位运算

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Bitwise Equations

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 652 Accepted Submission(s): 347

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

题意：给你一个数X和K，问你第K小的数Y使得X + Y = X | Y。

思路：显然对于X，若它的二进制位为0，那么Y对应的二进制位可以填0和1；反之Y对应的二进制位只能填0。

因此，只需要预处理X二进制的前i位的合法填数方案vnum[i]，这个过程中借助了num[i]——前i位不同的组合数，可能头一位是0。统计前缀和sum。接下来，一步步找好了，填好Y的二进制位，求出Y就可以了。

后来发现自己一开始就想复杂了。。。

先来个与题目无关的代码位运算的巧妙运用a+b：

<span style="font-size:10px;">#include<iostream>//c++#include<cmath>//数学公式#include<cstdlib>//malloc#include<cstring>#include<string>#include<cstdio>//输入输出#include<algorithm>//快排#include<queue>//队列#include<functional>//优先队列#include<stack>//栈#include<vector>//容器#include<map>//地图  if continueusing namespace std;int main(){int i,j,k,a,b,c;int x,y;cin>>a>>b;x=a|b;y=a&b;while(y){       <span style="white-space:pre"></span> <span style="white-space:pre"></span>x=x^y;       <span style="white-space:pre"></span>y<<=1;        <span style="white-space:pre"></span>a=x;        <span style="white-space:pre"></span>b=y;        <span style="white-space:pre"></span>x=a|b;        <span style="white-space:pre"></span>y=a&b;}cout<<x<<endl;return 0;}</span><span style="font-size: 18px;"></span>

本题代码如下：

<span style="font-size:10px;">#include<iostream>//c++#include<cmath>//数学公式#include<cstdlib>//malloc#include<cstring>#include<string>#include<cstdio>//输入输出#include<algorithm>//快排#include<queue>//队列#include<functional>//优先队列#include<stack>//栈#include<vector>//容器#include<map>//地图  if continuetypedef long long ll;const  int N=30;using namespace std;int main(){    //freopen("C:\\Users\\ch\\Desktop\\1.txt","r",stdin);//freopen("C:\\Users\\lenovo\\Desktop\\2.txt","w",stdout);int i,j,k;ll x,n;while(~scanf("%lld%lld",&x,&n)){    ll p=1,ans=0;    while(n)        {            while(x&p) p<<=1;            if(n&1)    ans|=p;            n>>=1;            p<<=1;        }        cout<<ans<<endl;}return 0;}</span><span style="font-size: 18px;"></span>

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