Uva 136-丑数 ugly number

原文链接： http://www.cnblogs.com/masking-timeflows/p/6670634.html

题目描述：

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>.

分析：

本题难点不多，直接使用队列完成。

优先队列priority_queque的详细用法，请戳（http://www.cnblogs.com/void/archive/2012/02/01/2335224.html）

代码：

 1 #include<iostream>
 2 #include<set>
 3 #include<queue>
 4 #include<vector>
 5 using namespace std;
 6 typedef long long ll;
 7 const int basic[3]={2,3,5};
 8 int main()
 9 {
10     priority_queue<ll,vector<ll>,greater<ll> >pq;
11     set<ll>q;
12     q.insert(1);
13     pq.push(1);
14     for(int i=1;;i++)
15     {
16         ll x=pq.top();
17         pq.pop();
18         if(i==1500)
19         {
20             cout<<"The 1500'th ugly number is "<<x<<".\n";
21             break;
22         }
23         for(int j=0;j<3;j++)
24         {
25             ll x2=basic[j]*x;
26             if(!q.count(x2))
27             {
28                 q.insert(x2);
29                 pq.push(x2);
30             }
31         }    
32     }
33     return 0;
34  } 

转载于:https://www.cnblogs.com/masking-timeflows/p/6670634.html