UVa136 Ugly Numbers（优先队列priority_queue）

Ugly Numbers

题目

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

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ...

shows the first 11 ugly numbers. By convention, 1 is included.

Write a program to find and print the 1500'th ugly number.

Input and Output

There is no input to this program. Output should consist of a single line as shown below, with <number> replaced by the number computed.

Sample output

The 1500'th ugly number is <number>.

这道题用到了优先队列。STL的queue头文件中提供了优先队列，用"priority_queue<int> s"方式定义。

用push()和pop()进行元素的入队和出队操作，

top()取队首元素。

priority_queue<int,vector<int>,greater<int> >pq表示越小的整数优先级越大。

看书上的解题的时候，一开始不明白为什么要循环1500次，想了一下，原来是因为每次都取的是最小的元素啊，取上1500次，就是第1500个丑数了。

还有觉得比较神奇的地方是，丑数和丑数相乘，结果依然是丑数。

然后用到了typedef，感觉挺新鲜~

渣渣一枚，坚持！

#include<iostream>
#include<vector>
#include<queue>
#include<set>
using namespace std;
typedef long long LL;
const int coeff[3]={2,3,5};

int main(){
    priority_queue<LL,vector<LL>,greater<LL> >pq;//优先队列 
    set<LL> s;//集合 
    pq.push(1);
    s.insert(1);
    for(int i=1;;i++){
        LL x=pq.top();
        pq.pop();
        if(i==1500){
            cout<<"The 1500'th ugly number is "<<x<<".
";
            break;
        }
        for(int j=0;j<3;j++){
            LL x2=x*coeff[j];
            if(!s.count(x2)){
                s.insert(x2);
                pq.push(x2);
            }
        }
    }
    return 0;
}