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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n. We define fa the following way:
* Initially fa = 0, M = 1;
* for every 2 ≤ i ≤ n if aM < ai then we set fa = fa + aM and then set M = i.
Calculate the sum of fa over all n! permutations of the array a modulo 109 + 7.
Note: two elements are considered different if their indices differ, so for every array a there are exactly n! permutations.
Input
The first line contains integer n (1 ≤ n ≤ 1 000 000) — the size of array a.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print the only integer, the sum of fa over all n! permutations of the array a modulo 109 + 7.
Examples
Input
2
1 3
Output
1
Input
3
1 1 2
Output
4
Note
For the second example all the permutations are:
* p = [1, 2, 3] : fa is equal to 1;
* p = [1, 3, 2] : fa is equal to 1;
* p = [2, 1, 3] : fa is equal to 1;
* p = [2, 3, 1] : fa is equal to 1;
* p = [3, 1, 2] : fa is equal to 0;
* p = [3, 2, 1] : fa is equal to 0.
Where p is the array of the indices of initial array a. The sum of fa is equal to 4.
Submitted Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
n = inp()
a = inpl(); a.sort()
c = Counter(a)
now = a[0]
mx = max(a)
A = 0
nn = 1
for i in range(1,n+1):
nn *= i; nn %= mod
res = 0
for i,x in enumerate(a):
if now != x:
A += c[now]
now = x
if mx == x: break
# print(x,A)
tmp = nn*x*pow(n-A,mod-2,mod)%mod
tmp %= mod
res += tmp
res %= mod
print(res)
``` | instruction | 0 | 38,310 | 12 | 76,620 |
No | output | 1 | 38,310 | 12 | 76,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n. We define fa the following way:
* Initially fa = 0, M = 1;
* for every 2 ≤ i ≤ n if aM < ai then we set fa = fa + aM and then set M = i.
Calculate the sum of fa over all n! permutations of the array a modulo 109 + 7.
Note: two elements are considered different if their indices differ, so for every array a there are exactly n! permutations.
Input
The first line contains integer n (1 ≤ n ≤ 1 000 000) — the size of array a.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).
Output
Print the only integer, the sum of fa over all n! permutations of the array a modulo 109 + 7.
Examples
Input
2
1 3
Output
1
Input
3
1 1 2
Output
4
Note
For the second example all the permutations are:
* p = [1, 2, 3] : fa is equal to 1;
* p = [1, 3, 2] : fa is equal to 1;
* p = [2, 1, 3] : fa is equal to 1;
* p = [2, 3, 1] : fa is equal to 1;
* p = [3, 1, 2] : fa is equal to 0;
* p = [3, 2, 1] : fa is equal to 0.
Where p is the array of the indices of initial array a. The sum of fa is equal to 4.
Submitted Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
n = inp()
a = inpl(); a.sort()
c = Counter()
now = a[0]
mx = max(a)
A = 0
nn = 1
for i in range(1,n+1):
nn *= i; nn %= mod
res = 0
for i,x in enumerate(a):
if now != x:
A += c[a[now]-1]
now = x
if mx == x: break
tmp = nn*x*pow(n-A,mod-2,mod)%mod
tmp %= mod
res += tmp
res %= mod
print(res)
``` | instruction | 0 | 38,311 | 12 | 76,622 |
No | output | 1 | 38,311 | 12 | 76,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0. | instruction | 0 | 38,602 | 12 | 77,204 |
Tags: combinatorics, data structures, math, sortings, two pointers
Correct Solution:
```
from heapq import *
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
j = 0
for i in range(n):
if a[i] > a[j]:
j += 1
print(j)
``` | output | 1 | 38,602 | 12 | 77,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0. | instruction | 0 | 38,603 | 12 | 77,206 |
Tags: combinatorics, data structures, math, sortings, two pointers
Correct Solution:
```
from collections import deque
n = int(input())
a = sorted([int(x) for x in input().split()])
res = 0
dq = deque()
for x in a:
if len(dq) and dq[0] < x:
res += 1
dq.popleft()
dq.append(x)
print(res)
``` | output | 1 | 38,603 | 12 | 77,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0. | instruction | 0 | 38,604 | 12 | 77,208 |
Tags: combinatorics, data structures, math, sortings, two pointers
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
n = int(input())
a = [int(i) for i in input().split()]
start = time.time()
ans = 0
t = list(set(a))
t.sort()
d = { i : 0 for i in t }
for i in range(n):
d[a[i]] += 1
b = d[t[0]]
for i in range(1, len(t)):
if b > d[t[i]]:
ans += d[t[i]]
b -= d[t[i]]
else:
ans += b
b = 0
b += d[t[i]]
print(ans)
finish = time.time()
#print(finish - start)
``` | output | 1 | 38,604 | 12 | 77,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0. | instruction | 0 | 38,605 | 12 | 77,210 |
Tags: combinatorics, data structures, math, sortings, two pointers
Correct Solution:
```
m,n=int(input()),sorted(input().split())
i=0
for j in range(0,m):
if n[j]>n[i]:
i+=1
print(i)
``` | output | 1 | 38,605 | 12 | 77,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0. | instruction | 0 | 38,606 | 12 | 77,212 |
