2)
def dtob(n):
return bin(n).replace("0b","")
# Driver Code
def solution():
#for _ in range(Int()):
x,y,m=Mint()
maxi=max(x,y)
mini=min(x,y)
ans=m-mini
if(maxi==0):
print(-1)
else:
ans=ans//maxi
print(ans)
#Call the solve function
if __name__ == "__main__":
solution()
``` | instruction | 0 | 92,989 | 5 | 185,978 |
No | output | 1 | 92,989 | 5 | 185,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Examples
Input
1 2 5
Output
2
Input
-1 4 15
Output
4
Input
0 -1 5
Output
-1
Note
In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).
In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).
Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
Submitted Solution:
```
from collections import *
from math import *
from sys import *
n,m,k = map(int,input().split())
if(n+m > min(n,m)):
ct = 0
#print(m,n)
if(n < 0):
ct = ceil(-n/m)
n = n + ct*m
elif(m < 0):
ct = ceil(-m/n)
m = m + ct*n
#print(m,n)
while(max(n,m) < k):
if(n < m):
n = n+m
else:
m = n+m
ct += 1
print(ct)
else:
if(max(n,m) < k):
print(-1)
else:
print(0)
``` | instruction | 0 | 92,990 | 5 | 185,980 |
No | output | 1 | 92,990 | 5 | 185,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an empty string S and a variable x whose initial value is 0.
Also, we have a string T consisting of `0` and `1`.
Now, Takahashi will do the operation with the following two steps |T| times.
* Insert a `0` or a `1` at any position of S of his choice.
* Then, increment x by the sum of the digits in the odd positions (first, third, fifth, ...) of S. For example, if S is `01101`, the digits in the odd positions are `0`, `1`, `1` from left to right, so x is incremented by 2.
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
T
Output
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Examples
Input
1101
Output
5
Input
0111101101
Output
26
Submitted Solution:
```
import math
import re
T=input()
x=0
Td=T
d = len(re.sub(r'0', '', Td))
x += math.ceil((d/2+1)*d/2)
spT = T.split('0')
lenT = list(map(len, spT))
tmpX = math.ceil(d/2)
bT = 0
aT = d
tmp = ''
for i in range(len(spT) - 1):
tmp += spT[i]
bT += lenT[i]
aT -= lenT[i]
if bT % 2==1:
bT += 1
if aT % 2 ==1:
tmpX += 1
x += tmpX
tmp += '0'
else:
tmp +='+'
tmp += spT[-1]
spT = tmp.split('+')
vT =[]
aT = bT
bT = 0
odT = 0
tmpOd = True
for i in range(len(spT)):
od = len(re.sub(r'0', '', spT[i][0 if tmpOd else 1::2]))
ev = len(re.sub(r'0', '', spT[i][1 if tmpOd else 0::2]))
odT += od
vT.append(od - ev)
if len(spT[i]) %2 == 1:
tmpOd = not tmpOd
for i in reversed(range(1, len(spT))):
odT = odT - vT[i]
x += odT
vT[i-1] -= vT[i]
print(x)
``` | instruction | 0 | 93,276 | 5 | 186,552 |
Yes | output | 1 | 93,276 | 5 | 186,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an empty string S and a variable x whose initial value is 0.
Also, we have a string T consisting of `0` and `1`.
Now, Takahashi will do the operation with the following two steps |T| times.
* Insert a `0` or a `1` at any position of S of his choice.
* Then, increment x by the sum of the digits in the odd positions (first, third, fifth, ...) of S. For example, if S is `01101`, the digits in the odd positions are `0`, `1`, `1` from left to right, so x is incremented by 2.
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
T
Output
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Examples
Input
1101
Output
5
Input
0111101101
Output
26
Submitted Solution:
```
T=input()
i=r=a=k=0
for c in T:a+=(i//2-k//2+(~k&1)*(len(T)-i))*(c=="1");i+=1;k-=(~k&1)*(c=='0')-1
print(a)
``` | instruction | 0 | 93,277 | 5 | 186,554 |
Yes | output | 1 | 93,277 | 5 | 186,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an empty string S and a variable x whose initial value is 0.
Also, we have a string T consisting of `0` and `1`.
Now, Takahashi will do the operation with the following two steps |T| times.
