message stringlengths 2 433k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7. | instruction | 0 | 86,786 | 12 | 173,572 |
Tags: brute force, constructive algorithms
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def solve(n, a):
ans = 0
for i in range(n):
for j in range(i, n):
for k in range(j, n):
ans = max(ans, a[i] | a[j] | a[k])
return ans
def main():
n = ri()
a = ria()
wi(solve(n, a))
if __name__ == '__main__':
main()
``` | output | 1 | 86,786 | 12 | 173,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7. | instruction | 0 | 86,787 | 12 | 173,574 |
Tags: brute force, constructive algorithms
Correct Solution:
```
n=int(input())
l=input().split()
li=[int(i) for i in l]
dp=[0 for i in range(n)]
if(n==1):
print(li[0])
elif(n==2):
print(li[0]|li[1])
else:
maxa=0
for i in range(n):
for j in range(i+1,n):
for k in range(j+1,n):
curr=(li[i]|li[j])|li[k]
if(curr>maxa):
maxa=curr
print(maxa)
``` | output | 1 | 86,787 | 12 | 173,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7. | instruction | 0 | 86,788 | 12 | 173,576 |
Tags: brute force, constructive algorithms
Correct Solution:
```
from sys import stdin,stdout
from math import gcd,sqrt
from collections import deque
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
P=lambda x:stdout.write(x)
hg=lambda x,y:((y+x-1)//x)*x
cu=lambda x:max(0,x-1)
cd=lambda x:min(r-1,x+1)
cl=lambda x:max(0,x-1)
cr=lambda x:min(c-1,x+1)
n=I()
a=list(R())
ans=0
for i in range(n):
for j in range(i,n):
for k in range(j,n):
ans=max(ans,(a[i] | a[j] | a[k]))
print(ans)
``` | output | 1 | 86,788 | 12 | 173,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7. | instruction | 0 | 86,789 | 12 | 173,578 |
Tags: brute force, constructive algorithms
Correct Solution:
```
import sys
# sys.setrecursionlimit(10**6)
input=sys.stdin.readline
# t=int(input())
# for t1 in range(t):
import math
n=int(input())
l=list(map(int,input().split(" ")))
ans=0
if(n<3):
for i in range(n):
ans=ans | l[i]
for i in range(n):
for j in range(i+1,n):
for k in range(j+1,n):
temp=l[i]|l[j]
temp=temp|l[k]
ans=max(ans,temp)
print(ans)
``` | output | 1 | 86,789 | 12 | 173,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7. | instruction | 0 | 86,790 | 12 | 173,580 |
Tags: brute force, constructive algorithms
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
def main(n,arr):
a=max(arr)
res=0
for i in range(min(3,n)):
res|=arr[i]
for k in range(n):
for i in range(k+1,n):
for j in range(i+1,n):
res=max(res,arr[k]|arr[i]|arr[j])
p_2=1
ans=0
for i in range(64):
if 1<<i &res:
ans+=p_2
p_2*=2
return ans
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
n=int(input())
arr=tuple(map(int,input().split() ))
print(main(n,arr))
``` | output | 1 | 86,790 | 12 | 173,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7. | instruction | 0 | 86,791 | 12 | 173,582 |
Tags: brute force, constructive algorithms
Correct Solution:
```
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
n=II()
aa=LI()
ans=0
if n==1:
ans=aa[0]
elif n==2:
for i in range(n):
for j in range(i):
ans = max(ans, aa[i] | aa[j])
else:
for i in range(n):
for j in range(i):
for k in range(j):
ans=max(ans,aa[i]|aa[j]|aa[k])
print(ans)
main()
``` | output | 1 | 86,791 | 12 | 173,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7. | instruction | 0 | 86,792 | 12 | 173,584 |
Tags: brute force, constructive algorithms
Correct Solution:
```
n = int(input())
arr = list(map(int,input().split()))
ans=0
for i in range(n):
for j in range(i,n):
for k in range(j,n):
ans = max( ans, arr[i] | arr[j] | arr[k] )
print(ans)
``` | output | 1 | 86,792 | 12 | 173,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def calc(arr):
v = max(1,len(arr)-2)
ans = 0
for i in range(m):
count = 0
for j in range(len(arr)):
if arr[j][i]=="1":
count += 1
if count>=v:
ans += 2**(m-i-1)
return ans
n = int(input())
a = sorted(list(map(int,input().split())))
a = a[::-1]
if n==1:
print (a[0])
exit()
if n==2:
print (a[0]|a[1])
exit()
ans = 0
for i in range(n):
for j in range(i+1,n):
for k in range(j+1,n):
ans = max(ans,a[i]|a[j]|a[k])
print (ans)
``` | instruction | 0 | 86,793 | 12 | 173,586 |
Yes | output | 1 | 86,793 | 12 | 173,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
Submitted Solution:
```
import sys
import collections as cc
import bisect as bi
input = sys.stdin.readline
I=lambda:list(map(int,input().split()))
n,=I()
l=I()
if n<3:
ans=0
for i in l:
ans|=i
print(ans)
else:
ans=0
for i in range(n):
for j in range(i+1,n):
for k in range(j+1,n):
ans=max(ans,l[i]|l[j]|l[k])
print(ans)
``` | instruction | 0 | 86,794 | 12 | 173,588 |
Yes | output | 1 | 86,794 | 12 | 173,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
Submitted Solution:
```
'''input
4
7 7 1 1
'''
import math
from itertools import combinations
def calc_final(nums):
count_bit = [0 for i in range(65)]
for i in range(65):
for j in range(len(nums)):
if nums[j] % 2 == 1:
count_bit[i] += 1
nums[j] //= 2
final = 0
po = 1
for i in range(65):
if count_bit[i] >= max(1, len(nums) - 2):
final += po
po *= 2
return final
n = int(input())
arr = list(map(int, input().split()))
if n <= 2:
ans = 0
for i in range(1, n + 1):
combs = combinations(arr, i)
for comb in combs:
# print(list(comb))
ans = max(ans, calc_final(list(comb)))
print(ans)
else:
ans = 0
for i in range(len(arr)):
ans = max(ans, arr[i])
for j in range(i + 1, len(arr)):
ans = max(ans, arr[i] | arr[j])
for k in range(j + 1, len(arr)):
ans = max(ans, arr[i] | arr[j] | arr[k])
print(ans)
``` | instruction | 0 | 86,795 | 12 | 173,590 |
Yes | output | 1 | 86,795 | 12 | 173,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
Submitted Solution:
```
import sys
import math
from collections import defaultdict,deque
import heapq
n=int(sys.stdin.readline())
arr=list(map(int,sys.stdin.readline().split()))
ans=0
for i in range(n):
for j in range(i,n):
for k in range(j,n):
ans=max(ans,arr[i]|arr[j]|arr[k])
print(ans)
``` | instruction | 0 | 86,796 | 12 | 173,592 |
Yes | output | 1 | 86,796 | 12 | 173,593 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import gcd
import heapq
raw_input = stdin.readline
pr = stdout.write
mod=998244353
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return tuple(map(int,stdin.read().split()))
range = xrange # not for python 3.0+
def fun(l):
ans=0
n=len(l)
for i in range(61):
x=1<<i
c=0
for j in l:
if j&x:
c+=1
if c>=max(1,n-2):
ans+=x
return ans
# main code
n=ni()
l=li()
d=defaultdict(set)
ans=0
for i in range(61):
x=1<<i
f=0
for j in range(n):
if l[j]&x:
f=1
#print x,l[j]
d[i].add(l[j])
if f:
ans=x
for i in d:
s1=[]
n1=0
for i1 in range(n):
if l[i1] not in d[i]:
s1.append(l[i1])
n1+=1
for j in range(n1):
temp=Counter(d[i])
temp[l[j]]+=1
ans=max(ans,fun(temp))
for k in range(j+1,n1):
temp[l[k]]+=1
ans=max(ans,fun(temp))
temp[l[k]]-=1
pn(ans)
``` | instruction | 0 | 86,797 | 12 | 173,594 |
No | output | 1 | 86,797 | 12 | 173,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
def main(n,arr):
a=max(arr)
res=0
for k in range(n):
for i in range(k+1,n):
for j in range(i+1,n):
res=max(res,arr[k]|arr[i]|arr[j])
p_2=1
ans=0
for i in range(64):
if 1<<i &res:
ans+=p_2
p_2*=2
return ans
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
n=int(input())
arr=tuple(map(int,input().split() ))
print(main(n,arr))
``` | instruction | 0 | 86,798 | 12 | 173,596 |
No | output | 1 | 86,798 | 12 | 173,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
Submitted Solution:
```
"""
NTC here
"""
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
profile = 0
pypy = 1
def iin(): return int(input())
def lin(): return list(map(int, input().split()))
def check(val):
x = val
ch = 0
while x:
if x%2==0:
return False
ch += 1
x//=2
return ch
def main():
n = iin()
a = lin()
b = [[] for i in range(61)]
for j, i in enumerate(a):
x = i
ch = 0
while x:
if x%2:
b[ch].append(j)
x //= 2
ch += 1
ans = -1
st = set()
ch = 1
# print(b)
for i in range(61):
if b[i]:
st.update(b[i])
ch1 = len(st)
# print(i, st, ch1)
a1 = 0
for i in st:
a1 |= a[i]
x = check(a1)
# print(x, a1)
if x:
ans = max(x-1, ans)
print(pow(2, ans+1)-1)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
if pypy:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
if profile:
import cProfile
cProfile.run('main()')
else:
main()
``` | instruction | 0 | 86,799 | 12 | 173,598 |
No | output | 1 | 86,799 | 12 | 173,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
Submitted Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# import numpy as np
sys.setrecursionlimit(int(pow(10,6)))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("out.txt", "w")
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
# @lru_cache(None)
t=1
# t=int(input())
for _ in range(t):
n=l()[0]
A=l()
B=[]
for i in range(n):
x=bin(A[i])[2:]
x="0"*(32-len(x))+x
B.append(x)
for i in range(32):
x=[B[j][i] for j in range(n)]
if "1" in x:
z=2**(32-i)-1
print(z)
exit()
``` | instruction | 0 | 86,800 | 12 | 173,600 |
No | output | 1 | 86,800 | 12 | 173,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers.
