计算机学院大学生程序设计竞赛（2015’12）Bitwise Equations

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本文探讨了一道关于位运算的编程题，题目要求找出满足特定条件的第K小正整数Y。通过分析输入X的二进制表示并结合K的值，文章提供了一种算法来高效解决这个问题。

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

3
5 1
5 5
2000000000 2000000000

Sample Output

2
18
16383165351936

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <list>
#define LL long long
#define INF 0x3f3f3f3f

using namespace std;

int a[110],b[110];
int ant[110];

int main()
{
    int T;
    LL x,k;
    cin>>T;
    while(T-- && cin>>x>>k)
    {
        LL ans = 0;
        int top1 = 0;
        int top2 = 0;
        while(x)
        {
            a[top1++] = x % 2;
            x /= 2;
        }
        while(k)
        {
            b[top2++] = k % 2;
            k /= 2;
        }
        int pos = 0;
        for(int i=0;i<top1;i++)
        {
            if(a[i] == 0)
            {
                if(pos != top2)
                    ant[i] = b[pos++];
                else
                    ant[i] = 0;
            }
            else
                ant[i] = 0;
        }
        int an = top1;
        while(pos < top2)
            ant[an++] = b[pos++];
        for(int i=an-1;i>=0;i--)
            ans = ans * 2 + ant[i];
        cout<<ans<<endl;
    }
    return 0;
}