Boxes

Boxes

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Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

There are N boxes arranged in a circle. The i-th box contains Ai 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≦105
• 1≦Ai≦109

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Input

The input is given from Standard Input in the following format:

N
A1 A2 … AN

Output

If it is possible to remove all the stones from the boxes, print YES. Otherwise, print NO.

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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.

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Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

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Sample Input 3

4
1 2 3 1

Sample Output 3

NO
分析：首先可以根据总和得出操作次数p，然后考虑相邻的两个数；
　　　a[i]-a[i+1]=(n-1)x-(p-x);（a[i]=n,a[i+1]=1贡献1-n，a[i]=y,a[i+1]=y+1贡献-1）；
　　　化简即a[i]-a[i+1]=nx-p;
　　　检验(a[i]-a[i+1]+p)%n==0即可；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e5+10;
using namespace std;
inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline void umax(ll &p,ll q){if(p<q)p=q;}
inline void umin(ll &p,ll q){if(p>q)p=q;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,a[maxn];
int main()
{
    int i,j;
    scanf("%d",&n);
    ll ret=0,ok=0;
    rep(i,1,n)scanf("%d",&a[i]),ret+=a[i],ok+=i;
    if(ret%ok!=0)return 0*puts("NO");
    ll x=ret/ok,p=0;
    a[n+1]=a[1];
    rep(i,1,n)
    {
        ll q=a[i]-a[i+1],qq=(q+x)/n;;
        if(qq*n-x!=q)return 0*puts("NO");
        if(qq<0||qq>x)return 0*puts("NO");
        p+=qq;
    }
    if(p!=x)return 0*puts("NO");
    puts("YES");
    return 0;
}