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;
}