Flip the Bits

最新推荐文章于 2021-04-08 17:18:26 发布

原创最新推荐文章于 2021-04-08 17:18:26 发布 · 472 阅读

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本文介绍了一个算法问题，即如何通过翻转二进制表示中的位来构造一个尽可能大的小于给定整数n的新数m。文章提供了一种有效的方法来计算达到这一目的所需的最小翻转次数，并分享了一个实用的C++代码实现。

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You are given a positive integer n. Your task is to build a number m by flipping the minimum number of bits in the binary representation of nsuch that m is less than n (m < n) and it is as maximal as possible. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 105) specifying the number of test cases.

Each test case consists of a single line containing one integer n (1 ≤ n ≤ 109), as described in the statement above.

Output

For each test case, print a single line containing the minimum number of bits you need to flip in the binary representation of n to build the number m.

Example

input

Copy

2
5
10

output

Copy

1
2

题解：就是计算n异或n-1的1的个数，看看变化几次，然后学到一个新函数__builtin_popcount(n^(n-1))来计算二进制中1的个数。

代码如下：

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define maxn 1007
#define N 100005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.000000001
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define Debug(x) cout<<x<<" "<<endl
using namespace std;
typedef long long ll;

int n;
int main()
{
    int t;
    cin>>t;
    while (t--)
    {
        cin>>n;
        cout<<__builtin_popcount(n^(n-1))<<endl;//取异或
    }
    return 0;
}