Problem 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, 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

There is no input to this program.

Output

Output should consist of a single line as shown below, with ‘<number>’ replaced by the number computed.

SampleOutput

The 1500’th ugly number is <number>.

题目大意：

丑数是指不能被2，3，5以外的其他素数整除的数。把丑数从小到大排列起来

结果如下：

1,2,3,4,5,6,8,9,10,12,15,…

求第1500个丑数

思路：

一般暴力求解那肯定会T，所以我们可以借助优先队列将其进行优化，注意的是需要用long long 型，这大家应该也都知道多嘴了ヾ(=ﾟ･ﾟ=)ﾉ喵♪。

具体看代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
#define ll long long
const int maxx=;
int main()
{
    ll a,b,c;
    //优先队列
    priority_queue<ll,vector<ll>,greater<ll> >q;
    q.push();//初始化为 1;
    int num=;//计数;
    ll ans;
    while(num<maxx)
    {
        ans=q.top();
        q.pop();
        num++;
        if(num==)
        {
            printf("The 1500'th ugly number is %d.\n", ans);
            break;
        }
        //如果ans 为ugly ，则2*ans,3*ans,5*ans都是丑数;
        a=ans*;
        b=ans*;
        c=ans*;
         // a 如果与 b或c是同一数的因数关系，那么该数一定在b或c中出现过
        // 因为b或c比a大
        if((a%)&&(a%))
        {
            q.push(a);
        }
        if((b%))
        {
            q.push(b);
        }
        q.push(c);
    }
    return ;
}

还有一种操作就是下面这种o(ﾟДﾟ)っ啥！，前提是已经知道结果：

#include<iostream>
using namespace std;
int main()
{
    cout<<"The 1500'th ugly number is 859963392.\n";
}