Tags: combinatorics, data structures, math, sortings, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
a.sort()
left = 0
for i in range(1, n):
if a[left] < a[i]:
left += 1
print(left)
``` | output | 1 | 38,606 | 12 | 77,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0. | instruction | 0 | 38,607 | 12 | 77,214 |
Tags: combinatorics, data structures, math, sortings, two pointers
Correct Solution:
```
#
import collections
from functools import cmp_to_key
#key=cmp_to_key(lambda x,y: 1 if x not in y else -1 )
import math
import sys
def getIntList():
return list(map(int, input().split()))
import bisect
try :
import numpy
dprint = print
dprint('debug mode')
except ModuleNotFoundError:
def dprint(*args, **kwargs):
pass
def makePair(z):
return [(z[i], z[i+1]) for i in range(0,len(z),2) ]
def memo(func):
cache={}
def wrap(*args):
if args not in cache:
cache[args]=func(*args)
return cache[args]
return wrap
@memo
def comb (n,k):
if k==0: return 1
if n==k: return 1
return comb(n-1,k-1) + comb(n-1,k)
N, = getIntList()
za = getIntList()
za.sort()
now = 0;
res = 0
for i in range(N):
while now<N and za[i] >=za[now]:
now+=1
if now<N:
now+=1
res+=1
print(res)
``` | output | 1 | 38,607 | 12 | 77,215 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0. | instruction | 0 | 38,608 | 12 | 77,216 |
Tags: combinatorics, data structures, math, sortings, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
a.sort()
i = 0
for j in range(n):
if a[i] < a[j]:
i = i + 1
print(i)
``` | output | 1 | 38,608 | 12 | 77,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0. | instruction | 0 | 38,609 | 12 | 77,218 |
Tags: combinatorics, data structures, math, sortings, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
b = [i for i in a]
a.sort()
b.sort()
c = 0
j = 0
for i in b:
if i > a[j]:
c += 1
j += 1
print (c)
``` | output | 1 | 38,609 | 12 | 77,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
Submitted Solution:
```
def IsThereMatch(k, n, data):
#[0, k) and [n - k, n)
checker = True
for i in range(k):
if(data[i] >= data[n - k + i]):
checker = False
break
return checker
def main():
n = int(input())
data = list(map(int, input().split()))
data.sort()
low, high = map(int, (0, n))
while high - low > 1:
middle = int((low + high) / 2)
if(IsThereMatch(middle, n, data)):
low = middle
else:
high = middle
print(low)
main()
``` | instruction | 0 | 38,610 | 12 | 77,220 |
Yes | output | 1 | 38,610 | 12 | 77,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
Submitted Solution:
```
from bisect import bisect_left
from collections import defaultdict
al=defaultdict(int)
a=int(input())
z=list(map(int,input().split()))
z.sort()
l=0
r=0
count=0
while(l<=r and r<len(z)):
if(z[r]>z[l]):
count+=1
l+=1
r+=1
else:
r+=1
print(count)
``` | instruction | 0 | 38,611 | 12 | 77,222 |
Yes | output | 1 | 38,611 | 12 | 77,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
Submitted Solution:
```
from sys import stdin
import math
from collections import defaultdict
#stdin = open('input.txt','r')
I = stdin.readline
n = int(I())
arr = [int(x) for x in I().split()]
ans,p1,p2 = 0,0,0
arr.sort()
while(p1<n and p2<n):
if(arr[p1] == arr[p2]):
p2+=1
else:
ans+=1
p1+=1
p2+=1
print(ans)
``` | instruction | 0 | 38,612 | 12 | 77,224 |
Yes | output | 1 | 38,612 | 12 | 77,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
a.sort()
i = 0
j = 1
while(i<n and j<n):
if a[i]<a[j]:
i += 1
j += 1
else:
j += 1
print(i)
``` | instruction | 0 | 38,613 | 12 | 77,226 |
Yes | output | 1 | 38,613 | 12 | 77,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
Submitted Solution:
```
x=input().strip('\n')
y=sorted([int(j) for j in input().strip('\n').split(' ')])
z=0
x=min(y)
for i in y:
if i>x:
z=z+1
print(z)
``` | instruction | 0 | 38,614 | 12 | 77,228 |
No | output | 1 | 38,614 | 12 | 77,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
Submitted Solution:
```
n = int(input())
lst = input().split()
length = len(lst)
for i in range(length):
lst[i] = int(lst[i])
sorted_lst = [[i, False] for i in lst]
sorted_lst.sort(key = lambda elem: elem[0])
answer = 0
for i in sorted_lst:
for j in lst:
if i[1] == False and i[0] > j:
answer += 1
i[1] = True
print(answer)
``` | instruction | 0 | 38,615 | 12 | 77,230 |
No | output | 1 | 38,615 | 12 | 77,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split(' ')))
z = {}
for i in range(n):
if a[i] not in z:
z[a[i]] = 1
else:
z[a[i]] +=1
ans = n
for i in range(n):
if z[a[i]]>1:
ans = ans - z[a[i]]
break
print(ans)
``` | instruction | 0 | 38,616 | 12 | 77,232 |
No | output | 1 | 38,616 | 12 | 77,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array [10, 20, 30, 40], we can permute it so that it becomes [20, 40, 10, 30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the array.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.
