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

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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 longusing 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

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