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>

long A[100000];
long d[100000];
long s;
int N;
int num;

long sum(int N){
	return N*(N+1)/2;
}

void input(){
	int i;
	s = 0;
	scanf("%d", &N);
	for( i = 0; i < N; i++ ){
		scanf("%ld", &A[i]);
		s += A[i];
	}
	num = s / sum(N);
	for( i = 0; i < N-1; i++ ){ d[i] = A[i+1] - A[i] - num; }
	d[N-1] = A[0] - A[N-1] - num;
	return;
}

int check(){
	int i, j = 0;
	for( i = 0; i < N; i++ ){
		if( d[i] != 0 && d[i] != -N ){ return 0; }
		if( ((-1) * d[i]) % N == 0 ){ j += ((-1)*d[i]) / N; }
	}
	if( j != num ){ return 0; }
	return 1;
}

int main(){
	input();
	if( num * sum(N) != s ){ printf("NO\n"); return 0; }
	if( check() == 0 ){ printf("NO\n"); return 0; }
	printf("YES\n");
	return 0;
}