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

3
5 1
5 5
2000000000 2000000000

 

Sample Output

2
18
16383165351936

 

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...