* Insert a `0` or a `1` at any position of S of his choice.
* Then, increment x by the sum of the digits in the odd positions (first, third, fifth, ...) of S. For example, if S is `01101`, the digits in the odd positions are `0`, `1`, `1` from left to right, so x is incremented by 2.
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
T
Output
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Examples
Input
1101
Output
5
Input
0111101101
Output
26
Submitted Solution:
```
T=input()
i=r=a=k=0
for c in T:a+=(i//2-k//2+(~k&1)*(len(T)-i))*(c=="1");i+=1;k-=(~k&1)*(c=='0')-1;
print(a)
``` | instruction | 0 | 93,278 | 5 | 186,556 |
Yes | output | 1 | 93,278 | 5 | 186,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an empty string S and a variable x whose initial value is 0.
Also, we have a string T consisting of `0` and `1`.
Now, Takahashi will do the operation with the following two steps |T| times.
* Insert a `0` or a `1` at any position of S of his choice.
* Then, increment x by the sum of the digits in the odd positions (first, third, fifth, ...) of S. For example, if S is `01101`, the digits in the odd positions are `0`, `1`, `1` from left to right, so x is incremented by 2.
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
T
Output
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Examples
Input
1101
Output
5
Input
0111101101
Output
26
Submitted Solution:
```
from itertools import groupby, accumulate
def solve(t):
grp = [(c, len(list(itr))) for c, itr in groupby(t)]
even_ones = 0
zeros_splitting_odd_ones = [0]
for c, cnt in grp:
if c == '1':
if cnt % 2 == 0:
even_ones += cnt
else:
even_ones += cnt - 1
zeros_splitting_odd_ones.append(0)
else:
zeros_splitting_odd_ones[-1] += cnt
# 初期状態で加算されるか
ad = int(zeros_splitting_odd_ones[0] % 2 == 0)
is_added_init = [ad]
for cnt in zeros_splitting_odd_ones[1:]:
ad ^= (cnt % 2) ^ 1
is_added_init.append(ad)
is_added_init.pop()
acc_added_init = list(accumulate(reversed(is_added_init))) # 自分より右で、初期状態で足される1の数
acc_added_init.reverse()
odd_ones = len(is_added_init)
ans = 0
if odd_ones > 0:
# 先頭の0を全て除くまでのスコア
zso = zeros_splitting_odd_ones[0]
zso1 = zso // 2
zso0 = zso - zso1
ans += acc_added_init[0] * zso0
ans += (odd_ones - acc_added_init[0]) * zso1
# 1に挟まれた0を1箇所1つずつ残して全て除くまでのスコア
ad = int(zeros_splitting_odd_ones[0] % 2 == 0)
for i in range(1, odd_ones):
zso = zeros_splitting_odd_ones[i] - 1
zso1 = zso // 2
zso0 = zso - zso1
fixed_ones = i
right_ones = odd_ones - fixed_ones
ans += fixed_ones * zso
if ad:
# 除外中の0区間の次の"1"が初期状態で足されるなら、区間を操作する直前にも足される
ans += acc_added_init[i] * zso0
ans += (right_ones - acc_added_init[i]) * zso1
else:
# 除外中の0区間の次の"1"が初期状態で足されないなら、区間を操作する直前には足される
ans += (right_ones - acc_added_init[i]) * zso0
ans += acc_added_init[i] * zso1
ad ^= zso % 2
# 末尾の0を全て除くまでのスコア
ans += zeros_splitting_odd_ones[-1] * odd_ones
# 1に挟まれた0を右から除いていくまでのスコア
for i in range(1, odd_ones):
left_ones = odd_ones - i
right_ones = i
ans += left_ones + (right_ones + 1) // 2
# 1だけの列を1つずつ除いていくスコア
all_ones = odd_ones + even_ones
ao1 = all_ones // 2
ao0 = all_ones - ao1
ans += (ao0 * (ao0 + 1) + ao1 * (ao1 + 1)) // 2
# 0を除いている間の、偶数個の1によるスコア
all_zeros = len(t) - all_ones
ans += all_zeros * (even_ones // 2)
return ans
t = input()
print(solve(t))
``` | instruction | 0 | 93,279 | 5 | 186,558 |
Yes | output | 1 | 93,279 | 5 | 186,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an empty string S and a variable x whose initial value is 0.