Output
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Examples
Input
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
0
Note
In the first sample, one of the best permutations is [1, 5, 5, 3, 10, 1, 1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
Submitted Solution:
```
def main():
n = int (input())
numeros = list(map(int,input().split()))
numeros.sort()
counter=0
current=numeros[0]
nNumbers=[]
# print(numeros)
for e in numeros:
if(e==current):
counter+=1
else:
nNumbers.append(counter)
current=e
counter=1
# print(nNumbers)
if(len(nNumbers)==0):
print("0")
else:
total=n - max(nNumbers)
# test = {i:numeros.count(i) for i in list(set(numeros))}
print(total)
main()
# def main():
# n = int(input())
# numeros = list(map(int,input().split()))
# numeros.sort(reverse=True)
# numeros.pop()
# tmp = list(set(numeros))
# if(len(tmp)>1):
# print(len(tmp))
# else:
# print(0)
# def main():
# n = int(input())
# numeros = list(map(int,input().split()))
# numeros.sort(reverse=True)
# # print(numeros)
# total=0
# indicesCambiados=[]
# for i in range(0,len(numeros)):
# for j in range(i,len(numeros)):
# if(numeros[i]>numeros[j] and not j in indicesCambiados):
# total+=1
# indicesCambiados.append(j)
# break
# print(total)
# # numeros.pop()
# # tmp = list(set(numeros))
# # if(len(tmp)>1):
# # print(len(tmp))
# # else:
# # print(0)
# def main():
# n = int(input())
# numeros = list(map(int,input().split()))
# numeros.sort(reverse=True)
# # print(numeros)
# total=0
# for i in range(0,len(numeros)//2):
# if(not numeros[i]== numeros[len(numeros)-i-1]):
# total+=1
# if(not len(numeros)%2==0):
# # print(n//2)
# # print(n//2+1)
# # print(n-1)
# if(numeros[n//2]<numeros[0] and numeros[n//2] >numeros[n-1]):
# total+=1
# print(total)
# main()
``` | instruction | 0 | 38,617 | 12 | 77,234 |
No | output | 1 | 38,617 | 12 | 77,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Drazil likes heap very much. So he created a problem with heap:
There is a max heap with a height h implemented on the array. The details of this heap are the following:
This heap contains exactly 2^h - 1 distinct positive non-zero integers. All integers are distinct. These numbers are stored in the array a indexed from 1 to 2^h-1. For any 1 < i < 2^h, a[i] < a[\left ⌊{i/2}\right ⌋].
Now we want to reduce the height of this heap such that the height becomes g with exactly 2^g-1 numbers in heap. To reduce the height, we should perform the following action 2^h-2^g times:
Choose an index i, which contains an element and call the following function f in index i:
<image>
Note that we suppose that if a[i]=0, then index i don't contain an element.
After all operations, the remaining 2^g-1 element must be located in indices from 1 to 2^g-1. Now Drazil wonders what's the minimum possible sum of the remaining 2^g-1 elements. Please find this sum and find a sequence of the function calls to achieve this value.
Input
The first line of the input contains an integer t (1 ≤ t ≤ 70 000): the number of test cases.
Each test case contain two lines. The first line contains two integers h and g (1 ≤ g < h ≤ 20). The second line contains n = 2^h-1 distinct positive integers a[1], a[2], …, a[n] (1 ≤ a[i] < 2^{20}). For all i from 2 to 2^h - 1, a[i] < a[\left ⌊{i/2}\right ⌋].
The total sum of n is less than 2^{20}.
Output
For each test case, print two lines.
The first line should contain one integer denoting the minimum sum after reducing the height of heap to g. The second line should contain 2^h - 2^g integers v_1, v_2, …, v_{2^h-2^g}. In i-th operation f(v_i) should be called.
Example
Input
2
3 2
7 6 3 5 4 2 1
3 2
7 6 5 4 3 2 1
Output
10
3 2 3 1
8
2 1 3 1 | instruction | 0 | 38,788 | 12 | 77,576 |
Tags: constructive algorithms, data structures, greedy, implementation
Correct Solution:
```
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
T = int(input())
for _ in range(T):
H, G = map(int, input().split())
A = [0] + list(map(int, input().split()))
N = len(A)
target_N = 2**G - 1
target_ans_len = 2**H - 2**G
Ans = []
Roots = [True] * (N+1)
idx_Roots = 1
while True:
idx = idx_Roots
st = []
while True:
idx_l = idx<<1
idx_r = idx_l+1
st.append((idx, A[idx]))
if idx_l >= N or A[idx_l] == A[idx_r] == 0:
A[idx] = 0
break
elif A[idx_l] > A[idx_r]:
A[idx] = A[idx_l]
idx = idx_l
else:
A[idx] = A[idx_r]
idx = idx_r
if st[-1][0] <= target_N:
for idx, a in st:
A[idx] = a
Roots[idx] = False
while not Roots[idx_Roots]:
idx_Roots += 1
else:
Ans.append(idx_Roots)
if len(Ans) == target_ans_len:
break
print(sum(A))
print(" ".join(map(str, Ans)))
``` | output | 1 | 38,788 | 12 | 77,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above. | instruction | 0 | 38,821 | 12 | 77,642 |
Tags: greedy, math
Correct Solution:
```
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys
input = sys.stdin.readline
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1;
return x;
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k;
res = 1;
for i in range(k):
res = res * (n - i);
res = res / (i + 1);
return int(res);
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
if hi == None:
hi = len(a);
while lo < hi:
mid = (lo+hi)//2;
if a[mid] < x:
lo = mid+1;
else:
hi = mid;
return lo;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret;
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
def solve():
n = int(input())
a = list(map(int, input().split()))
c = a.count(a[0])
if c == n:
print(n)
else:
print(1)
if __name__ == '__main__':
for _ in range(int(input())):
solve()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
``` | output | 1 | 38,821 | 12 | 77,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above. | instruction | 0 | 38,822 | 12 | 77,644 |
Tags: greedy, math
Correct Solution:
```
num_cases = int(input())
for case_num in range(num_cases):
n = int(input())
a = map(int, input().split())
if len(set(a)) == 1:
print(n)
else:
print(1)
``` | output | 1 | 38,822 | 12 | 77,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above. | instruction | 0 | 38,823 | 12 | 77,646 |
Tags: greedy, math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
if max(a)==min(a):
print(n)
else:
print(1)
``` | output | 1 | 38,823 | 12 | 77,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above. | instruction | 0 | 38,824 | 12 | 77,648 |
Tags: greedy, math
Correct Solution:
```
t=int(input())
a=[[]]*t
for i in range(t):
n=int(input())
a[i]=list(map(int,input().strip().split()))[:n]
for i in range(t):
a[i]=sorted(a[i])
