e5-7uva136 丑数

1、问题描述：
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>.

2、本题思路在于，对于任一丑数x，他的2,3,5倍也是丑数，我们已有最小丑数1，因此可以递推，我们由小到大选取丑数（选用优先队列），为了避免反复插入，我们选择用set来去除重复项，代码如下：

if(!s.count(x2)){
                s.insert(x2);
                q.push(x2);
            }

这是本题关键。

#include <iostream>
#include<queue>
#include<vector>
#include<set> 
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
typedef long long ll;
const int coef[3]={2,3,5};
int main(int argc, char** argv) {
    set<ll>s;
    priority_queue<ll,vector<ll>,greater<ll> >q;
    q.push(1);
    s.insert(1);
    for(int i=1;;i++){
        ll x=q.top();
        q.pop();
        if(i==1500){
            cout<<"The 1500'th ugly number is "<<x<<".\n";
            break;
        }
        for(int j=0;j<3;j++){
            ll x2=x*coef[j];
            if(!s.count(x2)){
                s.insert(x2);
                q.push(x2);
            }
        }
    }
    return 0;
}