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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

int a[100100],dif[100100],mx;
long long sum;
vector<int> v;
map<int,bool> mp;

main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
		sum+=a[i];
	}
	//printf("%lld %d\n",sum,(n*(n+1)/2));
	if(sum%(long long)((n*(n+1))/2)!=0)
	{
		printf("NO\n");
		return 0;
	}
	a[n]=a[0];
	n++;
	for(int i=0;i<n-1;i++)
	{
		dif[i]=a[i+1]-a[i];
		mx=max(mx,dif[i]);
	}
	n--;
	for(int i=0;i<n;i++)
	{
		if(dif[i]<mx)
		{
			if(mp[dif[i]]==0)
			{
				v.push_back(-dif[i]);
				mp[dif[i]]=1;
			}
		}
	}
	sort(v.begin(),v.end());
	long long mi=0;
	int cou=0;
	for(int i=0;i<n;i++)
	{
		a[i]-=mi;
		//printf("(%d,%d) ",mi,cou);
		mi+=cou;
		if(dif[i]<mx)
		{
			int ma=(lower_bound(v.begin(),v.end(),-dif[i])-v.begin())+1;
			mi+=ma;
			cou+=ma;
		}
	}
//	printf("\n");
	mi=0;
	cou=0;
	for(int i=n-1;i>=0;i--)
	{
		if(dif[i]<mx)
		{
			int ma=(lower_bound(v.begin(),v.end(),-dif[i])-v.begin())+1;
			mi+=ma*n;
			cou+=ma;
		}
		a[i]-=mi;
		//printf("(%d,%d) ",mi,cou);
		mi-=cou;
	}
	//printf("\n");
	int qu=-1;
	for(int i=0;i<n;i++)
	{
		if(a[i]<0)
		{
			printf("NO\n");
			return 0; 
		}
		else if(qu==-1)
		{
			qu=a[i];
		}
		else if(qu!=a[i])
		{
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");
}