The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1).
Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.
Help Ashish find the maximum value he can get by choosing some subsequence of a.
Input
The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.
The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).
Output
Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.
Examples
Input
3
2 1 3
Output
3
Input
3
3 1 4
Output
7
Input
1
1
Output
1
Input
4
7 7 1 1
Output
7
Note
For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.
For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.
For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.
For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def calc(arr):
v = max(1,len(arr)-2)
ans = 0
for i in range(m):
count = 0
for j in range(len(arr)):
if arr[j][i]=="1":
count += 1
if count>=v:
ans += 2**(m-i-1)
return ans
n = int(input())
a = sorted(list(map(int,input().split())))
if n==1:
print (a[0])
exit()
m = len(bin(a[-1])[2:])
g = []
for i in range(n):
b = bin(a[i])[2:]
b = "0"*(m-len(b))+b
g.append(list(b))
g = g[::-1]
c = []
for i in range(n):
if g[i][0]=="1" and len(c)!=1:
c.append(g[i])
# elif g[i][0]=="1":
# continue
else:
ind = i
break
ans = calc(c)
if len(c)==len(g):
print (ans)
exit()
# for i in g:
# print (*i)
# print ()
# for i in c:
# print (*i)
for i in range(ind,n):
c.append(g[i])
for j in range(i+1,n):
c.append(g[j])
ans = max(ans,calc(c))
c.pop()
ans = max(ans,calc(c))
c.pop()
print (ans)
``` | instruction | 0 | 86,801 | 12 | 173,602 |
No | output | 1 | 86,801 | 12 | 173,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
We hid from you a permutation p of length n, consisting of the elements from 1 to n. You want to guess it. To do that, you can give us 2 different indices i and j, and we will reply with p_{i} mod p_{j} (remainder of division p_{i} by p_{j}).
We have enough patience to answer at most 2 ⋅ n queries, so you should fit in this constraint. Can you do it?
As a reminder, a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 10^4) — length of the permutation.
Interaction
The interaction starts with reading n.
Then you are allowed to make at most 2 ⋅ n queries in the following way:
* "? x y" (1 ≤ x, y ≤ n, x ≠ y).
After each one, you should read an integer k, that equals p_x mod p_y.
When you have guessed the permutation, print a single line "! " (without quotes), followed by array p and quit.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Exit immediately after receiving "-1" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Hack format
In the first line output n (1 ≤ n ≤ 10^4). In the second line print the permutation of n integers p_1, p_2, …, p_n.
Example
Input
3
1
2
1
0
Output
? 1 2
? 3 2
? 1 3
? 2 1
! 1 3 2 | instruction | 0 | 86,818 | 12 | 173,636 |
Tags: constructive algorithms, interactive, math, two pointers
Correct Solution:
```
import sys
DEBUG = False
def debug(*args):
if not DEBUG:
return
print("\033[0;31m", end="", file=sys.stderr)
print(*args, file=sys.stderr)
print("\033[0m", end="", file=sys.stderr)
sys.stderr.flush()
def readInt():
line = input()
while line == "":
line = input()
result = int(line)
return result
cache = {}
def query(i, j):
if (i, j) not in cache:
print("? " + str(i + 1) + " " + str(j + 1), file=sys.stdout)
sys.stdout.flush()
if not DEBUG:
x = readInt()
debug("query", i, j, ":", x)
else:
x = REAL[i] % REAL[j]
debug("query", i, j, "\t", REAL[i], "%", REAL[j], ":", x)
cache[(i, j)] = x
return cache[(i, j)]
def answer(arr):
print("! " + " ".join(str(x) for x in arr), file=sys.stdout)
sys.stdout.flush()
debug("ans", arr)
def solve():
# Want the nth move to correspond with the nth bit.
# While reconstructing we just need to know whether to go right or down, so make sure the diagonals alternate bits
if DEBUG:
cache.clear()
N = len(REAL)
debug("Testing", N, REAL)
else:
N = readInt()
if N == 1:
answer([1])
exit()
ans = [-1 for i in range(N)]
last = 0
for i in range(1, N):
a = query(i, last)
b = query(last, i)
if a > b:
# last is larger, so a is a[i]
ans[i] = a
if DEBUG:
assert REAL[last] > REAL[i]
else:
ans[last] = b
if DEBUG:
assert REAL[last] < REAL[i]
last = i
for i in range(N):
if ans[i] == -1:
ans[i] = N
answer(ans)
assert len(cache) <= 2 * N
return ans
if DEBUG:
import random
random.seed(0)
for _ in range(1000):
N = 5
REAL = list(range(1, N + 1))
random.shuffle(REAL)
assert solve() == REAL
exit()
if __name__ == "__main__":
solve()
``` | output | 1 | 86,818 | 12 | 173,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
We hid from you a permutation p of length n, consisting of the elements from 1 to n. You want to guess it. To do that, you can give us 2 different indices i and j, and we will reply with p_{i} mod p_{j} (remainder of division p_{i} by p_{j}).
We have enough patience to answer at most 2 ⋅ n queries, so you should fit in this constraint. Can you do it?
As a reminder, a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 10^4) — length of the permutation.
Interaction
The interaction starts with reading n.
Then you are allowed to make at most 2 ⋅ n queries in the following way:
* "? x y" (1 ≤ x, y ≤ n, x ≠ y).
After each one, you should read an integer k, that equals p_x mod p_y.
When you have guessed the permutation, print a single line "! " (without quotes), followed by array p and quit.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Exit immediately after receiving "-1" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Hack format
In the first line output n (1 ≤ n ≤ 10^4). In the second line print the permutation of n integers p_1, p_2, …, p_n.
Example
Input
3
1
2
1
0
Output
? 1 2
? 3 2
? 1 3
? 2 1
! 1 3 2
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
n=int(input())
#find the max first
perm=[n for s in range(n)];setty=set(list(range(1,n+1)))
for s in range(1,n):
print("? "+str(s)+" "+str(s+1))
sys.stdout.flush()
int1=int(input())
print("? "+str(s+1)+" "+str(s))
sys.stdout.flush()
int2=int(input())
if int1<int2:
perm[s]=int2;setty.discard(int2)
else:
perm[s-1]=int1;setty.discard(int1)
for s in range(n):
if perm[s]==-1:
perm[s]=setty.pop();break
perm=[str(s) for s in perm]
print("! "+" ".join(perm))
sys.stdout.flush()
``` | instruction | 0 | 86,819 | 12 | 173,638 |
No | output | 1 | 86,819 | 12 | 173,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
We hid from you a permutation p of length n, consisting of the elements from 1 to n. You want to guess it. To do that, you can give us 2 different indices i and j, and we will reply with p_{i} mod p_{j} (remainder of division p_{i} by p_{j}).
We have enough patience to answer at most 2 ⋅ n queries, so you should fit in this constraint. Can you do it?
As a reminder, a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 10^4) — length of the permutation.
Interaction
The interaction starts with reading n.
Then you are allowed to make at most 2 ⋅ n queries in the following way:
* "? x y" (1 ≤ x, y ≤ n, x ≠ y).
After each one, you should read an integer k, that equals p_x mod p_y.
When you have guessed the permutation, print a single line "! " (without quotes), followed by array p and quit.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Exit immediately after receiving "-1" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Hack format
In the first line output n (1 ≤ n ≤ 10^4). In the second line print the permutation of n integers p_1, p_2, …, p_n.
Example
Input
3
1
2
1
0
Output
? 1 2
? 3 2
? 1 3
? 2 1
! 1 3 2
Submitted Solution:
```
n= int(input())
a= [0 ]*(n+1)
i=1
j=2
while(j<=n):
print('?',i,j,flush=True)
an=int(input())
print('?',j,i,flush=True)
bn =int(input())
if an>bn:
a[i]=an
i=j
else:
a[j]=bn
j+=1
print('?',1, 2, flush=True)
an = int(input())
print('?', 2 , 1, flush=True)
bn = int(input())
for i in range(1,n):
if a[i]==0:
a[i] = n*(n+1)//2 - sum(a)
break
print("!",end=" ")
print(*a[1:])
``` | instruction | 0 | 86,820 | 12 | 173,640 |
No | output | 1 | 86,820 | 12 | 173,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
We hid from you a permutation p of length n, consisting of the elements from 1 to n. You want to guess it. To do that, you can give us 2 different indices i and j, and we will reply with p_{i} mod p_{j} (remainder of division p_{i} by p_{j}).
We have enough patience to answer at most 2 ⋅ n queries, so you should fit in this constraint. Can you do it?
As a reminder, a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 10^4) — length of the permutation.
Interaction
The interaction starts with reading n.
Then you are allowed to make at most 2 ⋅ n queries in the following way:
* "? x y" (1 ≤ x, y ≤ n, x ≠ y).
