Codeforces Round #628 (Div. 2)

D. Ehab the Xorcist
Given 2 integers u and v, find the shortest array such that bitwise-xor of its elements is u, and the sum of its elements is v.
Input
The only line contains 2 integers u and v (0≤u,v≤1018).
Output
If there’s no array that satisfies the condition, print “-1”. Otherwise:
The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.
Examples
inputCopy
2 4
outputCopy
2
3 1
inputCopy
1 3
outputCopy
3
1 1 1
inputCopy
8 5
outputCopy
-1
inputCopy
0 0
outputCopy
0
Note
In the first sample, 3⊕1=2 and 3+1=4. There is no valid array of smaller length.
Notice that in the fourth sample the array is empty.
题意
输入两个数u和v，求是否存在一个最小的数组使得数组内所有数的异或和为u，和为v。

bitwise-xor 按位异或 
⊕          按位异或和符号

搞清楚这些就可以开始愉快(痛苦)的a题了

根据异或运算（1^1=0 ,00=0,10=1）的规则可以得出n个数异或运算的结果一定会小于等于n个数的和
所以得出如果u>v就无解
还有一种无解的情况就是u和v的奇偶性不同。如果异或和u是一个奇数，那么数组里面奇数的个数也是奇数个的(奇数个奇数的异或结果最低为一定是1，即奇数)，而奇数与奇数相加(偶数的个数随便多少个，因为偶数与偶数之和一定是偶数，偶数异或和最低位也一定是0)结果一定是奇数。u和v都是偶数（理由同上推导）
先判断u=v的情况，如果u=v=0输出0（样例），u=v！=0，结果就是u
最简单的一个答案就是数组个数为3，数组值分别为 u、(v-u)/2、（v-u）/2
因为(v-u)/2 ⊕（v-u）/2=0

因为要求最短，所以要判断一下数组个数为2满不满足条件，即把前两个合并一下

include

using namespace std;
int main()
{

long long int u,v; 
while(cin>>u>>v)
{ 
    if(u%2==v%2&&u<=v)//奇偶性判断以及u<=v 
    { 
        if(u==v)
        { 
            if(u==v&&u==0)printf("0\n");
            else printf("1\n%lld\n",u);
        }
        else 
        { 
            long long int w=(v-u)/2;
            long long int sum=u+w;//合并u和（v-u）/2 
            if((sum^w)==u&&sum+w==v)printf("2\n%lld %lld\n",w,sum);//(sum^w)这个括号不难少 
            else 
            { 
                printf("3\n%lld %lld %lld\n",u,w,w);
            }
        }
    }
    else
    printf("-1\n"); 
}
return 0;