UVa_136 - Ugly Numbers

Ugly Numbers

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 noinput to this program. Output should consist of a single line asshown below, with <number> replaced by the number computed.

Sample output

The 1500'th ugly number is <number>.

题意：

求出第1500个丑数（丑数是：一个数是丑数，改数只有2,3,5三个素数能整除，其余素数不能整除）

解题：

1. 设集合s是丑数的集合，如果x是s中的元素，那么2x,3x,5x也一定是丑数。

2. 基于上述准则，可以求出丑数的集合。用优先队列保存已经生成的丑数，每次取出队列中最小的丑数a,那么2a,3a,5a也是丑数，判定新生成的丑数是否重复即可

代码如下：

#include<iostream>
#include<cstdio>
#include<queue>
#include<set>
using namespace std;

typedef long long LL;
priority_queue<LL, vector<LL>, greater<LL> > pq;
set<LL> ugly;
int a[] = {2,3,5};

int main()
{
//freopen("572.txt","r",stdin);
int cnt = 0; //统计丑数的个数
pq.push(1); ugly.insert(1);
while(1)
{
LL min_ugly = pq.top(); pq.pop(); cnt++;
if(cnt==1500) { cout<<"The 1500'th ugly number is "<<min_ugly<<"."<<endl; break;}
if(cnt==1500) {cout<<min_ugly<<endl; break;}
for(int i=0;i<3;i++){
LL t = a[i]*min_ugly;
if(!ugly.count(t)) { pq.push(t); ugly.insert(t);} //新生成的丑数，不重复，就入队和插入集合
}
}
return 0;
}