Computer College Undergraduate Program Design Contest (2015 ') 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 integer Y, for positive th E following equation holds:x + Y =x | Y
Where ' | ' denotes the bitwise OR operator.
Input the ' the ' input contains an integer t (t <=) which means the number of test cases.
For each case, the 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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#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;
 }
}

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for what this time the topic is in English .... Qaq ...