After each one, you should read an integer k, that equals p_x mod p_y.
When you have guessed the permutation, print a single line "! " (without quotes), followed by array p and quit.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Exit immediately after receiving "-1" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Hack format
In the first line output n (1 ≤ n ≤ 10^4). In the second line print the permutation of n integers p_1, p_2, …, p_n.
Example
Input
3
1
2
1
0
Output
? 1 2
? 3 2
? 1 3
? 2 1
! 1 3 2
Submitted Solution:
```
from collections import Counter
import string
import math
import sys
# sys.setrecursionlimit(10**6)
from fractions import Fraction
def array_int():
return [int(i) for i in sys.stdin.readline().split()]
def vary(arrber_of_variables):
if arrber_of_variables==1:
return int(sys.stdin.readline())
if arrber_of_variables>=2:
return map(int,sys.stdin.readline().split())
def makedict(var):
return dict(Counter(var))
testcases=1
for _ in range(testcases):
n=vary(1)
i=0
j=n-1
arr=[1 for i in range(n)]
while i<=j:
print('?',i+1,j+1)
sys.stdout.flush()
k=vary(1)
print('?',j+1,i+1)
sys.stdout.flush()
k2=vary(1)
if k>k2:
arr[i]=k
i+=1
else:
arr[j]=k2
j-=1
print('!',*arr)
``` | instruction | 0 | 86,821 | 12 | 173,642 |
No | output | 1 | 86,821 | 12 | 173,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
We hid from you a permutation p of length n, consisting of the elements from 1 to n. You want to guess it. To do that, you can give us 2 different indices i and j, and we will reply with p_{i} mod p_{j} (remainder of division p_{i} by p_{j}).
We have enough patience to answer at most 2 ⋅ n queries, so you should fit in this constraint. Can you do it?
As a reminder, a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 10^4) — length of the permutation.
Interaction
The interaction starts with reading n.
Then you are allowed to make at most 2 ⋅ n queries in the following way:
* "? x y" (1 ≤ x, y ≤ n, x ≠ y).
After each one, you should read an integer k, that equals p_x mod p_y.
When you have guessed the permutation, print a single line "! " (without quotes), followed by array p and quit.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Exit immediately after receiving "-1" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
Hack format
In the first line output n (1 ≤ n ≤ 10^4). In the second line print the permutation of n integers p_1, p_2, …, p_n.
Example
Input
3
1
2
1
0
Output
? 1 2
? 3 2
? 1 3
? 2 1
! 1 3 2
Submitted Solution:
```
n = int(input())
ans = [0]*(n+1)
count = 0
i = 1
j = 2
while count!=(n-1):
print("?",i,j)
z1 = int(input())
print("?",j,i)
z2 = int(input())
if z1>z2:
ans[i] = z1
else:
ans[j] = z2
i+=1
j+=1
count+=1
for i in range(1,n+1):
if ans[i] == 0:
ans[i] = n
print('!',*ans[1:])
``` | instruction | 0 | 86,822 | 12 | 173,644 |
No | output | 1 | 86,822 | 12 | 173,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.
Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.
Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.
Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.
Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).
Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).
The numbers in the lines are separated by single spaces.
Output
On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Examples
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20 | instruction | 0 | 86,952 | 12 | 173,904 |
Tags: data structures, implementation
Correct Solution:
```
import io,os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n,m,k=map(int,input().split())
l=list(map(int,input().split()))
op=[]
for i in range(m):
op.append(list(map(int,input().split())))
kk=[]
for i in range(k):
kk.append(list(map(int,input().split())))
da1=[0]*(1+m)
for s,e in kk:
da1[s]+=1
if e+1 < m+1:
da1[e+1]-=1
r=0
for i in range(1,m+1):
r+=da1[i]
op[i-1][2]*=r
da2=[-1]+[l[0]]
for i in range(n-1):
da2.append(l[i+1]-l[i])
for s,e,w in op:
da2[s]+=w
if e+1 < n+1:
da2[e+1]-=w
r=0
for i in range(1,n+1):
r+=da2[i]
print(r,end=" ")
``` | output | 1 | 86,952 | 12 | 173,905 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.
Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.
Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.
Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.
Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).
Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).
The numbers in the lines are separated by single spaces.
Output
On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Examples
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20 | instruction | 0 | 86,953 | 12 | 173,906 |
Tags: data structures, implementation
Correct Solution:
```
if __name__ == '__main__':
n, m, k = map(int, input().split())
nums = list(map(int, input().split()))
operations = [tuple(map(int, input().split())) for _ in range(m)]
op_counter = [0] * (m+1)
# queries
for _ in range(k):
x, y = map(int, input().split())
op_counter[x-1] += 1
op_counter[y] -= 1
acc = 0
offset = [0]*(n+1)
for i in range(m):
l, r, d = operations[i]
acc += op_counter[i]
offset[l-1] += acc * d
offset[r] -= acc * d
acc = 0
for i in range(n):
acc += offset[i]
nums[i] += acc
print(' '.join(map(str, nums)))
``` | output | 1 | 86,953 | 12 | 173,907 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.
Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.
Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.
Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.
Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).
Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).
The numbers in the lines are separated by single spaces.
Output
On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Examples
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20 | instruction | 0 | 86,954 | 12 | 173,908 |
Tags: data structures, implementation
Correct Solution:
```
import sys
def answer(n, m, k, a, ops, q):
cnt_m = [0 for i in range(m+1)] # number of operations of each type to be performed. 1 to m, based on q.
for i in range(k):
l = q[i][0]
r = q[i][1]
cnt_m[l-1] += 1
cnt_m[r] -= 1
summ_a = [0 for i in range(n+1)]
s = 0
for i in range(m):
d = ops[i][2]
s += cnt_m[i]
summ = s * d
l = ops[i][0]
r = ops[i][1]
summ_a[l-1] += summ
summ_a[r] -= summ
s = 0
for i in range(n):
s += summ_a[i]
a[i] += s
return ' '.join(map(str, a))
def main():
n, m, k = map(int, sys.stdin.readline().split())
a = list(map(int, sys.stdin.readline().split()))
ops = [0 for i in range(m)]
q = [0 for i in range(k)]
for i in range(m):
ops[i] = list(map(int, sys.stdin.readline().split()))
for j in range(k):
q[j] = list(map(int, sys.stdin.readline().split()))
print(answer(n, m, k, a, ops, q))
return
main()
``` | output | 1 | 86,954 | 12 | 173,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.
Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.
Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.
Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.
Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).
Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).
The numbers in the lines are separated by single spaces.
Output
On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Examples
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20 | instruction | 0 | 86,955 | 12 | 173,910 |
Tags: data structures, implementation
Correct Solution:
```
from sys import *
rd = lambda: list(map(int, stdin.readline().split()))
n, m, k = rd()
a = rd()
b = [rd() for _ in range(m)]
x = [0]*(m+1)
y = [0]*(n+1)
for _ in range(k):
l, r = rd()
x[l-1] += 1
x[r] -= 1
s = 0
for i in range(m):
l, r, d = b[i]
s += x[i]
y[l-1] += s*d
y[r] -= s*d
s = 0
for i in range(n):
s += y[i]
a[i] += s
print(*a)
``` | output | 1 | 86,955 | 12 | 173,911 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.
Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.
Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.
Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.
Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).
Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).
The numbers in the lines are separated by single spaces.
Output
On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Examples
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20 | instruction | 0 | 86,956 | 12 | 173,912 |
Tags: data structures, implementation
Correct Solution:
```
n , m , k =map(int,input().split())
a = list(map(int,input().split()))[:n]
s = []
x = [0]*(m+1)
y = [0]*(n+1)
for i in range(m):
l,r,d = map(int,input().split())
s.append((l,r,d))
for i in range(k):
j , v = map(int,input().split())
x[j-1]+=1
x[v]-=1
cnt = 0
for i in range(m):
cnt+=x[i]
l,r,d = s[i]
y[l-1]+=cnt*d
y[r]-=cnt*d
cnt1 = 0
for i in range(n):
cnt1+=y[i]
a[i]+=cnt1
print(*a[:n])
``` | output | 1 | 86,956 | 12 | 173,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.
Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.
Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.
Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.
Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).
Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).
The numbers in the lines are separated by single spaces.
Output
On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Examples
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20 | instruction | 0 | 86,957 | 12 | 173,914 |
Tags: data structures, implementation
Correct Solution:
```
from itertools import accumulate
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
oper = [tuple(map(int, input().split())) for i in range(m)]
zapr = [tuple(map(int, input().split())) for i in range(k)]
count_ = [0 for i in range(m + 1)]
for el in zapr:
x, y = el
count_[x - 1] += 1
count_[y] -= 1
counter_ = list(accumulate(count_))[:-1]
a.append(0)
a_count = [a[0]]
for i, el in enumerate(a[1:]):
a_count.append(el - a[i])
for i, el in enumerate(oper):
l, r, d = el
d *= counter_[i]
a_count[l - 1] += d
a_count[r] -= d
a = list(accumulate(a_count))[:-1]
print(' '.join(map(str, a)))
``` | output | 1 | 86,957 | 12 | 173,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.
Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.
Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.
Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.
Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).
Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).
The numbers in the lines are separated by single spaces.
