Problem:
Score : 500 points

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 ≦ 10^5

1 ≦ A_i ≦ 10^9

Input

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

N
A_1 A_2 … A_N

Output

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

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.

Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

Sample Input 3

4
1 2 3 1

Sample Output 3

NO

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)


#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
int main(void)
{
	//変数の宣言
	int n;
	int a[100010];
	long long int sum;
	
	//よく使う変数
	int i,j,k,l;
	int flag=0;
	int ans=0;
	int count=0;
	int temp,temp1,temp2;
	int max,min;
	int len;
	//データの読み込み
	
	scanf("%d",&n);
	
	for(i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	a[n]=a[0];
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	if(n==1){
		printf("NO");
		return 0;
	}
	
	sum=0;
	for(i=0;i<n;i++){
		sum=sum+a[i];
	}
	
	if(sum%(n*(n+1)/2)!=0){
		printf("NO");
		return 0;
	}else{
		temp=(int)(sum/(n*(n+1)/2));
	}
	
	temp2=temp;
	for(i=0;i<n;i++){
		if((temp+a[i]-a[i+1])%n!=0){
			printf("NO");
			return 0;
		}else{
			temp2=temp2-(temp+a[i]-a[i+1])/n;
		}
	}
	
	
//	printf("計算部分終了\n");
	//出力
	
	if(temp2==0){
		printf("YES\n");
	}else{
		printf("NO\n");
	}
	
//	printf("結果の出力終了\n");
	
	return 0;
	
}