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

int main(){
  int n;
  int m;
  scanf("%d",&n);
  int onesum=0;
  for(m=1;m<=n;m++){
    onesum = onesum + m;}
  int a[n];
  int b[n];
  for(m=0;m<n;m++){
    b[m]=0;}
  int sum=0;
  int count;
  int flag=0;
  int p=1;
  int q=1;
  for(m=0;m<n;m++){
    scanf("%d",&a[m]);
    sum= sum + a[m];}
  count = sum / onesum;
  for(m=0;m<n;m++){
    b[a[m]%n]=1;}
  for(m=0;m<n;m++){
    p = p*b[m];
    if(a[m]<count){
      p=0;}
  }
  for(m=0;m<n;m++){
    if(a[m]%onesum != 0){
      q=0;}
  }
    
  if(sum%onesum!=0){
     p=0;}
  if((p==1)||(q==1)){
    printf("YES");}
  else{printf("NO");}
  return 0;
}