Output
On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Examples
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20 | instruction | 0 | 86,958 | 12 | 173,916 |
Tags: data structures, implementation
Correct Solution:
```
n,m,k = map(int,input().split())
a = list(map(int,input().split()))
ops = [list(map(int,input().split())) for _ in range(m)]
qus = [list(map(int,input().split())) for _ in range(k)]
starts = [0]*(m+1)
for l,r in qus:
starts[l-1] += 1
starts[r] -= 1
opcount = [0]*m
active = 0
for i in range(m):
active += starts[i]
opcount[i] = active
suffixOffset = [0]*(n+1)
for i in range(m):
l,r,d = ops[i]
suffixOffset[l-1] += opcount[i]*d
suffixOffset[r] -= opcount[i]*d
active = 0
for i in range(n):
active += suffixOffset[i]
a[i] += active
print(*a)
``` | output | 1 | 86,958 | 12 | 173,917 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.
Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.
Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.
Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.
Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).
Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).
The numbers in the lines are separated by single spaces.
Output
On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Examples
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20 | instruction | 0 | 86,959 | 12 | 173,918 |
Tags: data structures, implementation
Correct Solution:
```
def main():
n,m,k = map(int,input().split())
arr = list(map(int,input().split()))
ops = [0]*m
ops_arr = []
for i in range(m):
l,r,d = map(int,input().split())
ops_arr.append([l,r,d])
for i in range(k):
x,y = map(int,input().split())
x -= 1
y -= 1
ops[x] += 1
if y+1 < m:
ops[y+1] -= 1
for i in range(1,m):
ops[i] += ops[i-1]
#print(ops)
for i in range(m):
ops_arr[i][2] *= ops[i]
ans = [0]*n
for op in ops_arr:
l,r,d = op[0],op[1],op[2]
l -= 1
r -= 1
ans[l] += d
if r+1 < n:
ans[r+1] -= d
for i in range(1,n):
ans[i] += ans[i-1]
for i in range(n):
ans[i] += arr[i]
for i in ans:
print(i,end = ' ')
main()
``` | output | 1 | 86,959 | 12 | 173,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0 | instruction | 0 | 87,049 | 12 | 174,098 |
Tags: data structures
Correct Solution:
```
from math import inf, log2
class LazySegmentTree:
def __init__(self, array, func=max):
self.n = len(array)
self.size = 2**(int(log2(self.n-1))+1) if n != 1 else 1
self.func = func
self.default = 0 if self.func != min else inf
self.data = [self.default] * (2 * self.size)
self.lazy = [0] * (2 * self.size)
self.process(array)
def process(self, array):
self.data[self.size : self.size+self.n] = array
for i in range(self.size-1, -1, -1):
self.data[i] = self.func(self.data[2*i], self.data[2*i+1])
def push(self, index):
self.lazy[2*index] += self.lazy[index]
self.lazy[2*index+1] += self.lazy[index]
self.data[2 * index] += self.lazy[index]
self.data[2 * index + 1] += self.lazy[index]
self.lazy[index] = 0
def build(self, index):
"""Build data with the new changes!"""
index >>= 1
while index:
self.data[index] = self.func(self.data[2*index], self.data[2*index+1]) + self.lazy[index]
index >>= 1
def query(self, alpha, omega):
res = self.default
alpha += self.size
omega += self.size + 1
for i in range(len(bin(alpha)[2:])-1, 0, -1):
self.push(alpha >> i)
for i in range(len(bin(omega-1)[2:])-1, 0, -1):
self.push((omega-1) >> i)
while alpha < omega:
if alpha & 1:
res = self.func(res, self.data[alpha])
alpha += 1
if omega & 1:
omega -= 1
res = self.func(res, self.data[omega])
alpha >>= 1
omega >>= 1
return res
def update(self, alpha, omega, value):
alpha += self.size
omega += self.size + 1
l, r = alpha, omega
while alpha < omega:
if alpha & 1:
self.data[alpha] += value
self.lazy[alpha] += value
alpha += 1
if omega & 1:
omega -= 1
self.data[omega] += value
self.lazy[omega] += value
alpha >>= 1
omega >>= 1
self.build(l)
self.build(r-1)
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
n = int(input())
a = list(map(int, input().split()))
m = int(input())
st = LazySegmentTree(a, func=min)
for _ in range(m):
s = input().split()
if len(s) == 3:
a, b, c = map(int, s)
if a > b:
st.update(a, n-1, c)
st.update(0, b, c)
else:
st.update(a,b,c)
else:
a, b = map(int, s)
if a > b:
print(min(st.query(a,n-1), st.query(0,b)))
else:
print(st.query(a,b))
``` | output | 1 | 87,049 | 12 | 174,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0 | instruction | 0 | 87,050 | 12 | 174,100 |
Tags: data structures
Correct Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def solve():
n=int(input())
a=list(map(int,input().split()))
class st:
def __init__(self, size):
N = 1
h = 0
while N < size:
N <<= 1
h += 1
self.N = N
self.h = h
self.t = [float('inf')] * (2 * N)
self.d = [0] * N
def apply(self, p, value):
self.t[p] += value
if (p < self.N):
self.d[p] += value
def build(self, p):
t = self.t
d = self.d
while p > 1:
p >>= 1
t[p] = min(t[p<<1], t[p<<1|1]) + d[p]
def rebuild(self):
t = self.t
for p in reversed(range(1, self.N)):
t[p] = min(t[p<<1], t[p<<1|1])
def push(self, p):
d = self.d
for s in range(self.h, 0, -1):
i = p >> s
if d[i] != 0:
self.apply(i<<1, d[i])
self.apply(i<<1|1, d[i])
d[i] = 0
def inc(self, l, r, value):
if l > r:
return
l += self.N
r += self.N+1
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, value)
l += 1
if r & 1:
r -= 1
self.apply(r, value)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
if l > r:
return float('inf')
t = self.t
l += self.N
r += self.N+1
self.push(l)
self.push(r - 1)
res = float('inf')
while l < r:
if l & 1:
res = min(res, t[l])
l += 1
if r & 1:
r -= 1
res = min(t[r], res)
l >>= 1
r >>= 1
return res
se=st(n)
N=se.N
for i in range(n):
se.t[i+N]=a[i]
se.rebuild()
q=int(input())
for i in range(q):
b=list(map(int,input().split()))
if len(b)==2:
l=b[0]
r=b[1]
if l>r:
print(min(se.query(l,n-1),se.query(0,r)))
else:print(se.query(l,r))
else:
l=b[0]
r=b[1]
v=b[2]
if l>r:
se.inc(0,r,v)
se.inc(l,n-1,v)
else:se.inc(l,r,v)
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
``` | output | 1 | 87,050 | 12 | 174,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0 | instruction | 0 | 87,051 | 12 | 174,102 |
Tags: data structures
Correct Solution:
```
import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n=int(input())
a=list(map(int,input().split()))
class st:
def __init__(self, size):
N = 1
h = 0
while N < size:
N <<= 1
h += 1
self.N = N
self.h = h
self.t = [float('inf')] * (2 * N)
self.d = [0] * N
def apply(self, p, value):
self.t[p] += value
if (p < self.N):
self.d[p] += value
def build(self, p):
t = self.t
d = self.d
while p > 1:
p >>= 1
t[p] = min(t[p<<1], t[p<<1|1]) + d[p]
def rebuild(self):
t = self.t
for p in reversed(range(1, self.N)):
t[p] = min(t[p<<1], t[p<<1|1])
def push(self, p):
d = self.d
for s in range(self.h, 0, -1):
i = p >> s
if d[i] != 0:
self.apply(i<<1, d[i])
self.apply(i<<1|1, d[i])
d[i] = 0
def inc(self, l, r, value):
if l >= r:
return
l += self.N
r += self.N
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, value)
l += 1
if r & 1:
r -= 1
self.apply(r, value)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
if l >= r:
return float('inf')
t = self.t
l += self.N
r += self.N
self.push(l)
self.push(r - 1)
res = float('inf')
while l < r:
if l & 1:
res = min(res, t[l])
l += 1
if r & 1:
r -= 1
res = min(t[r], res)
l >>= 1
r >>= 1
return res
se=st(n)
N=se.N
for i in range(n):
se.t[i+N]=a[i]
se.rebuild()
q=int(input())
for i in range(q):
b=list(map(int,input().split()))
if len(b)==2:
l=b[0]
r=b[1]
if l>r:
print(min(se.query(l,n),se.query(0,r+1)))
else:print(se.query(l,r+1))
else:
l=b[0]
r=b[1]
v=b[2]
if l>r:
se.inc(0,r+1,v)
se.inc(l,n,v)
else:se.inc(l,r+1,v)
``` | output | 1 | 87,051 | 12 | 174,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0 | instruction | 0 | 87,052 | 12 | 174,104 |
Tags: data structures
Correct Solution:
```
import os, sys
from io import IOBase, BytesIO
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self,a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = '\n'
n=int(input())
a=list(map(int,input().split()))
class st:
def __init__(self, size):
N = 1
h = 0
while N < size:
N <<= 1
h += 1
self.N = N
self.h = h
self.t = [float('inf')] * (2 * N)
self.d = [0] * N
def apply(self, p, value):
self.t[p] += value
if (p < self.N):
self.d[p] += value
def build(self, p):
t = self.t
d = self.d
while p > 1:
p >>= 1
t[p] = min(t[p<<1], t[p<<1|1]) + d[p]
def rebuild(self):
t = self.t
for p in reversed(range(1, self.N)):
t[p] = min(t[p<<1], t[p<<1|1])
def push(self, p):
d = self.d
for s in range(self.h, 0, -1):
i = p >> s
if d[i] != 0:
self.apply(i<<1, d[i])
self.apply(i<<1|1, d[i])
d[i] = 0
def inc(self, l, r, value):
if l >= r:
return
l += self.N
r += self.N
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, value)
l += 1
if r & 1:
r -= 1