Also, we have a string T consisting of `0` and `1`.
Now, Takahashi will do the operation with the following two steps |T| times.
* Insert a `0` or a `1` at any position of S of his choice.
* Then, increment x by the sum of the digits in the odd positions (first, third, fifth, ...) of S. For example, if S is `01101`, the digits in the odd positions are `0`, `1`, `1` from left to right, so x is incremented by 2.
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
T
Output
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Examples
Input
1101
Output
5
Input
0111101101
Output
26
Submitted Solution:
```
s = input()
T = [int(s[i]) for i in range(len(s))]
x = 0
count_0 = s.count('0')
for i in range(count_0):
for j in range(0,len(T),2):
x += T[j]
T.remove(0)
count_1 = len(s)-count_0
if count_1 % 2 == 0:
m = count_1 // 2
x += m**2 + m
else:
m = (count_1+1) //2
x += m**2
print(x)
``` | instruction | 0 | 93,280 | 5 | 186,560 |
No | output | 1 | 93,280 | 5 | 186,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an empty string S and a variable x whose initial value is 0.
Also, we have a string T consisting of `0` and `1`.
Now, Takahashi will do the operation with the following two steps |T| times.
* Insert a `0` or a `1` at any position of S of his choice.
* Then, increment x by the sum of the digits in the odd positions (first, third, fifth, ...) of S. For example, if S is `01101`, the digits in the odd positions are `0`, `1`, `1` from left to right, so x is incremented by 2.
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
T
Output
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Examples
Input
1101
Output
5
Input
0111101101
Output
26
Submitted Solution:
```
t = input()
one = t.count("1")
zero = t.count("0")
if one % 2 == 0:
tmp = one // 2
ans = tmp * (tmp + 1)
else:
tmp = one // 2
ans = (tmp + 1) ** 2
even_one = 0
before_zero = []
zero_cnt = 0
for i, num in enumerate(t):
if i % 2 == 0 and num == "1":
even_one += 1
before_zero.append(zero_cnt)
if num == "0":
zero_cnt += 1
for i, num in enumerate(t):
if num == "0":
continue
if (i + before_zero[i]) % 2 == 0:
ans += (zero - before_zero[i]) + (before_zero[i]) // 2
else:
ans += (before_zero[i] + 1) // 2
print(ans)
``` | instruction | 0 | 93,282 | 5 | 186,564 |
No | output | 1 | 93,282 | 5 | 186,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has an empty string S and a variable x whose initial value is 0.
Also, we have a string T consisting of `0` and `1`.
Now, Takahashi will do the operation with the following two steps |T| times.
* Insert a `0` or a `1` at any position of S of his choice.
* Then, increment x by the sum of the digits in the odd positions (first, third, fifth, ...) of S. For example, if S is `01101`, the digits in the odd positions are `0`, `1`, `1` from left to right, so x is incremented by 2.
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
T
Output
Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.
Examples
Input
1101
Output
5
Input
0111101101
Output
26
Submitted Solution:
```
n =input()
while True:
for i in range(len(n)):
if i>n:
break
if i<n:
v=n[0::2]
b=sum(v)
b=b+1
print(b)
``` | instruction | 0 | 93,283 | 5 | 186,566 |
No | output | 1 | 93,283 | 5 | 186,567 |
Provide a correct Python 3 solution for this coding contest problem.
Find the number of sequences of N non-negative integers A_1, A_2, ..., A_N that satisfy the following conditions:
* L \leq A_1 + A_2 + ... + A_N \leq R
* When the N elements are sorted in non-increasing order, the M-th and (M+1)-th elements are equal.
Since the answer can be enormous, print it modulo 10^9+7.
Constraints
* All values in input are integers.
* 1 \leq M < N \leq 3 \times 10^5
* 1 \leq L \leq R \leq 3 \times 10^5
Input
Input is given from Standard Input in the following format:
N M L R
Output
Print the number of sequences of N non-negative integers, modulo 10^9+7.
