2015’12杭电校赛1005 Bitwise Equations（二进制找规律）

题解

有T（100）组数据， 对于每组数据，给定X,K（1<=X,K<=2e9）

让你求出，满足X+Y==X|Y的第K小的Y

很显然， X+Y==X|Y的Y，需要满足的性质是——

X二进制是1的位置，Y必须是0；X二进制是0的位置，Y必须是1。

代码

#include<cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
#define ll long long
using namespace std;

int t;
ll x, k, y;
ll a[100], b[100];

int main()
{
    scanf("%d", &t);
    while (t--) 
    {
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        scanf("%lld%lld", &x, &k);
        ll i, al, bl;
        for (i = 0; x; i++) 
        {
            a[i] = x & 1;
            x >>= 1;
        }
        al = i;
        for (i = 0; k; i++) 
        {
            b[i] = k & 1;
            k >>= 1;
        }
        bl = i;
        y = 0;
        ll j = 0;
        for (i = 0; j < bl; i++) 
        {
            if (a[i] == 1)
                a[i] = 0;
            else
                a[i] = b[j++];
            y += a[i] << i;
        }
        printf("%lld\n", y);
    }
    return 0;
}

题目

Bitwise Equations

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 713 Accepted Submission(s): 376

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