Average

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3032    Accepted Submission(s): 735
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 sodax 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 integerT, 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 integerm(0≤m≤n) in the second line denoting the number of operations needed. Then each of the followingm 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
Author [email protected]
Source
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題目大意：

    有N過杯子，每一個有ai個糖果，每相鄰兩個之間可以最多傳遞一個糖果，問能不能最後每個杯子的糖果一樣多，並出處傳遞過程。

解題思路：

    這題的最基本思路和之前CF的一道題很像，就是把不正常的狀態按照一個方向傳遞給相鄰的轉態，使得這個狀態正常。

    首先，如果糖果總數不能被等分的話就一定錯誤，其次每個杯子最多給出或接受兩個糖果，如果其糖果數與平均值差的絕對值超過2也一定不可以。

    由於杯子形成了一個圈，所以糖果的傳遞一共只有兩種轉態，向前穿一個和向後傳一個。我們就可以使用貪心的思路從左到右掃，每次儘量使得當前被子達到要求。這樣一共三種情況，如果本來就達到要求了就跳過；如果多了就向右傳遞一個，如果少了就讓右邊的傳來一個。

    由於是一個環，所以我們還需要掃第二次，重複剛才的操作（注意可能會出現類似回溯的情況，要把上次的操作抵消掉）。這樣我們得到的結果就一定是正確的。最後判斷一下經過操作有沒有達到要求，輸出即可。

AC程式碼：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f
#define mem(a,b) memset((a),(b),sizeof(a))

const int MAXN=100000+3;
LL N,cup[MAXN];
bool l[MAXN],r[MAXN];//是否左移、右移

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld",&N);
        bool ok=true;
        LL ave=0;
        for(int i=0;i<N;++i)
        {
            scanf("%lld",&cup[i]);
            ave+=cup[i];
            l[i]=r[i]=false;
        }
        if(ave%N)//不能整除，一定不可以
        {
            puts("NO");
            continue;
        }
        ave/=N;
        for(int i=0;i<N;++i)
            if(cup[i]>ave+2||cup[i]<ave-2)//超過ave+2或小於ave-2，一定不可以
            {
                ok=false;
                break;
            }
        if(!ok)
        {
            puts("NO");
            continue;
        }
        
        for(int i=0;i<N;++i)
        {
            if(cup[i]>ave&&r[i]==false&&cup[(i+1)%N]<=ave)//右移
            {
                --cup[i];
                ++cup[(i+1)%N];
                if(l[(i+1)%N]==true)//和上次操作相抵消
                    l[(i+1)%N]=false;
                else r[i]=true;
            }
            else if(cup[i]<ave&&l[(i+1)%N]==false&&cup[(i+1)%N]>=ave)//左移
            {
                ++cup[i];
                --cup[(i+1)%N];
                if(r[i]==true)//和上次操作相抵消
                    r[i]=false;
                else l[(i+1)%N]=true;
            }
        }
        //第二輪
        for(int i=0;i<N;++i)
        {
            if(cup[i]>ave&&r[i]==false&&cup[(i+1)%N]<=ave)//右移
            {
                --cup[i];
                ++cup[(i+1)%N];
                if(l[(i+1)%N]==true)
                    l[(i+1)%N]=false;
                else r[i]=true;
            }
            else if(cup[i]<ave&&l[(i+1)%N]==false&&cup[(i+1)%N]>=ave)//左移
            {
                ++cup[i];
                --cup[(i+1)%N];
                if(r[i]==true)
                    r[i]=false;
                else l[(i+1)%N]=true;
            }
        }
        
        LL num=0;
        for(int i=0;i<N;++i)
        {
            if(cup[i]!=ave)//有的杯子不符合要求
            {
                ok=false;
                break;
            }
            if(l[i])
                ++num;
            if(r[i])
                ++num;
        }
        if(!ok)
            puts("NO");
        else
        {
            puts("YES");
            printf("%lld\n",num);
            for(LL i=0;i<N;++i)
            {
                if(r[i])
                    printf("%lld %lld\n",i+1,(i+1)%N+1);
                if(l[i])
                    printf("%lld %lld\n",i+1,(i-1+N)%N+1);
            }
        }
    }
    
    return 0;
}