[AtCoder Beginner Contest 136]Max GCD

https://atcoder.jp/contests/abc136/tasks/abc136_e
分数：2100
Time Limit: 2 sec
Memory Limit: 1024 MB

Problem Statement

We have a sequence of $N$ integers: $A_1,A_2,⋯,A_N$.

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 $A_i$ and $−1$ to $A_j$ , 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 $y=xz$.
Constraints
$2≤N≤500$
$1≤A_i≤10^6$
$0≤K≤10^9$
All values in input are integers.

Input

Input is given from Standard Input in the following format:

$N$ $K$
$A_1, A_2 ⋯ A_{N−1} ,A_N$

Output

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

Sample Input

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2 10
3 5

Sample Output

8

Consider performing the following five operations:

Choose i=2,j=1. A becomes (2,6).
Choose i=2,j=1. A becomes (1,7).
Choose i=2,j=1. A becomes (0,8).
Choose i=2,j=1. A becomes (−1,9)
.Choose i=1,j=2. A becomes (0,8)
Then, 0=8×0 and 8=8×1, so 8 divides every element of A. We cannot reach the situation where 9 or greater integer divides every element of A.

题意：
给定 $n(n<=500)$个数字 $A_1$到 $A_n$还有最多操作次数 $k$
每次操作能够选出一对不同的 $i$和 $j$，让 $A_i$加一，并让 $A_j$减一
问最大的x，满足在 $k$次操作内，把所有 $A_i$变为 $k$的倍数。

题解：
设 $sum = \sum^n_{i=1} Ai$
因为我们的操作不会改变 $sum$但是一堆 $k$的倍数之和依旧是 $k$的倍数
那么答案一定是 $sum$的因数之一
那么考虑枚举 $sum$的因数，对答案进行check
设答案为 $x$
我们讲 $A_i$按照 $A_i$ $mod$ $x$从小到大排序
根据贪心，我们可以发现 $A_i$ $mod$ $x$比较小的是可以用减操作，比较大的用加操作。然后就可以头尾配对贪心取操作数了，最后判断操作数是否 $<=k$来判断解的合法性即可。

#include<bits/stdc++.h>
#define ll long long
#define pa pair<int,int>
using namespace std;
int n,k,A[504],sum;
int det[504];
bool check(int x){
    if(sum%x!=0)return 0;
    for(int i=1;i<=n;i++)det[i]=A[i]%x;
    sort(det+1,det+n+1);
    int l=1,r=n;
    ll opc=0;
    while(l<=r){
        if(det[l]%x==0){
            ++l;continue;
        }
        if(det[r]%x==0){
            --r;continue;
        }
        int delta=min(det[l]%x,x-(det[r]%x));
        det[l]-=delta;
        det[r]+=delta;
        opc+=delta;
    }
    return (opc<=k);
}
int main(){
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)scanf("%d",&A[i]);
    sum=0;
    for(int i=1;i<=n;i++)sum+=A[i];
    int ans=1;
    for(int k=sqrt(sum);k>=1;k--){
        if(check(k))ans=max(ans,k);
        if(check(sum/k))ans=max(ans,sum/k);
    }
    printf("%d\n",ans);
    return 0;
}