Average

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2153 Accepted Submission(s): 532
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 $n$ $(1 le n le 10^5)$, the number of soda.
The next line contains $n$ integers $a_1, a_2, dots, a_n$ $(0 le a_i le 10^9)$, where $a_i$ 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 $m$ $(0 le m le n)$ in the second line denoting the number of operations needed. Then each of the following $m$ lines contain two integers $x$ and $y$ $(1 le x, y le n)$, 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
Recommend wange2014

 

首先一個人有三種操作，每個人選擇的操作影響周邊2人.

如果無環的化可直接貪心。

所以枚舉人1和人2的關系，把環從1與2之間斷開

 

 

#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define mp make_pair
#define MAXN (100000+10)
typedef long long ll;
int n;
int a[MAXN],a2[MAXN];
void rel(int &a,int &b,int c)
{
	a=c-1,b=c+1;
	if (a==0) a=n;if (b>n) b=1;
}
pairans[MAXN];
int siz=0;
bool check(int a[])
{
	a[n+1]=a[1];
	Fork(i,2,n)
	{
		if (a[i]==-1) a[i]++,a[i+1]--,ans[siz++]=mp(i+1<=n?i+1:1,i);
		else if (a[i]==1) a[i]--,a[i+1]++,ans[siz++]=mp(i,i+1<=n?i+1:1);
	}
	a[1]=a[n+1];
	For(i,n) if (a[i]) return 0;
	return 1;
}
void pri()
{
	cout<>T;
	For(kcase , T)
	{
		cin>>n;
		For(i,n) scanf(%d,&a[i]);
		ll sum=0;
		For(i,n) sum+=a[i];
		if (sum%n){
			cout<2){
			cout<