sgu 551. Preparing Problem 優先隊列或數學方法？

551. Preparing Problem

Time limit per test: 1 second(s)
Memory limit: 262144 kilobytes

input: standard
output: standard

It is not easy to prepare a problem for a programming contest. Petya and Vasya decided that problem "A+B" needs at least n distinct solutions to be written. It doesn't matter how many solutions each of them will write, they need to write at least n solutions in total. We know that Petya needs t1 units of time to write a solution, and Vasya needs t2 units of time. They start to work simultaneously at time 0. Thus, for example, Petya finishes writing his first solution at time t1, his second solution at 2 · t1 and so on.

Petya and Vasya are working by the same algorithm. Each time Petya (Vasya) finishes writing a solution, he checks on how many solutions have already been written up to the current time moment t. Ready solutions are the solutions that have been fully written by this time. The solutions that were fully finished exactly at time t are also considered ready. If the number of such solutions is strictly less than n, then Petya (Vasya) starts writing the next solution. If a member of the jury began working on a problem, he doesn't stop working under any circumstances, and he will surely finish it.

Petya and Vasya realize that if they act on this algorithm, they will not necessarily write exactly n solutions in total. Maybe they'll write more solutions.

Considering that Petya and Vasya work non-stop, find, how many solutions they wrote in total and the moment when the latest solution was finished. The latest solution is one which was finished last.

Input

The only input line contains three integers n, t1 and t2 (1 ≤ n, t1, t2 ≤ 5000).

Output

Print two integers — m and f, where m is the number of written solutions, and f is the moment when the last solution was finished.

Example(s)

sample input	sample output
5 2 3	5 6

sample input	sample output
5 2 4	6 8

sample input	sample output
3 30 50	4 100

模擬生產過程優先隊列中取出用時少的，處理數據並決定是否再次加入隊列。

#include <bits/stdc++.h>

using namespace std;

struct node{
    int t;
    int s;
};

bool operator<(node x,node y){
    return x.t>y.t;
}

priority_queue<node> q;

int a[2];
void solve(){
    int n;
    scanf("%d%d%d",&n,&a[0],&a[1]);
    int ans=0;
    int tt=0;
    node x;
    x.s=0;
    x.t=a[0];
    q.push(x);
    x.s=1;
    x.t=a[1];
    q.push(x);
    while(!q.empty()){
        node temp=q.top();
        q.pop();
        ans++;
        if(q.top().t==temp.t&&(ans+1)==n){
            ans=n;
            tt=temp.t;
            break;
        }
        tt=temp.t;
        if(ans<n){
            temp.t+=a[temp.s];
            q.push(temp);
        }
    }
    printf("%d %d",ans,tt);
}

int main(){
    solve();
    return 0;
}