CodeForces - 627A

CodeForces - 627A
https://vjudge.net/problem/326413/origin
a+b == (a&b)<<1 +(a^b);
然后是位运算，如果对于这一位置，异或值为1时，有两种可能，由乘法原理，答案<<1。如果s==x，就会出现一方全0，一方全1的情况，所以-2.

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<stack>
#include<cstring>
#define inf 2147483647
#define ls rt<<1
#define rs rt<<1|1
#define lson ls,nl,mid,l,r
#define rson rs,mid+1,nr,l,r
#define N 100010
#define For(i,a,b) for(long long i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()

using namespace std;
long long s,x,k,ans=1,t;
void in(long long &x){
    long long y=1;
    char c=g();x=0;
    while(c<'0'||c>'9'){
        if(c=='-')y=-1;
        c=g();
    }
    while(c<='9'&&c>='0'){
        x=(x<<1)+(x<<3)+c-'0';c=g();
    }
    x*=y;
}
void o(long long x){
    if(x<0){
        p('-');
        x=-x;
    }
    if(x>9)o(x/10);
    p(x%10+'0');
}
int main(){
    in(s);in(x);
    t=x;
    k=(s-x)>>1;
    if(s<x||(s-x)%2==1){
        o(0);
        return 0;
    }
    while(x>0){
        if(x&1)
            ans<<=1;
        if((x&1)&&(k&1)){
            ans=0;
            break;
        }
        k>>=1;
        x>>=1;
    }
    if(s==t)
        ans-=2;
    o(ans);
    return 0;
}