nyoj1032——Save Princess——————【set应用】-优快云博客

本文介绍了一道趣味算法题目——“救公主”，该问题要求找出第n个仅包含质因数2、3或5的美丽数。通过使用优先队列和集合实现了一个高效的算法解决方案，确保了数字的独特性和正确顺序。

Save Princess

时间限制：1000 ms | 内存限制：65535 KB

难度：2

描述

Yesterday, the princess was kidnapped by a devil. The prince has to rescue our pretty princess.

"OK, if you want to save the beautiful princess, you must answer my questions correctly."the devil says.

"No problem!".

"I’ll ask you t questions. For each question, I’ll tell you an integer n, you must tell me the i th beatuiful number. If your answer is wrong, the princess and you will all die".

"But what is the characteristic of the beautiful number?" Pince asks.

"Beautiful numbers are numbers whose only prime factors are 2, 3 or 5. The sequence

1, 2, 3, 4, 5, 6, 8, 9, 10, ... shows the first 9 beautiful numbers.

By convention, 1 is included. "

Can you help the prince to save the princess?

输入

The input for each case is an integer n（1≤n≤5000) and it is terminated by a negative integer.

输出

For each test case, you should print an integer which represents the i th beautiful number.

样例输入

2
3
-1

样例输出

2
3


解题思路：用优先队列来存放美丽数，每次从队列中拿出最小的美丽数，由它扩展出三个美丽数，用set的去重功能来判断是否放入set和优先队列。依次从优先队列中出来的就是从小到大的美丽数。

#include<stdio.h>
#include<string.h>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long LL;
const int f[4]={2,3,5};
LL a[5500];
void prin(int n){

    priority_queue<LL,vector<LL>,greater<LL> >pq;
    set<LL>s;
    pq.push(1);
    s.insert(1);
    for(int i=1;;i++){

        LL tm=pq.top();
        a[i]=tm;
        pq.pop();
        if(i>n)
            break;
        for(int i=0;i<3;i++){

            LL x=tm*f[i];
            if(!s.count(x)){//返回元素x的个数

                s.insert(x);
                pq.push(x);
            }
        }
    }

}
int main(){

    int n;
    prin(5010);
    while( scanf("%d",&n)!=EOF&&n>0){

        printf("%lld\n",a[n]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/chengsheng/p/4385397.html