Examples
Input
4 2 3 7
Output
105
Input
2 1 4 8
Output
3
Input
141592 6535 89793 238462
Output
933832916 | instruction | 0 | 93,300 | 5 | 186,600 |
"Correct Solution:
```
MOD = 10**9+7
kaijo_memo = []
def kaijo(n):
if(len(kaijo_memo) > n): return kaijo_memo[n]
if(len(kaijo_memo) == 0): kaijo_memo.append(1)
while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
return kaijo_memo[n]
gyaku_kaijo_memo = []
def gyaku_kaijo(n):
if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
return gyaku_kaijo_memo[n]
def nCr(n,r):
if n == r: return 1
if n < r or r < 0: return 0
ret = 1
ret = ret * kaijo(n) % MOD
ret = ret * gyaku_kaijo(r) % MOD
ret = ret * gyaku_kaijo(n-r) % MOD
return ret
N,M,L,R = map(int,input().split())
P = [0 for i in range(R+1)]
Q = [0 for i in range(R+1)]
Z = [0 for i in range(R+1)]
for k in range(1,R+1):
u = (R-M*k)//k
for j in range(u+1):
P[k*j+M*k] += nCr(N-M,j)*(-1)**(j%2)
P[k*j+M*k] %= MOD
for k in range(R+1):
Q[k] += nCr(k+N,N)
Q[k] %= MOD
Z[k] += nCr(k+N,N)
Z[k] %= MOD
for k in range(R-M+1):
Q[k+M] -= nCr(k+N,N)
Q[k+M] %= MOD
def f(MAX):
ans = 0
for t in range(MAX+1):
ans += P[t]*Q[MAX-t]
ans %= MOD
return ans
def F(x):
return (Z[x]-nCr(N,M)*f(x))%MOD
print((F(R)-F(L-1))%MOD)
``` | output | 1 | 93,300 | 5 | 186,601 |
Provide a correct Python 3 solution for this coding contest problem.
Find the number of sequences of N non-negative integers A_1, A_2, ..., A_N that satisfy the following conditions:
* L \leq A_1 + A_2 + ... + A_N \leq R
* When the N elements are sorted in non-increasing order, the M-th and (M+1)-th elements are equal.
Since the answer can be enormous, print it modulo 10^9+7.
Constraints
* All values in input are integers.
* 1 \leq M < N \leq 3 \times 10^5
* 1 \leq L \leq R \leq 3 \times 10^5
Input
Input is given from Standard Input in the following format:
N M L R
Output
Print the number of sequences of N non-negative integers, modulo 10^9+7.
Examples
Input
4 2 3 7
Output
105
Input
2 1 4 8
Output
3
Input
141592 6535 89793 238462
Output
933832916 | instruction | 0 | 93,301 | 5 | 186,602 |
"Correct Solution:
```
MOD = 10**9+7
kaijo_memo = []
def kaijo(n):
if(len(kaijo_memo) > n): return kaijo_memo[n]
if(len(kaijo_memo) == 0): kaijo_memo.append(1)
while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
return kaijo_memo[n]
gyaku_kaijo_memo = []
def gyaku_kaijo(n):
if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
return gyaku_kaijo_memo[n]
def nCr(n,r):
if n == r: return 1
if n < r or r < 0: return 0
ret = 1
ret = ret * kaijo(n) % MOD
ret = ret * gyaku_kaijo(r) % MOD
ret = ret * gyaku_kaijo(n-r) % MOD
return ret
N,M,L,R = map(int,input().split())
P = [0 for i in range(R+1)]
for k in range(R+1):
u = (R-M*(k+1))//(k+1)
for j in range(u+1):
P[(k+1)*j+M*(k+1)] += nCr(N-M,j)*(-1)**(j%2)
P[(k+1)*j+M*(k+1)] %= MOD
for k in range(1,R+1):
u = (R-M*(k+1))//k
for j in range(u+1):
P[k*j+M*(k+1)] -= nCr(N-M,j)*(-1)**(j%2)
P[k*j+M*(k+1)] %= MOD
Q = [nCr(i+N,N) for i in range(R+1)]
def f(MAX):
ans = 0
for t in range(MAX+1):
ans += P[t]*Q[MAX-t]
ans %= MOD
return ans
def F(x):
return (Q[x]-nCr(N,M)*f(x))%MOD
print((F(R)-F(L-1))%MOD)
``` | output | 1 | 93,301 | 5 | 186,603 |
Provide a correct Python 3 solution for this coding contest problem.
