Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 500 points

Problem Statement

We have a sequence of N integers: A1,A2,⋯,AN.

You can perform the following operation between 0 and K times (inclusive):

Choose two integers i and j such that i≠j, each between 1 and N (inclusive). Add 1 to Ai and −1 to Aj, possibly producing a negative element.

Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y=xz.

题意很简单：对于给你的N的整数A1~AN 可以进行0~k次操作，从1~N中任取俩数,Ai和Aj(i!=j),Ai + 1 且 Aj - 1 , 最后使A1~AN的公因数最大

Constraints

2≤N≤500
1≤Ai≤10^6
0≤K≤10^9
All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A1 A2 ⋯⋯ AN−1 AN

Output

Print the maximum possible positive integer that divides every element of A after the operations.

Sample Input 1

2 3
8 20

Sample Output 1

77 will divide every element of AA if, for example, we perform the following operation:

Choose i=2,j=1. A becomes (7,21)(7,21).

We cannot reach the situation where 8 or greater integer divides every element of A.

Sample Input 2

2 10
3 5

Sample Output 2

Consider performing the following five operations:

Choose i=2,j=1. A becomes (2,6)(2,6).
Choose i=2,j=1. A becomes (1,7)(1,7).
Choose i=2,j=1. A becomes (0,8)(0,8).
Choose i=2,j=1. A becomes (−1,9)(−1,9).
Choose i=1,j=2. A becomes (0,8)(0,8).

Then, 0=8×0 and 8=8×1, so 88 divides every element of A. We cannot reach the situation where 9 or greater integer divides every element of A.

Sample Input 3

4 5
10 1 2 22

Sample Output 3

Sample Input 4

8 7
1 7 5 6 8 2 6 5

Sample Output 4

思路：

既然是A1~AN的所有数的因数，那么也是它们和sum的因数，所以枚举sum的所有因数，测试因数g,算出A1~AN对g的余数，并进行从小到大的排序，左边的数（即余数小的数）先减成g的倍数，右边的数（即余数大的数）对应加上减去的数，补成g的倍数

至于说为什么这样一定成功，我还没想明白

接下来是某个大佬的代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=500+9;
int n;
int a[maxn],lmt,b[maxn];bool check(int g){for(int i=1;i<=n;++i)b[i]=a[i]%g;sort(b+1,b+n+1);int tot=lmt,cost=0;int t=n;for(int i=1;i<=t;++i){tot-=b[i];if(tot<0)return false;cost+=b[i];//b[i]=0;while(cost>0){int d=g-b[t];//此时d就是b[t]距离g的倍数最近的距离，好好想下就明白了 if(d<=cost){//b[t]=0;cost-=d;--t;}else{b[t]+=cost;cost=0;}}}return true;
}int solve(){int sum=0;for(int i=1;i<=n;++i)sum+=a[i];int qt=sqrt(sum);vector <int> d;// divisorfor(int i=1;i<=qt;++i)if(sum%i==0){d.push_back(i);if(check(sum/i))return sum/i;}int tot=d.size();for(int i=tot-1;i>=0;--i){if(check(d[i]))return d[i];}
}void init(){cin>>n>>lmt;for(int i=1;i<=n;++i)cin>>a[i];
}
int main(){init();cout<<solve()<<"\n";return 0;
}

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