【Aizu2968】Non-trivial Common Divisor

题目链接

Non-trivial Common Divisor

题目描述

You are given a positive integer sequence (A) of length (N). You can remove any numbers from the sequence to make the sequence “friendly". A sequence is called friendly if there exists an integer (k) (>1) such that every number in the sequence is a multiple of (k). Since the empty sequence is friendly, it is guaranteed that you can make the initial sequence friendly.

You noticed that there may be multiple ways to make the sequence friendly. So you decide to maximize the sum of all the numbers in the friendly sequence. Please calculate the maximum sum of the all numbers in the friendly sequence which can be obtained from the initial sequence.

输入格式

The input consists of a single test case formatted as follows.
(N) (A_1) (vdots) (A_N)
The first line consists of a single integer (N) ((1 le N le 1000)). The (i+1)-st line consists of an integer (A_i) ((1 le A_i le 10^9) for (1 le i le N)).

输出格式

Print the maximum sum of all the numbers in the friendly sequence which can be obtained from the initial sequence.

样例输入输出

Input	Output
6 1 2 3 4 5 6	12
3 173 1733 111733	111733
4 1 1 1 1	0
10 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999	9999999990
1 999999999	999999999
10 28851 8842 9535 2311 25337 26467 12720 10561 8892 6435	56898

题解

题意：有(N)个数，从中选出(k)个数，使这些数有除(1)以外的公约数，求这(k)个数的和最大为多少。
我们考虑到一个数的因数是“对称”的。
那么，假设最大的数为(a)我们只要枚举公约数到(sqrt{a})。
但是我们会发现这么枚举，如果所有的数都是质数，那么答案的公约数肯定没有被枚举到，那么我们再枚举所有的数本身为公约数就好了。
ps：结尾不换行会PE
上代码：

#include<bits/stdc++.h>
using namespace std;
int n;
int a[1009];
bool p[40009];
int x[40009],lx;
long long ans;
bool cmp(int x,int y){return x<y;}
void pd(int x){
    if(x==1) return;
    long long sum=0;
    for(int j=1;j<=n;j++)
        if(a[j]%x==0) sum+=a[j];
    ans=max(ans,sum);
}
int main(){
    scanf("%d",&n);
    for(int j=1;j<=n;j++)
        scanf("%d",&a[j]);
    sort(a+1,a+n+1,cmp);
    for(int j=2;j*j<=a[n];j++){
        if(p[j]==0){
            x[++lx]=j;
            pd(j);
        }
        for(int i=1;i<=lx;i++){
            if(j*x[i]>40000) continue;
            p[j*x[i]]=1;
            if(j%x[i]==0) break;
        }
    }
    for(int j=1;j<=n;j++)
        pd(a[j]);
    printf("%lld
",ans);
    return 0;
}