C - Flip the Bits （二进制的翻转）

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 n such 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
2
5
10
Output
1
2
题意：题目意思是在数n的二进制上进行翻转，使得翻转后的数转换成十进制之后尽可能的大，但是比n小；他说的翻转次数最少应该没大有用，有用的是找到的是数是小于n的数中的最大的；
那就直接找n-1就好了，这个题只是问的翻转了几次；只要找到n与n-1的二进制有几个不一样的就是答案；通过写规律发现了十进制转换成的二进制的特点；比如10二进制是1010，9就是1001,10转换成9只是把10的二进制最后一个1换成0，把最后一个1之后的0都换成1就是9；

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int a[1000],b[1000];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
       int  p=0;int q=0;
        int m=n;
        while(m)
        {
            a[++p]=m%2;
            m=m/2;
        }
        int w=n-1;
        while(w)
        {
            b[++q]=w%2;
            w=w/2;
        }
        if(p-q==1)b[++q]=0;
        int sum=0;
        for(int i=p; i>=1; i--)
        {
            if(a[i]!=b[i])
            {
                sum++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}