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?
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.
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.
2
5
10
1
2
题目意思：将一个2进制的n中每个位翻转得到一个比n小且尽可能大的数，求输出翻转了几位。
解题思路：这道题该由我背锅，我当时先是翻译错了题意，后来稍微有一点眉目了，我又理解错了那个flip的意思，这里面的翻转并不是那种交换(swap那样的），而是像硬币正面换到反面那样的翻转，也就
是0与1的交换，根据题意可以推出想要得到一个既要比n小还有尽可能大的数，只有是n前面的那一个数n-1。所以就是根据n构造一个二进制的n-1,方法就是找到n的二进制中最后面的那一个1翻转为0，而最
后一个1之后的0全都翻转成1，统计所用的翻转次数即可。
1 #include<cstdio> 2 #include<cstring> 3 int main() 4 { 5 int t,n,j,k,i,count; 6 int a[32]; 7 scanf("%d",&t); 8 while(t--) 9 { 10 scanf("%d",&n); 11 memset(a,-1,sizeof(a)); 12 j=0; 13 count=0; 14 i=n; 15 while(i) 16 { 17 a[j]=i%2; 18 if(a[j]==0) 19 { 20 count++; 21 } 22 if(a[j]==1) 23 { 24 count++; 25 break; 26 } 27 i/=2; 28 j++; 29 } 30 printf("%d\n",count); 31 } 32 return 0; 33 }
原文：https://www.cnblogs.com/wkfvawl/p/9383334.html