Find the number of sequences of N non-negative integers A_1, A_2, ..., A_N that satisfy the following conditions:
* L \leq A_1 + A_2 + ... + A_N \leq R
* When the N elements are sorted in non-increasing order, the M-th and (M+1)-th elements are equal.
Since the answer can be enormous, print it modulo 10^9+7.
Constraints
* All values in input are integers.
* 1 \leq M < N \leq 3 \times 10^5
* 1 \leq L \leq R \leq 3 \times 10^5
Input
Input is given from Standard Input in the following format:
N M L R
Output
Print the number of sequences of N non-negative integers, modulo 10^9+7.
Examples
Input
4 2 3 7
Output
105
Input
2 1 4 8
Output
3
Input
141592 6535 89793 238462
Output
933832916 | instruction | 0 | 93,302 | 5 | 186,604 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
#############
# Libraries #
#############
import sys
input = sys.stdin.readline
import math
#from math import gcd
import bisect
import heapq
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache
#############
# Constants #
#############
MOD = 10**9+7
INF = float('inf')
AZ = "abcdefghijklmnopqrstuvwxyz"
#############
# Functions #
#############
######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]
######OUTPUT######
def P(arg): print(arg); return
def Y(): print("Yes"); return
def N(): print("No"); return
def E(): exit()
def PE(arg): print(arg); exit()
def YE(): print("Yes"); exit()
def NE(): print("No"); exit()
#####Shorten#####
def DD(arg): return defaultdict(arg)
#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)
######Combination######
kaijo_memo = []
def kaijo(n):
if(len(kaijo_memo) > n): return kaijo_memo[n]
if(len(kaijo_memo) == 0): kaijo_memo.append(1)
while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
return kaijo_memo[n]
gyaku_kaijo_memo = []
def gyaku_kaijo(n):
if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
return gyaku_kaijo_memo[n]
def nCr(n,r):
if n == r: return 1
if n < r or r < 0: return 0
ret = 1
ret = ret * kaijo(n) % MOD
ret = ret * gyaku_kaijo(r) % MOD
ret = ret * gyaku_kaijo(n-r) % MOD
return ret
######Factorization######
def factorization(n):
arr = []
temp = n
for i in range(2, int(-(-n**0.5//1))+1):
if temp%i==0:
cnt=0
while temp%i==0:
cnt+=1
temp //= i
arr.append([i, cnt])
if temp!=1: arr.append([temp, 1])
if arr==[]: arr.append([n, 1])
return arr
#####MakeDivisors######
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
return divisors
#####MakePrimes######
def make_primes(N):
max = int(math.sqrt(N))
seachList = [i for i in range(2,N+1)]
primeNum = []
while seachList[0] <= max:
primeNum.append(seachList[0])
tmp = seachList[0]
seachList = [i for i in seachList if i % tmp != 0]
primeNum.extend(seachList)
return primeNum
#####GCD#####
def gcd(a, b):
while b: a, b = b, a % b
return a
#####LCM#####
def lcm(a, b):
return a * b // gcd (a, b)
#####BitCount#####
def count_bit(n):
count = 0
while n:
n &= n-1
count += 1
return count
#####ChangeBase#####
def base_10_to_n(X, n):
if X//n: return base_10_to_n(X//n, n)+[X%n]
return [X%n]
def base_n_to_10(X, n):
return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))
def base_10_to_n_without_0(X, n):
X -= 1
if X//n: return base_10_to_n_without_0(X//n, n)+[X%n]
return [X%n]
#####IntLog#####
def int_log(n, a):
count = 0
while n>=a:
n //= a
count += 1
return count
#############
# Main Code #
#############
N,M,L,R = IL()
P = [0 for i in range(R+1)]
for k in range(R+1):
u = (R-M*(k+1))//(k+1)
for j in range(u+1):
P[(k+1)*j+M*(k+1)] += nCr(N-M,j)*(-1)**(j%2)
