Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
9 5
4
3 3
2
5 2
0
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
s=a+b x=a xor b
a+b=2*(a&b)+a xor b
想了半天,ppl说我能确定s-x是进位
想了一下,异或是半加法,没有进位,全加减半加,那么剩下的就是进位了。
异或为0的位,应该是全0或者全1,不用考虑
异或为1的位,应该是1 xor 0 或者 0 xor 1 。所以只要书异或中为1的位有多少就行了。
特判比较容易漏掉:s<x为0 ,全加法不可能小于半加法; s-x为奇数,或者是说最后一位为1,是不可能的,因为s-x的最后一位没有意义,倒数第二位才是最后一位相加的进位。xor为1,进位也为1,即AND为1,也是不可能的。
最后,如果s==x 或出现0的情况,答案减2
#include <iostream>
using namespace std;
int main()
{
long long s,x,t,a,ans=1;
cin>>s>>x;
t=x;
a=s-x;
a>>=1;
if(s<x || (s-x)%2==1)
cout<<0<<endl;
else
{
while(x!=0)
{
if(x&1)
ans*=2;
if(x&1 && a&1)
{
ans=0;
break;
}
x>>=1;
a>>=1;
}
if(s==t) ans-=2;
cout<<ans<<endl;
}
return 0;
}