UVA - 136 Ugly Numbers（丑数，STL优先队列+set）

Posted 2020-12-24 yinbiao

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>.

 

分析：

第一种方法：遍历每个数，判断是否为ugly，直到第1500个ugly为止（简单粗暴，没有效率可言，runtime 19s）

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
bool isugly(int n)
{
    while(n>1)
    {
        if(n%2==0)
            n/=2;
        else if(n%3==0)
            n/=3;
        else if(n%5==0)
            n/=5;
        else
            return 0;
    }
    return 1;
}
int main()
{
    int cnt=1;
    int num=2;
    while(cnt<1500)
    {
        if(isugly(num))
            cnt++;
        num++;
    }
    cout<<num-1<<endl;
    return 0;
}

第二种方法：利用STL优先队列从小到大生成各个ugly number。最小的丑数是1，而对于任意丑数x，2x，3x和5x也都是丑数。注意丑数判重。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<queue>
#include<set>
using namespace std;
typedef long long LL;
int f[3]={2,3,5};
int main()
{
    priority_queue<LL,vector<LL>,greater<LL> > qq;
    set<LL> s;
    qq.push(1);
    s.insert(1);
    for(int i=1;;i++)
    {
        LL x=qq.top();
        qq.pop();

        if(i==1500)
        {
            cout << "The 1500‘th ugly number is " << x << ".
";
            break;
        }
        for(int j=0;j<3;j++)
        {
            LL x2=x*f[j];
            if(!s.count(x2))
            {
                s.insert(x2);
                qq.push(x2);
            }
        }
    }
   // cout<<"The 1500‘th ugly number is 859963392."<<endl;
    return 0;
}