B-Boxes

　　http://agc010.contest.atcoder.jp/tasks/agc010_b

Problem Statement

There are N boxes arranged in a circle. The i-th box contains A_i stones.

Determine whether it is possible to remove all the stones from the boxes by repeatedly performing the following operation:

• Select one box. Let the box be the i-th box. Then, for each j from 1 through N, remove exactly j stones from the (i+j)-th box. Here, the (N+k)-th box is identified with the k-th box.

Note that the operation cannot be performed if there is a box that does not contain enough number of stones to be removed.

Constraints

• 1≦N≦10⁵
• 1≦A_i≦10⁹

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Input

The input is given from Standard Input in the following format:

N
A1 A2 … AN

Output

If it is possible to remove all the stones from the boxes, print YES. Otherwise, print NO.

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Sample Input 1

5
4 5 1 2 3

Sample Output 1

YES

All the stones can be removed in one operation by selecting the second box.

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Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

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Sample Input 3

4
1 2 3 1

Sample Output 3

NO
大致意思是说，假设初始化N个数字为0，选取某一个位置，分别增加1到N（如果到达数组末尾就从头开始）。

比如：

    0 0 0（选择第二个位置开始）

    3 1 2（选择第三个位置开始）

    5 4 3（选择第 X 个位置开始）

    .......

给出N个数字，看看是不是由以上操作得到的，是的话输出YES,不是输出NO;

对于数据，记录其和，则K=(n*(n+1))/2为操作次数，另d[i]=a[i]-a[i-1];必定有一个的d[i]=n-1;剩下的为1；

且有d[i]-(k-x)+(n-1)*x=0或k-d[i]=nx(x为相对于的d[i]来说异常操作的数)，所以，一定有k-d[i]>=0 && (k-d[i])%n==0,

也可以得到（k-d[i])/n的和为k,由此可解题。

#include<iostream>
using namespace std;
const int mod=1e5+5;
long long a[mod],b[mod];
int main()
{
    long long n,i,sum=0,count=0,k;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>a[i];
        b[i]=a[i]-a[i-1];
        sum+=a[i];
    }
    b[1]=a[1]-a[n];
    if(sum%(n*(n+1)/2))//如果k不为整数，直接输出NO
    {
        cout<<"NO"<<endl;
        return 0;
    }
    k=sum/(n*(n+1)/2);
    for(i=1;i<=n;i++)
    {
        if((k-b[i]<0) || (k-b[i])%n)
        {
            cout<<"NO"<<endl;
            return 0;
        }
        else count+=(k-b[i])/n;
    }
    if(count!=k) cout<<"NO"<<endl;
    else cout<<"YES"<<endl;
    return 0;
}