2016 Al-Baath University Training Camp Contest-1 E

ACM-SCPC-2017团队优化算法

最新推荐文章于 2020-05-25 16:17:09 发布

转载最新推荐文章于 2020-05-25 16:17:09 发布 · 213 阅读

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原文链接：http://www.cnblogs.com/yinghualuowu/p/6044085.html

本文介绍了一种针对ACM-SCPC-2017比赛的团队优化算法，旨在通过调整参赛者组合，确保每个团队人数相等，同时来自同一大学，最终达到减少参赛团队总数的目标。算法核心在于求取各大学参赛者人数的最大公约数，以确定最佳的团队规模。

Description

ACM-SCPC-2017 is approaching every university is trying to do its best in order to be the Champion, there are n universities, the ith of them has exactly ai contestants. No one knows what is the official number of contestants in each team yet, this year organizers are planing to minimize number of teams participating in the contest but they still want every contestant to participate, so they should select a number k so that each team will consist of exactly k contestant from the same university. Every contestant will belong to one team exactly, and number of these teams is as minimum as possible. Your job is to help the organizers in their task by writing a program that calculates two integers k, the number of contestants in each team, and m, the minimum number of teams will participate. 1 ≤ n ≤ 1000, 1 ≤ ai ≤ 106

Input

First line of input contains an integer T denotes number of test cases. Each test case contains two lines: the first line contains one integern denotes number of universities, while the second line contains n space-separated integers the ith of them denotes number of contestants from the ith university.

Output

For each test case, print one line containing two integers: the first one is k, the size of each team, and the second one is m, the minimum number of teams from all universities according to the previous conditions.

Example

input

output

5 6
2 15
题意：有n支队，每个队都有ai的队员，现在重新分配人数，每个队人数相同，且只有相同学校的才能组队，要求队伍数量最少
解法：相同学校的才能组队，也就是说每个队的人数必须都能整除ai，那么应该是求ai的gcd，队伍数量就是拿人数除以gcd

#include<bits/stdc++.h>
using namespace std;
int a[100000];
int main()
{
    int n,m;
    int t;
    cin>>t;
    while(t--)
    {
        int num;
        int sum=0;
        cin>>n;
        cin>>a[1];
        num=a[1];
        sum+=a[1];
        for(int i=2;i<=n;i++)
        {
            //int num;
            cin>>a[i];
            sum+=a[i];
            num=__gcd(a[i],num);
        }
        cout<<num<<" "<<sum/num<<endl;
    }
    return 0;

转载于:https://www.cnblogs.com/yinghualuowu/p/6044085.html