if a[i][0]==a[i][-1]:
print(len(a[i]))
else:
print(1)
``` | output | 1 | 38,824 | 12 | 77,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above. | instruction | 0 | 38,825 | 12 | 77,650 |
Tags: greedy, math
Correct Solution:
```
import math
t = int(input())
for q in range(t):
n = int(input())
L = [int(i) for i in input().split()]
first = 0
second = 0
for i in L:
if first == 0:
first = i
elif i != first:
second = i
break
if second == 0:
print(n)
else:
print(1)
``` | output | 1 | 38,825 | 12 | 77,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above. | instruction | 0 | 38,826 | 12 | 77,652 |
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
mini=min(a)
maxi=max(a)
if(mini==maxi):
print(n)
else:
print(1)
``` | output | 1 | 38,826 | 12 | 77,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above. | instruction | 0 | 38,827 | 12 | 77,654 |
Tags: greedy, math
Correct Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from bisect import bisect_left;
from bisect import bisect_right;
from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1;
return x;
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k;
res = 1;
for i in range(k):
res = res * (n - i);
res = res / (i + 1);
return int(res);
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret;
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
for _ in range(int(input())):
n = int(input()); a = int_array();
x = a.count(a[0]);
if x == n:
print(n);
else:
print(1);
``` | output | 1 | 38,827 | 12 | 77,655 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above. | instruction | 0 | 38,828 | 12 | 77,656 |
Tags: greedy, math
Correct Solution:
```
inp = lambda cast=int: [cast(x) for x in input().split()]
printf = lambda s='', *args, **kwargs: print(str(s).format(*args), flush=True, **kwargs)
t, = inp()
for _ in range(t):
n, = inp()
print(n if len(set(inp())) == 1 else 1)
``` | output | 1 | 38,828 | 12 | 77,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
if a.count(a[0]) == n:
print(n)
else:
print(1)
``` | instruction | 0 | 38,829 | 12 | 77,658 |
Yes | output | 1 | 38,829 | 12 | 77,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above.
Submitted Solution:
```
from sys import stdin
input = stdin.buffer.readline
for _ in range(int(input())):
n = int(input())
*a, = map(int, input().split())
if a == [a[0]] * n:
print(n)
else:
print(1)
``` | instruction | 0 | 38,830 | 12 | 77,660 |
Yes | output | 1 | 38,830 | 12 | 77,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above.
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
lis=list(map(int,input().split()))
lis.sort()
found=0
if n==1:
found=1
else:
for k in range(len(lis)-1):
if lis[k]!=lis[k+1]:
found=1
break
if found==0:
print(n)
else:
print(1)
``` | instruction | 0 | 38,831 | 12 | 77,662 |
Yes | output | 1 | 38,831 | 12 | 77,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above.
Submitted Solution:
```
def gcd(a, b):
if b == 0:
return a
return gcd(b, a%b)
# least common multiple
def lcm(a, b):
return (a*b) / gcd(a,b)
def read(func = int):
return map(func, input().split())
t = int(input())
while t:
n = int(input())
nums = list(read())
if (len(set(nums))) == 1:
print(len(nums))
else:
print(1)
t -= 1
``` | instruction | 0 | 38,832 | 12 | 77,664 |
Yes | output | 1 | 38,832 | 12 | 77,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above.
Submitted Solution:
```
# Author : devil9614 - Sujan Mukherjee
from __future__ import division, print_function
import os,sys
import math
import collections
from itertools import permutations
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
class my_dictionary(dict):
def __init__(self):
self = dict()
def add(self,key,value):
self[key] = value
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(100000000)
threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *,threading
#from itertools import permutations
#Copy 2D list m = [x[:] for x in mark] .. Avoid Using Deepcopy
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[1]
def sort2(l):return sorted(l, key=getKey,reverse=True)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(n-r))
def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return mid
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def ceil(x,y):
if x%y==0:
return x//y
else:
return x//y+1
def powerMod(x,y,p):
res = 1
x %= p
while y > 0:
if y&1:
res = (res*x)%p
y = y>>1
x = (x*x)%p
return res
def gcd(x, y):
while y:
x, y = y, x % y
return x
def isPrime(n) : # Check Prime Number or not
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def padded_bin_with_complement(x):
if x < 0:
return bin((2**16) - abs(x))[2:].zfill(16)
else:
return bin(x)[2:].zfill(16)
def binaryToDecimal(binary):
binary1 = binary
decimal, i, n = 0, 0, 0
while(binary != 0):
dec = binary % 10
decimal = decimal + dec * pow(2, i)
binary = binary//10
i += 1
print(decimal)
def CountFrequency(my_list):
freq = {}
for item in my_list:
if (item in freq):
freq[item] += 1
else:
freq[item] = 1
return freq
def pos(a):
b = [0]*len(a)
c = sorted(a)
for i in range(len(a)):
for j in range(len(a)):
if c[j] == a[i]:
b[i] = j
break
return b
def smallestDivisor(n):
# if divisible by 2
if (n % 2 == 0):
return 2
# iterate from 3 to sqrt(n)
i = 3
while(i * i <= n):
if (n % i == 0):
return i
i += 2
return n
def commonn(a,b,n):
c = []
for i in range(n):
if a[i] == b[i]:
c.append("-1")
else:
c.append(b[i])
return c
def primeFactors(n):
j = []
while n % 2 == 0:
j.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
j.append(int(n))
n = n / i
if n > 2:
j.append(int(n))
return j
def sumdigit(n):
n = str(n)
k = 0
for i in range(len(n)):
k+=int(n[i])
return k
def question(a,n):
j = a
for i in range(n):
if a[i] == "1":
j[i] = "0"
else:
j[i] = "1"
j[0,n-1].reverse()
return j
def main():
for _ in range(ii()):
n = ii()
a = li()
b = []
for i in range(1,n):
if a[0] != a[i]:
f = 1
else:
f = 0
if f:
print(1)
else:
print(n)
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
``` | instruction | 0 | 38,833 | 12 | 77,666 |
No | output | 1 | 38,833 | 12 | 77,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Aug 15 14:57:41 2020
@author: Manan Tyagi
"""
import sys
input=sys.stdin.buffer.readline
t=int(input())
while t:
t-=1
n=int(input())
a=list(map(int,input().split()))
c=0
i=0
while i <n-1:
if a[i]!=a[i+1]:
if a[i]+a[i+1] not in a:
c+=2
i+=2
else:
i+=1
else:
i+=1
print(n-c)
``` | instruction | 0 | 38,834 | 12 | 77,668 |
No | output | 1 | 38,834 | 12 | 77,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above.
