思路:一个经典的Nim博弈
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200000+50;
int a[maxn];
int main()
{
int n;
while(scanf("%d",&n)!=EOF && n)
{
int sum = 0;
for(int i = 1;i<=n;i++)
{
scanf("%d",&a[i]);
sum^=a[i];
}
if(sum==0)
printf("No\n");
else
{
printf("Yes\n");
int res;
for(int i = 1;i<=n;i++)
{
res = a[i]^sum;
if(res<a[i])
printf("%d %d\n",a[i],res);
}
}
}
}