self.apply(r, value)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
if l >= r:
return float('inf')
t = self.t
l += self.N
r += self.N
self.push(l)
self.push(r - 1)
res = float('inf')
while l < r:
if l & 1:
res = min(res, t[l])
l += 1
if r & 1:
r -= 1
res = min(t[r], res)
l >>= 1
r >>= 1
return res
se=st(n)
N=se.N
for i in range(n):
se.t[i+N]=a[i]
se.rebuild()
q=int(input())
for i in range(q):
b=list(map(int,input().split()))
if len(b)==2:
l=b[0]
r=b[1]
if l>r:
print(min(se.query(l,n),se.query(0,r+1)))
else:print(se.query(l,r+1))
else:
l=b[0]
r=b[1]
v=b[2]
if l>r:
se.inc(0,r+1,v)
se.inc(l,n,v)
else:se.inc(l,r+1,v)
``` | output | 1 | 87,052 | 12 | 174,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0 | instruction | 0 | 87,053 | 12 | 174,106 |
Tags: data structures
Correct Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def solve():
n=int(input())
a=list(map(int,input().split()))
class st:
def __init__(self, size):
N = 1
h = 0
while N < size:
N <<= 1
h += 1
self.N = N
self.h = h
self.t = [float('inf')] * (2 * N)
self.d = [0] * N
def apply(self, p, value):
self.t[p] += value
if (p < self.N):
self.d[p] += value
def build(self, p):
t = self.t
d = self.d
while p > 1:
p >>= 1
t[p] = min(t[p<<1], t[p<<1|1]) + d[p]
def rebuild(self):
t = self.t
for p in reversed(range(1, self.N)):
t[p] = min(t[p<<1], t[p<<1|1])
def push(self, p):
d = self.d
for s in range(self.h, 0, -1):
i = p >> s
if d[i] != 0:
self.apply(i<<1, d[i])
self.apply(i<<1|1, d[i])
d[i] = 0
def inc(self, l, r, value):
if l >= r:
return
l += self.N
r += self.N
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, value)
l += 1
if r & 1:
r -= 1
self.apply(r, value)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
if l >= r:
return float('inf')
t = self.t
l += self.N
r += self.N
self.push(l)
self.push(r - 1)
res = float('inf')
while l < r:
if l & 1:
res = min(res, t[l])
l += 1
if r & 1:
r -= 1
res = min(t[r], res)
l >>= 1
r >>= 1
return res
se=st(n)
N=se.N
for i in range(n):
se.t[i+N]=a[i]
se.rebuild()
q=int(input())
for i in range(q):
b=list(map(int,input().split()))
if len(b)==2:
l=b[0]
r=b[1]
if l>r:
print(min(se.query(l,n),se.query(0,r+1)))
else:print(se.query(l,r+1))
else:
l=b[0]
r=b[1]
v=b[2]
if l>r:
se.inc(0,r+1,v)
se.inc(l,n,v)
else:se.inc(l,r+1,v)
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
``` | output | 1 | 87,053 | 12 | 174,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0 | instruction | 0 | 87,054 | 12 | 174,108 |
Tags: data structures
Correct Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def solve():
n=int(input())
a=list(map(int,input().split()))
class st:
def __init__(self, size):
n = size
h = int(math.log2(n-1))+1 if n>1 else 0
N = 2**h
self.t = [float('inf')]*(2*N)
self.d = [0]*N
# N = 1
# h = 0
# while N < size:
# N <<= 1
# h += 1
self.N = N
self.h = h
self.t = [float('inf')] * (2 * N)
self.d = [0] * N
def apply(self, p, value):
self.t[p] += value
if (p < self.N):
self.d[p] += value
def build(self, p):
t = self.t
d = self.d
while p > 1:
p >>= 1
t[p] = min(t[p<<1], t[p<<1|1]) + d[p]
def rebuild(self):
t = self.t
for p in reversed(range(1, self.N)):
t[p] = min(t[p<<1], t[p<<1|1])
def push(self, p):
d = self.d
for s in range(self.h, 0, -1):
i = p >> s
if d[i] != 0:
self.apply(i<<1, d[i])
self.apply(i<<1|1, d[i])
d[i] = 0
def inc(self, l, r, value):
if l > r:
return
l += self.N
r += self.N+1
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, value)
l += 1
if r & 1:
r -= 1
self.apply(r, value)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
if l > r:
return float('inf')
t = self.t
l += self.N
r += self.N+1
self.push(l)
self.push(r - 1)
res = float('inf')
while l < r:
if l & 1:
res = min(res, t[l])
l += 1
if r & 1:
r -= 1
res = min(t[r], res)
l >>= 1
r >>= 1
return res
se=st(n)
N=se.N
for i in range(n):
se.t[i+N]=a[i]
se.rebuild()
q=int(input())
for i in range(q):
b=list(map(int,input().split()))
if len(b)==2:
l=b[0]
r=b[1]
if l>r:
print(min(se.query(l,n-1),se.query(0,r)))
else:print(se.query(l,r))
else:
l=b[0]
r=b[1]
v=b[2]
if l>r:
se.inc(0,r,v)
se.inc(l,n-1,v)
else:se.inc(l,r,v)
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
``` | output | 1 | 87,054 | 12 | 174,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0 | instruction | 0 | 87,055 | 12 | 174,110 |
Tags: data structures
Correct Solution:
```
import sys,os,io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
inf = float('inf')
class LazySegmentTree:
def __init__(self, data, default=inf, func=min):
"""initialize the lazy segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self._lazy = [0] * (2 * _size)
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __len__(self):
return self._len
def _push(self, idx):
"""push query on idx to its children"""
# Let the children know of the queries
q, self._lazy[idx] = self._lazy[idx], 0
self._lazy[2 * idx] += q
self._lazy[2 * idx + 1] += q
self.data[2 * idx] += q
self.data[2 * idx + 1] += q
def _update(self, idx):
"""updates the node idx to know of all queries applied to it via its ancestors"""
for i in reversed(range(1, idx.bit_length())):
self._push(idx >> i)
def _build(self, idx):
"""make the changes to idx be known to its ancestors"""
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1]) + self._lazy[idx]
idx >>= 1
def add(self, start, stop, value):
"""lazily add value to [start, stop)"""
start = start_copy = start + self._size
stop = stop_copy = stop + self._size
while start < stop:
if start & 1:
self._lazy[start] += value
self.data[start] += value
start += 1
if stop & 1:
stop -= 1
self._lazy[stop] += value
self.data[stop] += value
start >>= 1
stop >>= 1
# Tell all nodes above of the updated area of the updates
self._build(start_copy)
self._build(stop_copy - 1)
def query(self, start, stop, default=inf):
"""func of data[start, stop)"""
start += self._size
stop += self._size
# Apply all the lazily stored queries
self._update(start)
self._update(stop - 1)
res = default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "LazySegmentTree({0})".format(self.data)
n = int(input())
a = [int(i) for i in input().split()]
q = int(input())
st = LazySegmentTree(a)
for i in range (q):
qi = [int(i) for i in input().split()]
if len(qi)==3:
if qi[0]>qi[1]:
st.add(qi[0],n,qi[2])
st.add(0,qi[1]+1,qi[2])
else:
st.add(qi[0],qi[1]+1, qi[2])
else:
if qi[0]>qi[1]:
ans = min(st.query(qi[0],n), st.query(0,qi[1]+1))
else:
ans = st.query(qi[0],qi[1]+1)
print(ans)
``` | output | 1 | 87,055 | 12 | 174,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0 | instruction | 0 | 87,056 | 12 | 174,112 |
Tags: data structures
Correct Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def solve():
n=int(input())
a=list(map(int,input().split()))
class st:
def __init__(self, size):
n = size
h = int(math.log2(n-1))+1 if n>1 else 0
N = 2**h
self.t = [float('inf')]*(2*N)
self.d = [0]*N
self.N = N
self.h = h
def apply(self, p, value):
self.t[p] += value
if (p < self.N):
self.d[p] += value
def build(self, p):
t = self.t
d = self.d
while p > 1:
p >>= 1
t[p] = min(t[p<<1], t[p<<1|1]) + d[p]
def rebuild(self):
t = self.t
for p in reversed(range(1, self.N)):
t[p] = min(t[p<<1], t[p<<1|1])
def push(self, p):
d = self.d
for s in range(self.h, 0, -1):
i = p >> s
if d[i] != 0:
self.apply(i<<1, d[i])
self.apply(i<<1|1, d[i])
d[i] = 0
def inc(self, l, r, value):
if l > r:
return
l += self.N
r += self.N+1
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, value)
l += 1
if r & 1:
r -= 1
self.apply(r, value)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
if l > r:
return float('inf')
t = self.t
l += self.N
r += self.N+1
self.push(l)
self.push(r - 1)
res = float('inf')
while l < r:
if l & 1:
res = min(res, t[l])
l += 1
if r & 1:
r -= 1
res = min(t[r], res)
l >>= 1
r >>= 1
return res
se=st(n)
N=se.N
for i in range(n):
se.t[i+N]=a[i]
se.rebuild()
q=int(input())
for i in range(q):
b=list(map(int,input().split()))
if len(b)==2:
l=b[0]
r=b[1]
if l>r:
print(min(se.query(l,n-1),se.query(0,r)))
else:print(se.query(l,r))
else:
l=b[0]
r=b[1]
v=b[2]
if l>r:
se.inc(0,r,v)
se.inc(l,n-1,v)
else:se.inc(l,r,v)
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
``` | output | 1 | 87,056 | 12 | 174,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0
Submitted Solution:
```
from math import inf, log2
class LazySegmentTree:
def __init__(self, array, func=max):
self.n = len(array)
self.size = 2**(int(log2(self.n-1))+1) if n != 1 else 1
self.func = func
self.default = 0 if self.func != min else inf
self.data = [self.default] * (2 * self.size)
self.lazy = [0] * (2 * self.size)
self.process(array)
def process(self, array):
self.data[self.size : self.size+self.n] = array
for i in range(self.size-1, -1, -1):
self.data[i] = self.func(self.data[2*i], self.data[2*i+1])
def push(self, index):
"""Push the information of the root to it's children!"""
