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

int main(){
    int n, i, time=0;
    scanf(" %d", &n);
    int a[n];
    for(i=0; i<n; i++){
        scanf(" %d", &a[i]);
    }
    long long int cur=0, dev = n*(n+1)/2;
    for(i=0; i<n; i++){
        cur+=a[i];
        if(cur>=dev){
            int hoge = cur/dev;
            time+=hoge;
            cur-=dev*hoge;
        }
    }
    if(cur!=0){
        printf("NO");
        return 0;
    }
    int x, y, z=0;
    for(i=0; i<n-1; i++){
        x=a[i+1]-a[i];
        y=time-x;
        if(y%n != 0 || a[i]<time){
            printf("NO");
            return 0;
        }
        else{
            z+=y/n;
        }
    }
    x=a[0]-a[n-1];
    y=time-x;
    if(y%n != 0){
        printf("NO");
        return 0;
    }
    else{
        z+=y/n;
    }
    if(z==time){
        printf("YES");
    }
    else{
        printf("NO");
    }
    return 0;
}