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Boxes

2024-09-03 02:55:55   219人阅读

Boxes

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

Input

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

NA1 A2 … AN

Output

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

Sample Input 1

54 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

56 9 12 10 8

Sample Output 2

YES

Sample Input 3

41 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;}

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