self.lazy[2*index] += self.lazy[index]
self.lazy[2*index+1] += self.lazy[index]
self.data[2 * index] += self.lazy[index]
self.data[2 * index + 1] += self.lazy[index]
self.lazy[index] = 0
def build(self, index):
"""Build data with the new changes!"""
index >>= 1
while index:
self.data[index] = self.func(self.data[2*index], self.data[2*index+1]) + self.lazy[index]
index >>= 1
def query(self, alpha, omega):
"""Returns the result of function over the range (inclusive)!"""
res = self.default
alpha += self.size
omega += self.size + 1
for i in range(len(bin(alpha)[2:])-1, 0, -1):
self.push(alpha >> i)
for i in range(len(bin(omega-1)[2:])-1, 0, -1):
self.push((omega-1) >> i)
while alpha < omega:
if alpha & 1:
res = self.func(res, self.data[alpha])
alpha += 1
if omega & 1:
omega -= 1
res = self.func(res, self.data[omega])
alpha >>= 1
omega >>= 1
return res
def update(self, alpha, omega, value):
"""Increases all elements in the range (inclusive) by given value!"""
alpha += self.size
omega += self.size + 1
l, r = alpha, omega
while alpha < omega:
if alpha & 1:
self.data[alpha] += value
self.lazy[alpha] += value
alpha += 1
if omega & 1:
omega -= 1
self.data[omega] += value
self.lazy[omega] += value
alpha >>= 1
omega >>= 1
self.build(l)
self.build(r-1)
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
n = int(input())
a = list(map(int, input().split()))
m = int(input())
st = LazySegmentTree(a, func=min)
for _ in range(m):
s = input().split()
if len(s) == 3:
a, b, c = map(int, s)
if a > b:
st.update(a, n-1, c)
st.update(0, b, c)
else:
st.update(a,b,c)
else:
a, b = map(int, s)
if a > b:
print(min(st.query(a,n-1), st.query(0,b)))
else:
print(st.query(a,b))
``` | instruction | 0 | 87,057 | 12 | 174,114 |
Yes | output | 1 | 87,057 | 12 | 174,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0
Submitted Solution:
```
from typing import TypeVar, Generic, Callable, List
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
T = TypeVar('T')
S = TypeVar('S')
class StarrySkyTree(Generic[T, S]):
__slots__ = ['size', 'node', 'lazy', 'unit', 'lazy_unit',
'op', 'upd', 'lazy_upd', 'subt_size']
def __init__(self, size: int, default: T, unit: T, lazy_unit: S,
op: Callable[[T, T], T]) -> None:
self.size = size2 = 1 << (len(bin(size)) - 2)
self.unit, self.lazy_unit = unit, lazy_unit
self.node = [default] * (size2 * 2)
self.lazy = [lazy_unit] * (size2 * 2)
self.op = op
self.subt_size = subt = [0] * size2 + [1] * size + [0] * (size2 - size)
for i in range(size2 * 2 - 1, 1, -1):
subt[i >> 1] += subt[i]
def build(self, a: List[T]) -> None:
node = self.node
node[self.size:self.size + len(a)] = a
for i in range(self.size - 1, 0, -1):
node[i] = self.op(node[i << 1], node[(i << 1) + 1])
def find(self, left: int, right: int) -> T:
left, right = left + self.size, right + self.size
node, lazy = self.node, self.lazy
self.__propagate(self.__enum_index(left, right))
result = self.unit
while left < right:
if left & 1:
if lazy[left] == self.lazy_unit:
result = self.op(node[left], result)
else:
# node + lazy
result = self.op(node[left] + lazy[left], result)
left += 1
if right & 1:
if lazy[right - 1] == self.lazy_unit:
result = self.op(node[right - 1], result)
else:
# node + lazy
result = self.op(node[right - 1] + lazy[right - 1], result)
left, right = left >> 1, right >> 1
return result
def update(self, left: int, right: int, value: S) -> None:
left, right = left + self.size, right + self.size
node, lazy, l_unit = self.node, self.lazy, self.lazy_unit
_l, _r = left, right
while _l < _r:
if _l & 1:
# update lazy
lazy[_l] += value
_l += 1
if _r & 1:
# update lazy
lazy[_r - 1] += value
_l, _r = _l >> 1, _r >> 1
for i in self.__enum_index(left, right):
node[i] = self.op(
node[i * 2] if lazy[i * 2] == l_unit else node[i * 2] + lazy[i * 2],
node[i * 2 + 1] if lazy[i * 2 + 1] == l_unit else node[i * 2 + 1] + lazy[i * 2 + 1])
def __enum_index(self, left: int, right: int) -> List[int]:
flag, idx = 0, []
while left < right:
if flag & 1:
idx.append(left)
if flag & 2:
idx.append(right)
flag |= (left & 1) | ((right & 1) * 2)
left, right = left >> 1, right >> 1
while left:
idx.append(left)
left >>= 1
return idx
def __propagate(self, index_list):
node, lazy, l_unit = self.node, self.lazy, self.lazy_unit
for i in reversed(index_list):
if lazy[i] != l_unit:
# refresh node
node[i] += lazy[i]
if i < self.size:
# update lazy
lazy[i * 2] += lazy[i]
lazy[i * 2 + 1] += lazy[i]
lazy[i] = l_unit
n = int(input())
segt = StarrySkyTree[int, int](n, 10**9, 10**18, 0, min)
segt.build(list(map(int, input().split())))
t = int(input())
ans = []
for _ in range(t):
q = tuple(map(int, input().split()))
if len(q) == 3:
if q[0] > q[1]:
segt.update(q[0], n + 1, q[2])
segt.update(0, q[1] + 1, q[2])
else:
segt.update(q[0], q[1] + 1, q[2])
else:
if q[0] > q[1]:
ans.append(min(segt.find(q[0], n + 1), segt.find(0, q[1] + 1)))
else:
ans.append(segt.find(q[0], q[1] + 1))
print(*ans, sep='\n')
``` | instruction | 0 | 87,058 | 12 | 174,116 |
Yes | output | 1 | 87,058 | 12 | 174,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0
Submitted Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def solve():
class st:
def __init__(self, n):
self.ntrl = float('inf')
h = int(math.log2(n-1))+1 if n>1 else 0
N = 2**h
self.t = [self.ntrl]*(2*N)
self.d = [0]*N
self.N = N
self.h = h
def merge(self,a,b):
return min(a,b)
def apply(self, p, value):
self.t[p] += value
if (p < self.N):
self.d[p] += value
def build(self, p):
t = self.t
d = self.d
while p > 1:
p >>= 1
t[p] = self.merge(t[p<<1], t[p<<1|1]) + d[p]
def rebuild(self):
t = self.t
for p in reversed(range(1, self.N)):
t[p] = self.merge(t[p<<1], t[p<<1|1])
def push(self, p):
d = self.d
for s in range(self.h, 0, -1):
i = p >> s
if d[i] != 0:
self.apply(i<<1, d[i])
self.apply(i<<1|1, d[i])
d[i] = 0
def inc(self, l, r, value):
if l > r:
return
l += self.N
r += self.N+1
l0, r0 = l, r
while l < r:
if l & 1:
self.apply(l, value)
l += 1
if r & 1:
r -= 1
self.apply(r, value)
l >>= 1
r >>= 1
self.build(l0)
self.build(r0 - 1)
def query(self, l, r):
if l > r:
return self.ntrl
t = self.t
l += self.N
r += self.N+1
self.push(l)
self.push(r - 1)
res = self.ntrl
while l < r:
if l & 1:
res = self.merge(res, t[l])
l += 1
if r & 1:
r -= 1
res = self.merge(t[r], res)
l >>= 1
r >>= 1
return res
n,=aj()
a=aj()
se=st(n)
N=se.N
for i in range(n):
se.t[i+N]=a[i]
se.rebuild()
q=int(input())
for i in range(q):
b=list(map(int,input().split()))
if len(b)==2:
l=b[0]
r=b[1]
if l>r:
print(min(se.query(l,n-1),se.query(0,r)))
else:print(se.query(l,r))
else:
l=b[0]
r=b[1]
v=b[2]
if l>r:
se.inc(0,r,v)
se.inc(l,n-1,v)
else:se.inc(l,r,v)
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
``` | instruction | 0 | 87,059 | 12 | 174,118 |
Yes | output | 1 | 87,059 | 12 | 174,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0
Submitted Solution:
```
import sys
from math import inf, log2
class LazySegmentTree:
def __init__(self, array, func=min):
self.n = len(array)
self.size = 2**(int(log2(self.n-1))+1) if n != 1 else 1
self.func = func
self.default = 0 if self.func != min else inf
self.data = [self.default] * (2 * self.size)
self.lazy = [0] * (2 * self.size)
self.process(array)
def process(self, array):
self.data[self.size : self.size+self.n] = array
for i in range(self.size-1, -1, -1):
self.data[i] = self.func(self.data[2*i], self.data[2*i+1])
def push(self, index):
self.lazy[2*index] += self.lazy[index]
self.lazy[2*index+1] += self.lazy[index]
self.data[2 * index] += self.lazy[index]
self.data[2 * index + 1] += self.lazy[index]
self.lazy[index] = 0
