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

Bitwise Equations

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 633    Accepted Submission(s): 335

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

 

总是望着曾经的空间发呆，那些说好不分开的朋友不在了，转身，陌路。 熟悉的，安静了， 安静的，离开了， 离开的，陌生了， 陌生的，消失了， 消失的，陌路了。

#include <string.h>#include <iostream>using namespace std;long long rets=0;class jisuan{public:    long long kthPlusOrSolution(int, int);    string getBin(int);};string jisuan::getBin(int x){    string ret="";    if(x==0)    {        ret="0";        return ret;    }    while(x>0)    {        ret=char(x%2+'0')+ret;        x/=2;    }    return ret;}long long jisuan::kthPlusOrSolution(int x, int k){    string strx, strk;    strx=getBin(x);    strk=getBin(k);    int lx=strx.length()-1;    int lk=strk.length()-1;    string ret="";    while(lx>=0 && lk>=0)    {        if(strx[lx]=='1')        {            ret="0"+ret;            --lx;        }        else        {            ret=strk[lk]+ret;            --lx;            --lk;        }    }    while(lk>=0)        ret=strk[lk--]+ret;    for(int i=0;i<(int)ret.length();i++)        rets=(rets<<1)+ret[i]-'0';    return rets;}int main(){    int n,x,k;    jisuan bit;    cin>>n;    while(n--)    {        rets=0;        cin>>x>>k;        bit.kthPlusOrSolution(x,k);        cout<<rets<<endl;    }}

@执念  "@☆但求“❤”安★ 下次我们做的一定会更好。。。。
为什么这次的题目是英文的。。。。QAQ...