UVA 136 Ugly Number（优先队列）

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>.

思路：

题目求第1500个丑数（因数只有2，3，5的数，包括1）。借用STL的优先队列，假设队列中有若干个丑数（初始值为1），每次取出队列中最小的丑数，判断其是否是第1500个。若不是，则用其乘以2，3，5，再入队（保证不重复，借用set.count来判断）。依此类推，当循环到i为1500的时候取出的丑数就是第1500个（因为每次取出的都是对立中最小的）。

#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <queue>
#include <vector>
#define ll long long

using namespace std;

int main()
{
priority_queue<ll,vector<ll>,greater<ll> > q;
set<ll> s;
int a[3]={2,3,5};
q.push(1);
s.insert(1);

for(int i=1;;i++)
{
ll t=q.top();
q.pop();
if(i==1500)
{
cout<<"The 1500'th ugly number is "<<t<<"."<<endl;
break;
}
for(int j=0;j<3;j++)
{
if(!s.count(a[j]*t))
{
q.push(a[j]*t);
s.insert(a[j]*t);
}
}
}
return 0;
}