hdu-

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Bitwise Equations

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 645 Accepted Submission(s): 343

Problem Description

You are given two positive integers X and K. Return the K-th smallest positive integer Y, for which the following equation holds: X + Y =X | Y
Where '|' denotes the bitwise OR operator.

Input

The first line of the input contains an integer T (T <= 100) which means the number of test cases.
For each case, there are two integers X and K (1 <= X, K <= 2000000000) in one line.

Output

For each case, output one line containing the number Y.

Sample Input

35 15 52000000000 2000000000

Sample Output

21816383165351936
解体思路：这道题目可以这样想，就是求出x的二进制，k的二进制，x的二进制序列中只有等于零处可以填充k的二进制数，要将k的二进制数填完。填好后，再转换成十进制后就是答案。
代码如下：
#include<stdio.h>#include<cmath>int a[40];int b[40];int main(){int t;int x,k,num,d,l;long long m;scanf("%d",&t);while(t--){d=0;l=0;num=0;m=0;scanf("%d%d",&x,&k);    while(x){    if(x%2==0){    a[++d]=l;//d表示一共有多少的零     }    x=x/2;    l++;//从0位开始计算     }    while(k){    b[++num]=k%2;    k=k/2;    }    for(int i=1;i<=d&&i<=num;i++)    m+=b[i]*pow(2,a[i]);    if(num>d){    for(int i=d+1;i<=num;i++){    m+=b[i]*pow(2,l);    l++;    }    }       printf("%lld\n",m);}return 0;}

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