Codeforces Round #682 (Div. 2)——D. Powerful Ksenia

Ksenia has an array a consisting of n positive integers a1,a2,…,an.

In one operation she can do the following:

choose three distinct indices i, j, k, and then
change all of ai,aj,ak to ai⊕aj⊕ak simultaneously, where ⊕ denotes the bitwise XOR operation.
She wants to make all ai equal in at most n operations, or to determine that it is impossible to do so. She wouldn’t ask for your help, but please, help her!

Input
The first line contains one integer n (3≤n≤105) — the length of a.

The second line contains n integers, a1,a2,…,an (1≤ai≤109) — elements of a.

Output
Print YES or NO in the first line depending on whether it is possible to make all elements equal in at most n operations.

If it is possible, print an integer m (0≤m≤n), which denotes the number of operations you do.

In each of the next m lines, print three distinct integers i,j,k, representing one operation.

If there are many such operation sequences possible, print any. Note that you do not have to minimize the number of operations.

Examples
inputCopy
5
4 2 1 7 2
outputCopy
YES
1
1 3 4
inputCopy
4
10 4 49 22
outputCopy
NO
Note
In the first example, the array becomes [4⊕1⊕7,2,4⊕1⊕7,4⊕1⊕7,2]=[2,2,2,2,2].

题意： 给你一些数，你可以有一个操作，可以选择三个数然后把他们的值变成他们异或后的值。问你是否可以在n次操作内把所有的数变成一样的。

思路： 对于所有的奇数n来说，是必定可以的，对于所有的偶数n来说，如果他的所有的值异或起来为0就是可以的。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[100007];
int main()
{int n;cin>>n;int sum=0;for(int i=1;i<=n;i++){cin>>a[i];sum=sum^a[i];}int flag=0;if(n%2==0&&sum!=0){printf("NO\n");return 0;}else{printf("YES\n%d\n",(n-1)/2*2-1);for(int i=1;i<n-1;i+=2){printf("%d %d %d\n",i,i+1,i+2);}for(int i=1;i<n-3;i+=2){printf("%d %d %d\n",i,i+1,n);}}return 0;
}