P[(k+1)*j+M*(k+1)] %= MOD
for k in range(1,R+1):
u = (R-M*(k+1))//k
for j in range(u+1):
P[k*j+M*(k+1)] -= nCr(N-M,j)*(-1)**(j%2)
P[k*j+M*(k+1)] %= MOD
"""
@lru_cache(maxsize=None)
def P(t):
ans = 0
for k in range(t+1):
u = t-M*(k+1)
if u>=0:
if u%(k+1) == 0:
ans += nCr(N-M,u//(k+1))*(-1)**((u//(k+1))%2)
ans %= MOD
for k in range(t+1):
u = t-M*(k+1)
if u>=0:
if k!=0:
if u%k == 0:
ans -= nCr(N-M,u//k)*(-1)**((u//k)%2)
ans %= MOD
return ans
"""
@lru_cache(maxsize=None)
def Q(t):
ans = nCr(t+N,N)
return ans
def f(MAX):
ans = 0
for t in range(MAX+1):
ans += P[t]*Q(MAX-t)
ans %= MOD
return ans
def F(x):
return (Q(x)-nCr(N,M)*f(x))%MOD
print((F(R)-F(L-1))%MOD)
``` | output | 1 | 93,302 | 5 | 186,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of sequences of N non-negative integers A_1, A_2, ..., A_N that satisfy the following conditions:
* L \leq A_1 + A_2 + ... + A_N \leq R
* When the N elements are sorted in non-increasing order, the M-th and (M+1)-th elements are equal.
Since the answer can be enormous, print it modulo 10^9+7.
Constraints
* All values in input are integers.
* 1 \leq M < N \leq 3 \times 10^5
* 1 \leq L \leq R \leq 3 \times 10^5
Input
Input is given from Standard Input in the following format:
N M L R
Output
Print the number of sequences of N non-negative integers, modulo 10^9+7.
Examples
Input
4 2 3 7
Output
105
Input
2 1 4 8
Output
3
Input
141592 6535 89793 238462
Output
933832916
Submitted Solution:
```
#include <bits/stdc++.h>
using namespace std;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
constexpr long long TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
template <class T, class U>
void chmin(T& t, const U& u) {
if (t > u) t = u;
}
template <class T, class U>
void chmax(T& t, const U& u) {
if (t < u) t = u;
}
template <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
os << "(" << p.first << "," << p.second << ")";
return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
os << "{";
for (int i = 0; i < (v.size()); i++) {
if (i) os << ",";
os << v[i];
}
os << "}";
return os;
}
const long long MOD = TEN(9) + 7;
const int MX = TEN(6);
long long inv[MX], fact[MX], ifact[MX];
void init() {
inv[1] = 1;
for (int i = 2; i < MX; ++i) {
inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
}
fact[0] = ifact[0] = 1;
for (int i = 1; i < MX; ++i) {
fact[i] = fact[i - 1] * i % MOD;
ifact[i] = ifact[i - 1] * inv[i] % MOD;
}
}
long long comb(int n, int r) {
if (n < 0 || r < 0 || r > n) return 0;
return fact[n] * ifact[r] % MOD * ifact[n - r] % MOD;
}
long long solve_sub(int N, int M, int S, int r, int l) {
long long v = 0;
for (int i = 0; i <= N - M; ++i) {
long long rem = S - (long long)r * M - (long long)(l + 1) * i;
if (rem < 0) break;
long long t = comb(rem + N, N);
if (i % 2 == 1) {
t *= -1;
}
t = t * comb(N - M, i) % MOD;
if (t < 0) t += MOD;
v = (v + t) % MOD;
}
v = v * comb(N, M) % MOD;
return v;
}
long long solve(int N, int M, int S) {
long long all = comb(S + N, N);
long long ng = 0;
for (int x = 1; (long long)x * M <= S; ++x) {
long long a =
solve_sub(N, M, S, x, x - 1) - solve_sub(N, M, S, x + 1, x - 1);
ng = (ng + a) % MOD;
if (ng < 0) ng += MOD;
}
return (all - ng + MOD) % MOD;
}
int main() {
int N, M, L, R;
cin >> N >> M >> L >> R;
init();
long long ans = (solve(N, M, R) - solve(N, M, L - 1) + MOD) % MOD;
cout << ans << endl;
return 0;
}
``` | instruction | 0 | 93,303 | 5 | 186,606 |
No | output | 1 | 93,303 | 5 | 186,607 |
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