题目描述

There are N boxes arranged in a circle. The i-th box contains Ai 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≤10

⁵

1≤Ai≤10

⁹

输入

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

N

A1 A2 … AN

输出

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

样例输入

54 5 1 2 3

样例输出

YES

提示

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

#include 
       
        #define ll long longusing namespace std;const int maxn=1e5+10;ll a[maxn];int main(){    ll i,j,k,m,n;    cin>>n;    ll cnt=0,ans=0,sum=0;    for(i=1; i<=n; i++)    {        cin>>a[i];        sum+=a[i];    }    if((2*sum)%(n*(n+1))!=0)    {        cout<<"NO"<
       

首先几个数的和一定是1到n的和的倍数

不是则no

根据是几倍判断有多少个1到n

然后找规律