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
arr = [int(x) for x in input().split()]
if n == 1:
print(n)
continue
if n == 2:
if arr[0] == arr[1]:
print(n)
continue
else:
print(1)
continue
tset = set(arr)
if tset == 1:
print(n)
print(1)
``` | instruction | 0 | 38,835 | 12 | 77,670 |
No | output | 1 | 38,835 | 12 | 77,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lord Omkar has permitted you to enter the Holy Church of Omkar! To test your worthiness, Omkar gives you a password which you must interpret!
A password is an array a of n positive integers. You apply the following operation to the array: pick any two adjacent numbers that are not equal to each other and replace them with their sum. Formally, choose an index i such that 1 ≤ i < n and a_{i} ≠ a_{i+1}, delete both a_i and a_{i+1} from the array and put a_{i}+a_{i+1} in their place.
For example, for array [7, 4, 3, 7] you can choose i = 2 and the array will become [7, 4+3, 7] = [7, 7, 7]. Note that in this array you can't apply this operation anymore.
Notice that one operation will decrease the size of the password by 1. What is the shortest possible length of the password after some number (possibly 0) of operations?
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the password.
The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 10^9) — the initial contents of your password.
The sum of n over all test cases will not exceed 2 ⋅ 10^5.
Output
For each password, print one integer: the shortest possible length of the password after some number of operations.
Example
Input
2
4
2 1 3 1
2
420 420
Output
1
2
Note
In the first test case, you can do the following to achieve a length of 1:
Pick i=2 to get [2, 4, 1]
Pick i=1 to get [6, 1]
Pick i=1 to get [7]
In the second test case, you can't perform any operations because there is no valid i that satisfies the requirements mentioned above.
Submitted Solution:
```
# A. Omkar and Password
t = int(input())
for _ in range(t):
n = int(input())
l = list(map(int, input().split()))
for i in range(int(n/2), -1,-1):
try:
if l[i]!=l[i+1]:
l[i]=l[i]+l[i+1]
l.pop(i+1)
print(l)
except:
pass
print(len(l))
``` | instruction | 0 | 38,836 | 12 | 77,672 |
No | output | 1 | 38,836 | 12 | 77,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions. | instruction | 0 | 38,837 | 12 | 77,674 |
Tags: bitmasks, data structures, divide and conquer, dp, greedy, math, sortings, strings, trees
Correct Solution:
```
import sys,collections as cc
input = sys.stdin.buffer.readline
I = lambda : list(map(int,input().split()))
n,=I()
l=I()
an=ban=0
ar=cc.defaultdict(list)
ar[0]=l
xo=1<<30
for i in range(30,-1,-1):
a=b=0
pr=cc.defaultdict(list)
for j in ar:
ff=j
j=ar[j]
aa=bb=0
oo=zz=0
for x in range(len(j)):
if j[x]&xo:
aa+=zz
oo+=1
pr[2*ff+0].append(j[x])
else:
bb+=oo
zz+=1
pr[2*ff+1].append(j[x])
a+=aa
b+=bb
ar=pr
if a<b:
an+=xo
ban+=a
else:
ban+=b
xo//=2
print(ban,an)
``` | output | 1 | 38,837 | 12 | 77,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions. | instruction | 0 | 38,838 | 12 | 77,676 |
Tags: bitmasks, data structures, divide and conquer, dp, greedy, math, sortings, strings, trees
Correct Solution:
```
n = int(input())
A = list(map(int, input().split()))
que = [A, 1]
ans = 0
ans1 = 0
p = 29
for p in range(29, -1, -1):
X1 = 0
Y1 = 0
r = 2 ** p
q1 = que.pop()
q2 = 0
que1 = []
for i in range(q1):
x = 0
y = 0
X = 0
Y = 0
A1 = []
A2 = []
a = que.pop()
for j in a:
if j < r:
X += x
y += 1
A1 += [j]
else:
x += 1
Y += y
A2 += [j - r]
if A1:
q2 += 1
que1 += [A1]
if A2:
q2 += 1
que1 += [A2]
X1 += X
Y1 += Y
if X1 > Y1:
ans += r
ans1 += min(X1, Y1)
que += que1
que += [q2]
print(ans1, ans)
``` | output | 1 | 38,838 | 12 | 77,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions. | instruction | 0 | 38,839 | 12 | 77,678 |
Tags: bitmasks, data structures, divide and conquer, dp, greedy, math, sortings, strings, trees
Correct Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n=int(input())
a=list(map(int,input().split()))
current=0
fail=0
for i in range(31,-1,-1):
yescount=0
nocount=0
tmp=2**i
dic={}
for j in range(n):
r=a[j]//tmp
if r%2==0 and r+1 in dic:
yescount+=dic[r+1]