def build(self, index):
index >>= 1
while index:
self.data[index] = self.func(self.data[2*index], self.data[2*index+1]) + self.lazy[index]
index >>= 1
def query(self, alpha, omega):
res = self.default
alpha += self.size
omega += self.size + 1
for i in range(len(bin(alpha)[2:])-1, 0, -1):
self.push(alpha >> i)
for i in range(len(bin(omega-1)[2:])-1, 0, -1):
self.push((omega-1) >> i)
while alpha < omega:
if alpha & 1:
res = self.func(res, self.data[alpha])
alpha += 1
if omega & 1:
omega -= 1
res = self.func(res, self.data[omega])
alpha >>= 1
omega >>= 1
return res
def update(self, alpha, omega, value):
alpha += self.size
omega += self.size + 1
l, r = alpha, omega
while alpha < omega:
if alpha & 1:
self.data[alpha] += value
self.lazy[alpha] += value
alpha += 1
if omega & 1:
omega -= 1
self.data[omega] += value
self.lazy[omega] += value
alpha >>= 1
omega >>= 1
self.build(l)
self.build(r-1)
debug=True
lines = sys.stdin.read().split("\n")
lines.pop(-1)
debug=False
n = int(lines.pop(0))
nodes = [int(x) for x in lines.pop(0).split(" ")]
ops = int(lines.pop(0))
segtree = LazySegmentTree(nodes)
counter=1
for line in lines:
values = line.split(" ")
if len(values) == 2:
a,b = values
a=int(a)
b=int(b)
if (debug):
print("===================================")
print("Consulta de Minimo N "+str(counter)+":")
print("A es "+str(a))
print("B es "+str(b))
print("Resultado:",end="")
if a <b:
print(segtree.query(a,b))
elif a==b:
print(segtree.query(a%n,a%n))
else:
if abs(a-b)== 1:
print(segtree.query(0, n-1))
else:
print(min(segtree.query(a%n,n-1), segtree.query(0,b) ) )
counter+=1
else:
a,b,v = values
a=int(a)
b=int(b)
v=int(v)
if (debug):
print("==========")
#print(segtree)
print("Update de Rango:")
print("A es "+str(a))
print("B es "+str(b))
print("V es "+str(v))
if a<=b:
segtree.update(a,b,v)
else:
segtree.update(a,n-1,v)
segtree.update(0,b,v);
``` | instruction | 0 | 87,060 | 12 | 174,120 |
Yes | output | 1 | 87,060 | 12 | 174,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0
Submitted Solution:
```
from math import ceil, log2
def inc(segment, lazy, fr, to, value, L, R, index):
if(R < L):
return INF
if(lazy[index] != 0):
segment[index] += lazy[index]
if(L != R):
lazy[index*2+1] += lazy[index]
lazy[index*2+2] += lazy[index]
lazy[index] = 0
if(to < L or R< fr):
return min(INF, segment[index])
elif(fr <= L and to >= R):
segment[index]+= value
if(L!= R):
lazy[index*2 +1] += value
lazy[index*2 +2] += value
return segment[index]
else:
mid = (L +R) //2
u =inc(segment, lazy, fr, to, value, L, mid, index*2+1)
v = inc(segment, lazy, fr, to, value, mid+ 1, R, index*2+2)
segment[index] = min(u,v)
return segment[index]
def RMQ(segment, lazy, fr, to, L, R , index):
# -- L ---fr --to-R
if(L >R):
return INF
if(lazy[index] != 0):
segment[index] += lazy[index]
if(L != R):
lazy[index*2+1] += lazy[index]
lazy[index*2+2] += lazy[index]
lazy[index] = 0
if(to < L or R < fr):
return INF
elif(fr <= L and to >= R):
return segment[index]
else:
mid = (L + R) //2
u = RMQ(segment, lazy, fr, to, L,mid , 2* index+ 1)
v = RMQ(segment, lazy, fr, to, mid + 1,R , 2* index+ 2)
segment[index] = min(u, v)
return segment[index]
def buildTree(segment, a, L ,R, index):
if(L ==R) :
segment[index] = a[L]
return segment[index]
mid= (L + R) //2
u = buildTree(segment, a,L ,mid , 2* index + 1 )
v = buildTree(segment,a, mid + 1 ,R, 2* index + 2)
segment[index] = min(u, v)
return segment[index]
if __name__ == '__main__':
INF = 1e9
n = int(input())
a = list(map(int, input().split()))
# for i in range(0, n-1, 1):
# a.append(a[i])
# print(a)
size = (2*( 2** (ceil(log2( len(a) ) )) ) - 1)
segment = [INF] * size
lazy = [0] * size
buildTree(segment,a, 0 , len(a) -1, 0)
# print(segment)
Q = int(input())
for _ in range(Q):
info = list(map(int, input().split()))
if(len(info) == 2):
fr,to = info[0], info[1]
# ans = INF
if(fr > to):
# print('0sdfdf', to, len(a)-1) ## 3 fr -- 0to
u = RMQ(segment, lazy, fr, len(a) -1, 0, len(a) -1 , 0)
# print('1dfdffd',0, fr)
v = RMQ(segment, lazy, 0, to, 0, len(a) -1 , 0)
ans = min(u, v)
print(ans)
else:
ans = RMQ(segment, lazy, fr, to, 0, len(a) -1 , 0)
print(ans)
# print(ans)
else:
fr, to , value = info[0],info[1], info[2]
if(fr > to):
# print('0 dfdf', to, len(a)-1)
u = inc(segment, lazy, fr, len(a) -1, value, 0, len(a) -1, 0)
# print('1df df' ,0, fr)
v = inc(segment, lazy, 0, to, value, 0, len(a) -1, 0)
else:
inc(segment, lazy, fr, to, value, 0, len(a) -1, 0)
``` | instruction | 0 | 87,061 | 12 | 174,122 |
No | output | 1 | 87,061 | 12 | 174,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0
Submitted Solution:
```
useless = input()
array = input().split(' ')
operation_amount = input()
operations = list()
for i in range(int(operation_amount)):
operations.append(input().split(' '))
def result(operations, array):
transformed_array = list(map(int, array))
results = list()
for operation in operations:
if len(operation) == 2:
results.append(req(operation[0], operation[1], transformed_array))
else:
add(operation[0], operation[1], operation[2], transformed_array)
return results
def add(start, end, value, array):
if start == end:
array[int(start)] += int(value)
elif int(start) < int(end):
for i in range(int(start), int(end) + 1):
array[i] += int(value)
else:
for i in range(int(start), len(array) + int(end) + 1):
array[i % len(array)] += int(value)
def req(start, end, array):
if start == end:
return array[int(start)]
elif int(start) < int(end):
return min(array[int(start):int(end) + 1])
else:
print(array[int(end):-len(array) - int(start)],
len(array), -len(array) + int(start), int(end))
return min(array[i % len(array)] for i in range(int(start), len(array) + int(end) + 1))
for i in result(operations, array):
print(i)
``` | instruction | 0 | 87,062 | 12 | 174,124 |
No | output | 1 | 87,062 | 12 | 174,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0
Submitted Solution:
```
class stree:
def __init__(self, n):
self.n = n
self.tree = [0]*(4*n+1)
self.delta = [0]*(4*n+1)
def prop(self, i):
self.delta[2*i] += self.delta[i]
self.delta[2*i+1] += self.delta[i]
self.delta[i] = 0
def update(self, i):
self.tree[i] = min(self.tree[2*i]+self.delta[2*i], self.tree[2*i+1]+self.delta[2*i+1])
def incr(self, node, i, j, l, r, val):
if i > r or j < l:
return
if i <= l and j >= r:
self.delta[node] += val
#print(val,"added to", node)
return
self.prop(node)
mid = (l+r)//2
self.incr(2*node, i, j, l, mid, val)
self.incr(2*node+1, i, j, mid+1, r, val)
self.update(node)
def query(self, node, i, j, l, r):
if i > r or j < l:
return 10**7
if i <= l and j >= r:
return self.delta[node]+self.tree[node]
self.prop(node)
mid = (l+r)//2
left = self.query(2*node, i, j, l, mid)
right = self.query(2*node+1, i, j, mid+1, r)
return min(left, right)
n = int(input().strip())
arr = list(map(int, input().strip().split()))
i = 0
st = stree(n)
while i<n:
st.incr(1, i, i, 0, n-1, arr[i])
i += 1
q = int(input().strip())
#print(st.tree," and ", st.delta)
while q:
a = list(map(int, input().strip().split()))
if len(a) == 2:
if a[0] <= a[1]:
print(st.query(1, a[0], a[1], 0, n-1))
else:
print(min(st.query(1, a[0], n-1, 0, n-1), st.query(1, 0, a[1], 0, n-1)))
else:
if a[0] <= a[1]:
st.incr(1, a[0], a[1], 0, n-1, a[2])
else:
st.incr(1, a[0], n-1, 0, n-1, a[2])
st.incr(1, 0, a[1], 0, n-1, a[2])
q -= 1
``` | instruction | 0 | 87,063 | 12 | 174,126 |
No | output | 1 | 87,063 | 12 | 174,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
* inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
* rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).
Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.
Write program to process given sequence of operations.
Input
The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.
Output
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Examples
Input
4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1
Output
1
0
0
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**30, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] > k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
n=int(input())
a=list(map(int,input().split()))
q=int(input())
tree=[0]*(3*n)
lazy=[0]*(3*n)
def push(v):
tree[2*v]+=lazy[v]
lazy[2*v]=lazy[v]
tree[2 * v+1] += lazy[v]
lazy[2 * v+1] = lazy[v]
lazy[v]=0
def build(a,st,end,ind):
if st==end:
tree[ind]=a[st]
else:
mid=(st+end)//2
build(a,st,mid,2*ind)
build(a,mid+1,end,2*ind+1)
tree[ind]=min(tree[2*ind],tree[2*ind+1])
def update(a,l,r,st,end,ind,val):
if l>r:
return
if st==end:
tree[ind]+=val
elif l==st and r==end:
tree[ind]+=val
lazy[ind]+=val
else:
push(ind)
mid = (st + end) // 2
update(a,l,min(r,mid),st,mid,2*ind,val)
update(a, max(l, mid+1),r,mid+1,end, 2 * ind+1,val)
tree[ind]=min(tree[2*ind],tree[2*ind+1])
def query(a,l,r,st,end,ind):
if l>r:
return 999999
if st==end:
return tree[ind]
elif l==st and r==end:
return tree[ind]
else:
push(ind)
mid = (st + end) // 2
return min(query(a,l,min(r,mid),st,mid,2*ind),query(a, max(l, mid+1),r,mid+1,end, 2 * ind+1))
build(a,0,n-1,1)
for i in range(q):
l=list(map(int,input().split()))
if len(l)==2:
le,ri=l
if le>ri:
w=query(a,le,n-1,0,n-1,1)
w1=query(a,0,ri,0,n-1,1)
print(min(w,w1))
else:
w=query(a,le,ri,0,n-1,1)
print(w)
else:
le,ri,val=l
if le > ri:
update(a,le,n-1,0,n-1,1,val)
update(a, 0, ri, 0, n - 1,1, val)
else:
update(a, le, ri, 0, n - 1,1, val)
``` | instruction | 0 | 87,064 | 12 | 174,128 |
No | output | 1 | 87,064 | 12 | 174,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 ≤ n ≤ 3 ⋅ 10^5) — the length of the first array.
The second line contains n integers a_1, a_2, ⋅⋅⋅, a_n~(1 ≤ a_i ≤ 10^9) — elements of the array A.
The third line contains a single integer m~(1 ≤ m ≤ 3 ⋅ 10^5) — the length of the second array.
The fourth line contains m integers b_1, b_2, ⋅⋅⋅, b_m~(1 ≤ b_i ≤ 10^9) - elements of the array B.
Output
Print a single integer — the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3 | instruction | 0 | 87,541 | 12 | 175,082 |
Tags: greedy, two pointers
Correct Solution:
```
import sys
f=sys.stdin
out=sys.stdout
n=int(f.readline().rstrip('\r\n'))
a=list(map(int,f.readline().rstrip('\r\n').split()))
m=int(f.readline().rstrip('\r\n'))
b=list(map(int,f.readline().rstrip('\r\n').split()))
i=n-1
j=m-1
c=0
while i>=0 and j>=0:
if a[i]==b[j]:
c+=1
i-=1
j-=1
elif a[i]<b[j]:
a[i-1]+=a[i]
i-=1
else:
b[j-1]+=b[j]
j-=1
if i==-1 and j==-1:
out.write(str(c))
else:
out.write("-1")
``` | output | 1 | 87,541 | 12 | 175,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 ≤ n ≤ 3 ⋅ 10^5) — the length of the first array.
The second line contains n integers a_1, a_2, ⋅⋅⋅, a_n~(1 ≤ a_i ≤ 10^9) — elements of the array A.
The third line contains a single integer m~(1 ≤ m ≤ 3 ⋅ 10^5) — the length of the second array.
The fourth line contains m integers b_1, b_2, ⋅⋅⋅, b_m~(1 ≤ b_i ≤ 10^9) - elements of the array B.
Output
Print a single integer — the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3 | instruction | 0 | 87,542 | 12 | 175,084 |
Tags: greedy, two pointers
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
m=int(input())
b=list(map(int,input().split()))
ans=0
i=0
j=0
while(i!=n and j!=m):
if a[i]==b[j]:
ans+=1
i+=1
j+=1
elif a[i]<b[j]:
if i+1<len(a):
a[i+1]+=a[i]
i+=1
else:
if j+1<len(b):
b[j+1]+=b[j]
j+=1
if i!=n or j!=m:
print(-1)
else:
print(ans)
``` | output | 1 | 87,542 | 12 | 175,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 ≤ n ≤ 3 ⋅ 10^5) — the length of the first array.
The second line contains n integers a_1, a_2, ⋅⋅⋅, a_n~(1 ≤ a_i ≤ 10^9) — elements of the array A.
The third line contains a single integer m~(1 ≤ m ≤ 3 ⋅ 10^5) — the length of the second array.
The fourth line contains m integers b_1, b_2, ⋅⋅⋅, b_m~(1 ≤ b_i ≤ 10^9) - elements of the array B.
Output
Print a single integer — the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3 | instruction | 0 | 87,543 | 12 | 175,086 |
Tags: greedy, two pointers
Correct Solution:
```
#------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
from collections import deque
n=int(input())
l=list(map(int,input().split()))
m=int(input())
l1=list(map(int,input().split()))
t=0
p=0
d=dict()
y=0
k=1
f=0
k1=1
while(True):
if l[t]==l1[p]:
d.update({y:l[t]})
y+=1
t+=k
p+=k1
k=1
k1=1
else:
if l[t]<l1[p]:
if t+k>=len(l):
f=1
break
else:
l[t]+=l[t+k]
k+=1
elif l1[p]<l[t]:
if p+k1>=len(l1):
f=1
break
else:
l1[p]+=l1[p+k1]
k1+=1
if t>=len(l) or p>=len(l1):
break
if f==1 or t!=n or p!=m:
print(-1)
else:
print(len(d))
``` | output | 1 | 87,543 | 12 | 175,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 ≤ n ≤ 3 ⋅ 10^5) — the length of the first array.
The second line contains n integers a_1, a_2, ⋅⋅⋅, a_n~(1 ≤ a_i ≤ 10^9) — elements of the array A.
The third line contains a single integer m~(1 ≤ m ≤ 3 ⋅ 10^5) — the length of the second array.
The fourth line contains m integers b_1, b_2, ⋅⋅⋅, b_m~(1 ≤ b_i ≤ 10^9) - elements of the array B.
Output
Print a single integer — the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3 | instruction | 0 | 87,544 | 12 | 175,088 |
Tags: greedy, two pointers
Correct Solution:
```
from fractions import gcd
import math
n = int(input())
a= [int(i) for i in input().split()]
m = int(input())
b= [int(i) for i in input().split()]
nn = []
mm = []
if sum(a)!=sum(b):
print(-1)
exit()
ptr_a = 0
ptr_b = 0
sum_a = 0
sum_b = 0
while ptr_a<n and ptr_b<m:
if sum_a==0 and sum_b==0 and a[ptr_a]==b[ptr_b]:
mm.append(a[ptr_a])
nn.append(a[ptr_a])
ptr_a+=1
ptr_b+=1
else:
if sum_a==0 and sum_b==0:
sum_a = a[ptr_a]
sum_b = b[ptr_b]
if sum_a < sum_b:
ptr_a+=1
sum_a+=a[ptr_a]
elif sum_a > sum_b:
ptr_b+=1
sum_b+=b[ptr_b]
else:
ptr_a+=1
ptr_b+=1
nn.append(sum_b)
mm.append(sum_b)
sum_a=0
sum_b = 0
# print(nn,mm,ptr_a,ptr_b,sum_a,sum_b)
print(len(nn))
``` | output | 1 | 87,544 | 12 | 175,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 ≤ n ≤ 3 ⋅ 10^5) — the length of the first array.
The second line contains n integers a_1, a_2, ⋅⋅⋅, a_n~(1 ≤ a_i ≤ 10^9) — elements of the array A.
The third line contains a single integer m~(1 ≤ m ≤ 3 ⋅ 10^5) — the length of the second array.
The fourth line contains m integers b_1, b_2, ⋅⋅⋅, b_m~(1 ≤ b_i ≤ 10^9) - elements of the array B.
Output
Print a single integer — the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3 | instruction | 0 | 87,545 | 12 | 175,090 |
Tags: greedy, two pointers
Correct Solution:
```
ka=int(input())
a=list(map(int,input().split()))
kb=int(input())
b=list(map(int,input().split()))
def sum1(a):
k=0
for i in a:
k+=a
return k
def www(a ,b, k1, k2):
ka=k1
kb=k2
sa=a[ka]
sb=b[kb]
while sa!=sb:
if sa>sb:
kb+=1
sb+=b[kb]
else:
ka+=1
sa+=a[ka]
return ka+1,kb+1
k1=0
k2=0
if sum(a)!=sum(b):
print('-1')
else:
ans=0
while k1<len(a):
k1,k2=www(a,b,k1,k2)
ans+=1
print(ans)
``` | output | 1 | 87,545 | 12 | 175,091 |
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