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 <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main() {
  int i, j;
  int n;
  int data[100001];
  int dif[100001];
  //  long int times[100001];
  long long int sum = 0;
  int times;
  int tme[100001];
  long long int one;
  int sum2 = 0;
  int flag = 0;

  scanf("%d", &n);

  for (i = 0; i < n; i++) {
    scanf("%d", & data[i]);
    sum += data[i];
  }

  one = n * (n + 1) / 2;
  if (sum % one != 0) {
    flag = 1;
  } else {
    times = sum / one;
    for (i = 0; i < n-1; i++) {
      dif[i] = data[i+1] - data[i];
    }
    dif[n-1] = data[0] - data[n - 1];

    for (i = 0; i < n; i++) {
      if ((times - dif[i]) % n != 0) {
        flag = 1;
        break;
      } else {
        tme[i] = (times - dif[i]) / n;
        if(tme[i]<0){
          flag = 1;
          break;
        }
      }
    }
    if (flag == 0) {
      for (i = 0; i < n; i++) {
        sum2 += tme[i];
      }
    }
   }
  if (flag == 0) {
    printf("YES");
/*
    if (sum2 == times) {
      printf("YES");

    } else {
      printf("NO");
    }
*/
  } else {
    printf("NO");
  }

  return 0;
}