elif r%2==1 and r-1 in dic:
nocount+=dic[r-1]
if r in dic:
dic[r]+=1
else:
dic[r]=1
if yescount>nocount:
fail+=nocount
current+=tmp
else:
fail+=yescount
print(fail, current)
``` | output | 1 | 38,839 | 12 | 77,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions. | instruction | 0 | 38,840 | 12 | 77,680 |
Tags: bitmasks, data structures, divide and conquer, dp, greedy, math, sortings, strings, trees
Correct Solution:
```
import sys,collections as cc
input = sys.stdin.buffer.readline
I = lambda : list(map(int,input().split()))
n,=I()
l=I()
an=ban=0
ar=cc.defaultdict(list)
ar[0]=l
xo=1<<30
for i in range(30,-1,-1):
a=b=0
pr=cc.defaultdict(list)
for j in ar:
ff=j
j=ar[j]
aa=bb=0
oo=zz=0
for x in range(len(j)):
if j[x]&xo:
aa+=zz
oo+=1
pr[2*ff+0].append(j[x])
else:
bb+=oo
zz+=1
pr[2*ff+1].append(j[x])
a+=aa
b+=bb
ar=pr
if a<b:
an+=xo
ban+=a
else:
ban+=b
xo//=2
"""
for i in l:
k=format(i,'b')
k='0'*(4-len(k))+k
x=format(i^an,'b')
x='0'*(4-len(x))+x
print(k,x,i^an)
"""
print(ban,an)
``` | output | 1 | 38,840 | 12 | 77,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions. | instruction | 0 | 38,841 | 12 | 77,682 |
Tags: bitmasks, data structures, divide and conquer, dp, greedy, math, sortings, strings, trees
Correct Solution:
```
from bisect import *
from collections import *
from math import *
from heapq import *
from typing import List
from itertools import *
from operator import *
from functools import *
import sys
'''
@lru_cache(None)
def fact(x):
if x<2:
return 1
return fact(x-1)*x
@lru_cache(None)
def per(i,j):
return fact(i)//fact(i-j)
@lru_cache(None)
def com(i,j):
return per(i,j)//fact(j)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def power2(n):
while not n&1:
n>>=1
return n==1
'''
'''
for i in range(t):
#n,m=map(int,input().split())
n,m=1,0
graph=defaultdict(set)
for i in range(m):
u,v=map(int,input().split())
graph[u-1].add(v-1)
visited=[0]*n
ans=[]
def delchild(u):
for child in graph[u]:
visited[child]=1
ans.append(child+1)
for i in range(n):
if not visited[i]:
children=graph[i]
if len(children)==1:
u=children.pop()
visited[u]=1
delchild(u)
elif len(children)==2:
u=children.pop()
v=children.pop()
if u in graph[v]:
delchild(v)
visited[v]=1
elif v in graph[u]:
delchild(u)
visited[u]=1
else:
delchild(u)
delchild(v)
visited[u]=visited[v]=1
print(len(ans))
sys.stdout.flush()
print(' '.join(map(str,ans)))
sys.stdout.flush()
import time
s=time.time()
e=time.time()
print(e-s)
'''
n=int(input())
ans=x=0
arr=[list(map(int,input().split()))]
for k in range(29,-1,-1):
a,b=0,0
mask=1<<k
tmp=[]
for ar in arr:
one=zero=0
tmp1=[]
tmp2=[]
for ai in ar:
cur=ai&mask
if cur==0:
b+=one
zero+=1
tmp1.append(ai)
else:
a+=zero
one+=1
tmp2.append(ai)
if len(tmp1)>1:
tmp.append(tmp1)
if len(tmp2)>1:
tmp.append(tmp2)
if b>a:
x|=mask
ans+=a
else:
ans+=b
arr=tmp
print(ans,x)
``` | output | 1 | 38,841 | 12 | 77,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions. | instruction | 0 | 38,842 | 12 | 77,684 |
Tags: bitmasks, data structures, divide and conquer, dp, greedy, math, sortings, strings, trees
Correct Solution:
```
# Not my code
# https://codeforces.com/contest/1416/submission/94027802
import sys,io,os
from collections import defaultdict
inp = [int(x) for x in sys.stdin.buffer.read().split()]
n = inp.pop(0)
A = inp
t = 0.0
x = 0
for v in range(1, 31)[::-1]:
u = d = 0.0
r = defaultdict(int)
w = 1 << v - 1
for a in A:
p = a >> v
b = a & w
if b:
r[p] += 1
elif p in r:
d += 2 * r[p]
r[~p] += 1
for p in r:
if p >= 0 and ~p in r:
rp = r[p] + 0.0
cp = r[~p] + 0.0
u += cp * (cp - 1) - rp * (rp - 1) - (cp - rp) * (cp - rp - 1)
if d > u - d:
x += w
d = u - d
t += d
print(int(t) // 2, x)
``` | output | 1 | 38,842 | 12 | 77,685 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions. | instruction | 0 | 38,843 | 12 | 77,686 |
Tags: bitmasks, data structures, divide and conquer, dp, greedy, math, sortings, strings, trees
Correct Solution:
```
# region fastio # from https://codeforces.com/contest/1333/submission/75948789
import sys, io, os
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.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(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#endregion
