最小公倍数（lcm与gcd）

http://acm.hdu.edu.cn/showproblem.php?pid=6576

题意：n个仓库m个工人，给出n个仓库的每天每个工人做ai件事，问如何分配m个工人到n个仓库，使得每一个仓库每天做相等数量的事。

 

解法：求出n个仓库的ai的lcm，lcm/a[i]为该仓库所需最小工人数，相加起来判断是否为m的约数。最后结果为最小工人数乘以m/sum。

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i ,j , n) for(int i = j ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 1000000007
#define PI acos(-1)
using namespace std;
typedef long long ll ;
ll a[1009];
ll gcd(ll a , ll b){
    if(a%b == 0) return b;
    else return gcd(b , a%b);
}
ll lcm(ll a , ll b){
    return a / gcd(a , b) * b;
}
 
int main()
{
    ll n , m;
    while(cin >> n >> m)
    {
        int sum = 0 ;
        int l = 1;
        rep(i , 0 , n){
            scanf("%lld" , &a[i]);
            l = lcm(l , a[i]);
        }
        rep(i , 0 , n){
            sum += l / a[i]  ;
        }
        if(m % sum == 0){
            cout << "Yes" << endl;
            int b = m / sum ;
            rep(i , 0 , n-1){
                cout << l/a[i]*b << " ";
            }
            cout << l / a[n-1] * b << endl;
        }else{
            cout << "No" << endl;
        }
    }
 
    return 0 ;
}