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";

}

巴特西

Ugly Numbers UVA - 136（优先队列+vector）

Problem Description

Input

Output

SampleOutput

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