N = int(input())
A = list(map(int, input().split()))
ans_count = 0
ans_x = 0
C = [A[:]]
for b in range(31, 0, -1):
C_new = []
D = {}
for a in A:
if a>>b not in D:
D[a>>b] = [a>>b-1&1]
else:
D[a>>b].append(a>>b-1&1)
count_inv_0 = count_inv_1 = 0
for v in D.values():
if len(v) <= 1:
continue
cnt_1 = 0
cnt_0 = 0
cnt_inv_0 = 0
cnt_inv_1 = 0
for zero_one in v:
if zero_one:
cnt_1 += 1
cnt_inv_1 += cnt_0
else:
cnt_0 += 1
cnt_inv_0 += cnt_1
count_inv_0 += cnt_inv_0
count_inv_1 += cnt_inv_1
if count_inv_0 <= count_inv_1:
ans_count += count_inv_0
else:
ans_count += count_inv_1
ans_x |= 1<<b-1
print(ans_count, ans_x)
``` | output | 1 | 38,843 | 12 | 77,687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions. | instruction | 0 | 38,844 | 12 | 77,688 |
Tags: bitmasks, data structures, divide and conquer, dp, greedy, math, sortings, strings, trees
Correct Solution:
```
n=int(input())
l=input().split()
li=[int(i) for i in l]
xori=0
ans=0
mul=1
for i in range(32):
hashi1=dict()
hashi0=dict()
inv1=0
inv2=0
for j in li:
if(j//2 in hashi1 and j%2==0):
inv1+=hashi1[j//2]
if(j//2 in hashi0 and j%2==1):
inv2+=hashi0[j//2]
if(j%2):
if j//2 not in hashi1:
hashi1[j//2]=1
else:
hashi1[j//2]+=1
else:
if j//2 not in hashi0:
hashi0[j//2]=1
else:
hashi0[j//2]+=1
if(inv1<=inv2):
ans+=inv1
else:
ans+=inv2
xori=xori+mul
mul*=2
for j in range(n):
li[j]=li[j]//2
print(ans,xori)
``` | output | 1 | 38,844 | 12 | 77,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
n = int(input())
ll = list(map(int,input().split()))
wynik = 0
pot = 1
inwersje = 0
m = max(ll)
while pot * 2 <= m:
pot *= 2
listy = [ll]
while pot >= 1:
inww = 0
chujnww = 0
nowe = []
for l in listy:
jedna = []
druga = []
inw = 0
jedynki = 0
zera = 0
chujnw = 0
for i in range(len(l)):
if l[i]^pot < l[i]:
jedna.append(l[i])
chujnw += zera
jedynki += 1
else:
inw += jedynki
zera += 1
druga.append(l[i])
if jedna:
nowe.append(jedna)
if druga:
nowe.append(druga)
inww += inw
chujnww += chujnw
if inww > chujnww:
wynik += pot
inwersje += min(inww, chujnww)
listy = nowe
pot //= 2
print(inwersje, wynik)
``` | instruction | 0 | 38,845 | 12 | 77,690 |
Yes | output | 1 | 38,845 | 12 | 77,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
Submitted Solution:
```
import os
from sys import stdin, stdout
class Input:
def __init__(self):
self.lines = stdin.readlines()
self.idx = 0
def line(self):
try:
return self.lines[self.idx].strip()
finally:
self.idx += 1
def array(self, sep = ' ', cast = int):
return list(map(cast, self.line().split(sep = sep)))
def known_tests(self):
num_of_cases, = self.array()
for case in range(num_of_cases):
yield self
def unknown_tests(self):
while self.idx < len(self.lines):
yield self
def problem_solver():
oo = float('Inf')
mem = None
def backtrack(a, L, R, B = 31):
if B < 0 or L == R:
return
mask = 1 << B
l = L
r = L
inv0 = 0
inv1 = 0
zero = 0
one = 0
la = []
ra = []
while l < R:
while r < R and (a[l] & mask) == (a[r] & mask):
if (a[r] & mask):
la.append(a[r])
else:
ra.append(a[r])
r += 1
w = r - l
if (a[l] & mask) == 0:
zero += w
inv0 += w * one
else:
one += w
inv1 += w * zero
l = r
backtrack(la, 0, len(la), B - 1)
backtrack(ra, 0, len(ra), B - 1)
mem[B][0] += inv0
mem[B][1] += inv1
'''
'''
def solver(inpt):
nonlocal mem
n, = inpt.array()
a = inpt.array()
mem = [[0, 0] for i in range(32)]
backtrack(a, 0, n)
inv = 0
x = 0
for b in range(32):
if mem[b][0] <= mem[b][1]:
inv += mem[b][0]
else:
inv += mem[b][1]
x |= 1 << b
print(inv, x)
'''Returns solver'''
return solver
try:
solver = problem_solver()
for tc in Input().unknown_tests():
solver(tc)
except Exception as e:
import traceback
traceback.print_exc(file=stdout)
``` | instruction | 0 | 38,846 | 12 | 77,692 |
Yes | output | 1 | 38,846 | 12 | 77,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
cur = [a]
s = sum(a)
ans = 0
cnt = 0
i = 1 << 31
while i:
i >>= 1
fcnt = 0
bcnt = 0
nex = []
for arr in cur:
aa, bb = [], []
zcnt = 0
ocnt = 0
for ai in arr:
if ai & i:
ocnt += 1
fcnt += zcnt
aa.append(ai)
else:
zcnt += 1
bcnt += ocnt
bb.append(ai)
if aa:
nex.append(aa)
if bb:
nex.append(bb)
if bcnt > fcnt:
ans |= i
cnt += fcnt
