HDU6576 Worker

Worker

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1120    Accepted Submission(s): 275

Problem Description

Avin meets a rich customer today. He will earn 1 million dollars if he can solve a hard problem. There are n warehouses and m workers. Any worker in the i-th warehouse can handle ai orders per day. The customer wonders whether there exists one worker assignment method satisfying that every warehouse handles the same number of orders every day. Note that each worker should be assigned to exactly one warehouse and no worker is lazy when working.

Input

The first line contains two integers n (1 ≤ n ≤ 1, 000), m (1 ≤ m ≤ 10^18). The second line contains n integers. The i-th integer ai (1 ≤ ai ≤ 10) represents one worker in the i-th warehouse can handle ai orders per day.

Output

If there is a feasible assignment method, print "Yes" in the first line. Then, in the second line, print n integers with the i-th integer representing the number of workers assigned to the i-th warehouse.
Otherwise, print "No" in one line. If there are multiple solutions, any solution is accepted.

Sample Input

2 6
1 2
2 5
1 2

Sample Output

Yes
4 2
No

Source

2019CCPC-江西省赛（重现赛）- 感谢南昌大学

做这题要注意求最小公倍数的写法，以及用long long整型，剩下的都是赤裸裸的硬算。

还要注意代码有一处求当前输入的所有数据的最小公倍数，具体代码上有标注。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){
    if(b==0) return a;
    return gcd(b,a%b);
}
ll lcm(ll a,ll b){
    return a*b/gcd(b,a%b);
}
int main(){
    ll n,m;
    ll a[1010];
    ll sum;
    while(scanf("%lld%lld",&n,&m)!=EOF){
        sum=0;
        ll x=1;
        for(ll i=0;i<n;i++){
            scanf("%lld",&a[i]);
            x=lcm(max(x,a[i]),min(x,a[i]));//这一处用的挺巧妙的。 
        }
        for(ll i=0;i<n;i++){
            sum+=(x/a[i]);
        }
        if(m%sum!=0) printf("No
");
        else {
        ll c=m/sum;
        printf("Yes
");
        printf("%lld",c*(x/a[0]));
        for(ll i=1;i<n;i++){
            printf(" %lld",c*(x/a[i]));
          }
          cout<<endl;
        }
    }
    return 0; 
}