Ugly Numbers(1.5.8)

Ugly Numbers(1.5.8)

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64

Description

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, ...
shows the first 10 ugly numbers. By convention, 1 is included.
Given the integer n,write a program to find and print the n'th ugly number.

Input

Each line of the input contains a postisive integer n (n <= 1500).Input is terminated by a line with n=0.

Output

For each line, output the n’th ugly number .:Don’t deal with the line with n=0.

Sample Input

1
2
9
0

Sample Output

1
2
10

题意：

把质因子仅仅有2,3,5的数称为Ugly数，求第k大的Ugly数。（1是第一个）

思路：

定义优先队列，数越小优先级越高。先出队列，然后*2 *3 *5进队列（注意消去反复的数）

#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;
#define ll __int64
struct node  
{
	ll num;
	friend bool operator < (const node &a,const node &b)//定义优先级
	{
		return a.num>b.num;
	}
};
ll bfs(ll n) 
{
	ll sum;
	node x,y,z;
	sum=0;
	priority_queue<node>que;
		x.num=1;z.num=0; 
		que.push(x);// 1进队列
	while(1)
	{
		x=que.top();
		que.pop();//优先级最高的先出队列
		if(x.num==z.num) continue;//与之前反复的数不计
		sum++;
		if(sum==n) return x.num;//找到第n个
		y.num=x.num*2;
		que.push(y);
		y.num=x.num*3;
		que.push(y);
		y.num=x.num*5;
		que.push(y);
		z.num=x.num;
	}
	return 0;
}
int main()
{
	ll num;
	while(scanf("%I64d",&num)&&num)
	{
		printf("%I64d
",bfs(num));
	}
	return 0;
}