else:
cnt += bcnt
cur = nex
print(cnt, ans)
``` | instruction | 0 | 38,847 | 12 | 77,694 |
Yes | output | 1 | 38,847 | 12 | 77,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
a=list(map(int,input().split()))
groups=[a]
tot=0
ans=0
for pos in range(32)[::-1]:
bit=1<<pos
invs=0
flipped_invs=0
nxt_groups=[]
for group in groups:
bin=[]
big=[]
small=[]
for x in group:
if x&bit:
bin.append(1)
big.append(x)
else:
bin.append(0)
small.append(x)
if big:
nxt_groups.append(big)
if small:
nxt_groups.append(small)
zeros=0
ones=0
for x in bin[::-1]:
if x==0:
zeros+=1
flipped_invs+=ones
else:
ones+=1
invs+=zeros
groups=nxt_groups
if flipped_invs<invs:
ans|=bit
tot+=min(flipped_invs,invs)
print(tot,ans)
``` | instruction | 0 | 38,848 | 12 | 77,696 |
Yes | output | 1 | 38,848 | 12 | 77,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def countInversions(arr):
arr = list(map(lambda xx:[xx],arr))
ans = 0
while len(arr) != 1:
arr1 = []
for i in range(0,len(arr),2):
if i == len(arr)-1:
arr1.append(arr[-1])
continue
arr1.append([])
a,b = 0,0
for _ in range(len(arr[i])+len(arr[i+1])):
if a == len(arr[i]):
arr1[-1].append(arr[i+1][b])
b += 1
elif b == len(arr[i+1]):
arr1[-1].append(arr[i][a])
a += 1
elif arr[i][a] <= arr[i+1][b]:
arr1[-1].append(arr[i][a])
a += 1
else:
arr1[-1].append(arr[i+1][b])
b += 1
ans += len(arr[i])-a
arr = arr1
return ans
num = 0
def rec(a,val):
global num
old,new = [0,0],[0,0]
# inversions ; bigger nums till now
a1,a2 = [],[]
for i in a:
if i&val:
old[1] += 1
new[0] += new[1]
a1.append(i)
else:
new[1] += 1
old[0] += old[1]
a2.append(i)
if val != 1:
if len(a1):
x1,x2 = rec(a1,val>>1)
else:
x1,x2 = [0,0],[0,0]
if len(a2):
y1,y2 = rec(a2,val>>1)
else:
y1,y2 = [0,0],[0,0]
if x1[0]+y1[0] > x2[0]+y2[0]:
num |= val>>1
return old,new
def main():
global num
n = int(input())
a = list(map(int,input().split()))
r = max(a).bit_length()-1
if r < 0:
print(0,0)
exit()
val = 1<<r
old,new = rec(a,val)
if old[0] > new[0]:
num |= val
a = list(map(lambda xx:xx^num,a))
print(countInversions(a),num)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 38,849 | 12 | 77,698 |
No | output | 1 | 38,849 | 12 | 77,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
Submitted Solution:
```
import sys
z=sys.stdin.readline
n=int(z())
a=[*map(int,z().split())]
v=2**4
w=2*v
t=x=0
while v:
u=d=0
s,r,c={},{},{}
for i in a:
p=i//w
b=i%w>=v
if b:r[p]=r.get(p,0)+1
else:s[p]=s.get(p,0)+r.get(p,0)
c[p]=c.get(p,0)+1
for p in c:
sp,rp,cp=s.get(p,0),r.get(p,0),c.get(p,0)
d+=sp
u+=(cp*(cp-1))//2-(rp*(rp-1))//2-((cp-rp)*(cp-rp-1))//2-sp
if d>u:
x+=v
u,d=d,u
t+=d
w=v
v//=2
print(t,x)
``` | instruction | 0 | 38,850 | 12 | 77,700 |
No | output | 1 | 38,850 | 12 | 77,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).
An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.
You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.
Input
First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.
Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
Output
Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.
Examples
Input
4
0 1 3 2
Output
1 0
Input
9
10 7 9 10 7 5 5 3 5
Output
4 14
Input
3
8 10 3
Output
0 8
Note
In the first sample it is optimal to leave the array as it is by choosing x = 0.
In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:
* i = 2, j = 3;
* i = 2, j = 4;
* i = 3, j = 4;
* i = 8, j = 9.
In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
Submitted Solution:
```
# Not my code
# https://codeforces.com/contest/1416/submission/94027802
import sys,io,os
from collections import defaultdict
inp = [int(x) for x in sys.stdin.buffer.read().split()]
n = inp.pop(0)
A = inp
t = 0.0
x = 0
for v in range(1, 31)[::-1]:
u = d = 0.0
r = defaultdict(int)
w = 1 << v - 1
for a in A:
p = a >> v
b = a & w
if b:
r[p] += 1
elif p in r:
d += 2 * r[p]
r[~p] += 1
for p in r:
if p & 1 and ~p in r:
rp = r[p] + 0.0
cp = r[~p] + 0.0
u += cp * (cp - 1) - rp * (rp - 1) - (cp - rp) * (cp - rp - 1)
if d > u - d:
x += w
d = u - d
t += d
print(int(t) // 2, x)
``` | instruction | 0 | 38,851 | 12 | 77,702 |
No | output | 1 | 38,851 | 12 | 77,703 |
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