各种情况特判,然后枚举前两个点之间的关系

Average

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Special Judge

Problem Description

There are n soda
sitting around a round table. soda are numbered from 1 to n and i-th
soda is adjacent to (i+1)-th
soda, 1-st
soda is adjacent to n-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 x and y can
do one of the following operations only once:

1. x-th
soda gives y-th
soda a candy if he has one;

2. y-th
soda gives x-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 T,
indicating the number of test cases. For each test case:

The first contains an integer n (1≤n≤105),
the number of soda.

The next line contains n integers a1,a2,…,an (0≤ai≤109),
where ai denotes
the candy i-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 m (0≤m≤n) in
the second line denoting the number of operations needed. Then each of the following m lines
contain two integers x and y (1≤x,y≤n),
which means that x-th
soda gives y-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

Source

2015 Multi-University Training Contest 6

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月08日 星期六 09时35分46秒
File Name     :HDOJ5353.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

const int maxn=100100;

int n;
int a[maxn],b[maxn],f[maxn];
LL s;

void print()
{
	puts("YES");
	int cnt=0;
	for(int i=0;i<n;i++)
		if(f[i]) cnt++;
	printf("%d\n",cnt);
	for(int i=0;i<n;i++)
	{
		int nx=i+1;
		if(nx==n) nx=0;
		if(f[i]==1)
		{
			printf("%d %d\n",i+1,nx+1);
		}
		else if(f[i]==-1)
		{
			printf("%d %d\n",nx+1,i+1);
		}
	}
}

bool func(int x)
{
	memcpy(b,a,sizeof(a));
	memset(f,0,sizeof(f));

	f[0]=x;
	b[0]-=x; b[1]+=x;
	b[n]=b[0];

	for(int i=1;i<n;i++)
	{
		if(abs(b[i])>=2) return false;
		if(b[i]==1)
		{
			f[i]=1;
			b[i]-=1;
			b[i+1]+=1;
		}
		else if(b[i]==-1)
		{
			f[i]=-1;
			b[i]+=1;
			b[i+1]-=1;
		}
	}
	return true;
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d",&n);
		s=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",a+i);
			s+=a[i];
		}
		if(s%n!=0) { puts("NO"); continue; }
		s=s/n;
		bool fg=true;
		bool zero=true;
		for(int i=0;i<n&&fg;i++)
		{
			a[i]-=s;
			if(a[i]!=0) zero=false;
			if(!(a[i]>=-2&&a[i]<=2)) fg=false;
		}

		if(fg==false) { puts("NO"); continue; }
		if(zero==true) { puts("YES\n0"); continue; }

		if(func(0)) print();
		else if(func(1)) print();
		else if(func(-1)) print();
		else puts("NO");
	}

    return 0;
}

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