Problem Description

There are  soda sitting around a round table. soda are numbered from  to  and -th soda is adjacent to -th soda, -st soda is adjacent to -th soda.

Each soda has some candies in their hand. And they want to make the number of candies the same by doing some taking and giving operations. More specifically, every two adjacent soda  and  can do one of the following operations only once:
1. -th soda gives -th soda a candy if he has one;
2. -th soda gives -th soda a candy if he has one;
3. they just do nothing.

Now you are to determine whether it is possible and give a sequence of operations.

 

Input

There are multiple test cases. The first line of input contains an integer , indicating the number of test cases. For each test case:

The first contains an integer  , the number of soda.
The next line contains  integers  , where  denotes the candy -th soda has.

 

Output

For each test case, output "YES" (without the quotes) if possible, otherwise output "NO" (without the quotes) in the first line. If possible, then the output an integer   in the second line denoting the number of operations needed. Then each of the following  lines contain two integers  and  , which means that -th soda gives -th soda a candy.

 

Sample Input

3
6
1 0 1 0 0 0
5
1 1 1 1 1
3
1 2 3

 

Sample Output

NO
YES
0
YES
2
2 1
3 2


这题要注意可能会有从n绕回1的，或者从1绕到n的。感觉情况比較多。
。必须注意2的情况。。
#include<cstdio>
#include<cmath>
#include<iostream>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn=100005;
ll T,n,m,a[maxn],sum,tot;
bool flag;
int f[maxn][2];

int main()
{
    scanf("%lld",&T);
    while (T--)
    {
        scanf("%lld",&n);
        sum=0;    flag=true;
        for (int i=0;i<n;i++) 
        {
            scanf("%lld",&a[i]);
            sum+=a[i];
        }
        if (sum%n) flag=false;
        else 
        {
            sum=sum/n;
            tot=0;
            for (int i=0;i<n;i++) a[i]-=sum;
            memset(f,0,sizeof(f));
            for (int i=0,j,k;i<n+n;i++)
            {
                j=i%n;    k=(i+1)%n;
                if (a[j]>0&&a[k]<=0&&!f[j][1]) { 
                    a[k]++;
                    a[j]--;
                    if (!f[k][0]) f[j][1]=1;
                      else f[k][0]=f[j][1]=0; 
                }
                else if (a[j]<0&&a[k]>=0&&!f[k][0]) {  
                    a[k]--;
                    a[j]++;
                    if (!f[j][1]) f[k][0]=1;
                      else f[j][1]=f[k][0]=0; 
                }
            }
            for (int i=0;i<n;i++) if (a[i]) {flag=false; break;}
             for (int i=0;i<n;i++) tot+=f[i][0]+f[i][1];
            if (n==2&&tot==2) flag=false;
        }
        if (flag)
        {
            printf("YES\n%d\n",tot);
            for (int i=1,j,k;i<=n;i++)
             {
                j=i-1;    if (j==0) j=n;
                k=i+1;  if (k>n) k=1;
                if (f[i-1][0]) printf("%d %d\n",i,j);
                if (f[i-1][1]) printf("%d %d\n",i,k);
             }
        }
        else printf("NO\n");
    }
    return 0;
}