【CodeForces 627A】XOR Equation（位运算）

最新推荐文章于 2021-01-30 11:43:25 发布

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本文探讨了给定两数之和及XOR值时，如何找出所有可能的有序数对。通过位运算推导出解决方案，并附带示例说明。

A. XOR Equation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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)?

Input

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.

Output

Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.

Examples

Input

9 5

Output

Input

3 3

Output

Input

5 2

Output

Note

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).

题目大意：给出两个数 a，b 的和 s ，XOR x，求有多少个 a，b

思路：表示渣渣还不会位运算，看了好久题解才懂

For any two integers a and b, we have $\text{[math]}$ , where $\text{[math]}$ is the xor and a&b is the bitwise AND. This is because $\text{[math]}$ is non-carrying binary addition. Thus, we can find a&b = (s - x) / 2 (if this is not an integer, there are no solutions).

Now, for each bit, we have 4 cases: $\text{[math]}$ , and $\text{[math]}$ . If $\text{[math]}$ , then ai = bi, so we have one possibility:ai = bi = ai&bi. If $\text{[math]}$ , then we must have ai&bi = 0 (otherwise we print 0), and we have two choices: ai = 1 and bi = 0 or vice versa. Thus, we can return 2n, where n is the number of one-bits in x. (Remember to subtract 2 for the cases a = 0 or b = 0 if necessary.)

a + b = ( a ^ b ) + ( a & b ) * 2 有了这个关系，可以枚举 x 的每一位，若 x 的第 i 位是 1，则 a=0,b=1 或 a=1,b=0 有两种情况，此时 a & b=0 ，如果 a & b = 0，那么 ans<<1 ，如果 a & b = 1 则不成立，结果为 0 。若 x 的第 i 位是 0 ，则 a=1,b=1 ( a & b = 1）或 a=0,b=0 （ a & b = 0 ) 因为 a & b的值是确定的，所以此时只有 1 种情况即 a 和 b 的第 i 位是固定的。首先 s - x 必须是非负的并且必须是偶数，如果 s == x，会有解（0，s）和（s，0），与题目的 a，b 非负矛盾，所以减2，（ 0 ^ a = a）

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int main()
{
    ll s,x;
    cin>>s>>x;
    ll n=s-x;
    if (n < 0 || n%2){
        cout<<"0"<<endl;
        return 0;
    }
    ll m=n>>1,ans=1;
    while(x){
        if(x&1 && m&1){
            cout<<"0"<<endl;
            return 0;
        }
        if(x&1) ans<<=1;
        x>>=1;
        m>>=1;
    }
    if(!n) ans-=2;
    cout<<ans<<endl;
    return 0;
}