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>

int main(int argc, char const *argv[])
{
	int n;
	int *a;
	scanf("%d",&n);
	a = malloc(n*sizeof(int));
	long long sum = 0;
	int sum_ = n *(n+1)/2;
	// printf("sum_:%d\n",sum_);
	int common =1000;
	for(int i = 0;i<n;i++){
		scanf("%d",a+i);
		if(a[i]/sum_ < common ) common = a[i] / sum_ ;
	}
	// printf("common = %d\n",common);
	for(int i = 0; i < n; i++){
		a[i] -= common * sum_;
		sum += a[i];
	}

	int j = 0;
	int max_id=0;
	if(sum%sum_ != 0){
		printf("NO\n");
		return 0;
	}
	while(1){
		for(int i = 0; i<n;i++){
			if(a[max_id] < a[i]) max_id = i;
		}

		if(a[max_id] == 0){
			printf("YES\n");
			return 0;
		}

		for(int i = 0; i < n;i++ ){
			j = (n + max_id - i)%n;
			if(a[j] >= n-i){
			 a[j] -= n-i;
			}
			else{
				printf("NO\n");
				return 0;
			}
		}
	}


	return 0;
}