题目描述
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
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
输入
The input is given from Standard Input in the following format:
N
A1 A2 … AN
N
A1 A2 … AN
输出
If it is possible to remove all the stones from the boxes, print YES. Otherwise, print NO.
样例输入
5
4 5 1 2 3
样例输出
YES
提示
All the stones can be removed in one operation by selecting the second box.
题解
假设a1有x1个。。。an有xn个,然后全部加起来就是n*(n+1)/2*(x1+x2+...xn)=(a1+a2+...an),求出(x1+x2+...xn)来,然后相邻的两个a相减就可求出
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> #define ll long long using namespace std; const int maxn=1e5+10; ll a[maxn]; int main() { ll i,j,k,m,n; cin>>n; ll cnt=0,ans=0,sum=0; for(i=1; i<=n; i++) { cin>>a[i]; sum+=a[i]; } if((2*sum)%(n*(n+1))!=0) { cout<<"NO"<<endl; return 0; } else { ans=(2*sum)/(n*(n+1)); for(i=1; i<=n; i++) { if(i==n) { if((a[i]-a[1]+ans)%n!=0||(a[i]-a[1]+ans)<0) { cout<<"NO"<<endl; return 0; } } else { if((a[i]-a[i+1]+ans)%n!=0||(a[i]-a[i+1]+ans)<0) { cout<<"NO"<<endl; return 0; } } } cout<<"YES"<<endl; } return 0; }