Bitwise Equations

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

3

5 1

5 5

2000000000 2000000000

 

Sample Output

2

18

16383165351936

 

 

题目大意：X + Y =X | Y。给出X和一个数K，问能使该式成立的第k小的Y是多少。

题目分析：要想使上式成立，X和Y必须没有交集。所以，此问题其实是求X的补集中第K小的子集是多少。

 

代码如下：

# include<iostream>
# include<cstdio>
# include<cmath>
# include<string>
# include<vector>
# include<list>
# include<set>
# include<map>
# include<queue>
# include<cstring>
# include<algorithm>
using namespace std;

# define LL long long
# define REP(i,s,n) for(int i=s;i<n;++i)
# define CL(a,b) memset(a,b,sizeof(a))
# define CLL(a,b,n) fill(a,a+n,b)

const double inf=1e30;
const int INF=1<<30;
const int N=1000;

unsigned long long n,k;

const unsigned long long a=0xffffffffffffffff;
unsigned long long ans;

void dfs(unsigned x)
{
    if(x<=0) return ;
    int t=x;
    int base=0;
    while(t){
        t/=2;
        ++base;
    }
    unsigned long long p=1;
    int id;
    for(int cnt=0,i=0;i<64&&cnt<=base;++i){
        id=i;
        if(n&(p<<i)){
            ++cnt;
            if(cnt==base) ans|=(p<<i);
        }
    }
    dfs(x-(p<<(base-1)));
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        cin>>n>>k;
        n=~n;
        ans=0;
        dfs(k);
        cout<<ans<<endl;
    }
    return 0;
}