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最新推荐文章于 2021-03-04 14:56:10 发布

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本文介绍了一个寻找第1500个丑数的C++程序实现。丑数是指只包含2、3、5这三个质因数的正整数。程序使用了优先队列和集合来避免重复计算，最终输出指定位置的丑数。

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水题。

/*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.
Sample Output
The 1500'th ugly number is <number>.*/
#include<iostream>
#include<vector>
#include<map>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
int co[]={2,3,5};
int main()
{
    priority_queue<ll,vector<ll>,greater<ll> > pq;
    set<ll>s;
    pq.push(1);
    for(int i=1;;i++)
    {
        ll x=pq.top();pq.pop();
        if(i==1500)
        {
            cout<<"The 1500'th ugly number is "<<x<<"."<<endl;
            break;
        }
        for(int j=0;j<3;j++)
        {
            ll x2=x*co[j];
            if(!s.count(x2)){s.insert(x2);pq.push(x2);}

        }

    }
    return 0;
}