poj1338(Ugly Numbers)

weixin_34200628

于 2018-04-02 20:13:00 发布

阅读量82

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/ACRykl/p/8696839.html

本文介绍了一种通过预计算的方式找到第N个丑数的方法。丑数定义为仅包含质因数2、3和5的正整数。文章提供了一个C++实现的例子，并详细解释了如何生成丑数序列。

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, ...
shows the first 10 ugly numbers. By convention, 1 is included.
Given the integer n,write a program to find and print the n'th ugly number.

Input

Each line of the input contains a postisive integer n (n <= 1500).Input is terminated by a line with n=0.

Output

For each line, output the n’th ugly number .:Don’t deal with the line with n=0.

Sample Input

Sample Output

1
2
10

分析：打表，好笨的方法=_=.

#include<cstdio>
#include<algorithm>
using namespace std;
long long a[30000],cnt=0;
long long Pow(int x,int k)
{
    if(k==0) return 1;
    if(k==1) return x;
    long long y=Pow(x,k/2);
    y*=y;
    if(k%2) y*=x;
    return y;
}
void f()
{
    long long N=(long long)17592186044416;
    for(int i=0;i<=44;i++)
    {
        long long ans2=Pow(2,i);
        for(int j=0;j<30;j++)
        {
            long long ans3=Pow(3,j);
            for(int k=0;k<20;k++)
            {
                long long ans5=Pow(5,k);
                if(ans5<=N/ans2/ans3) 
                    a[cnt++]=ans2*ans3*ans5;
            }
        }
    }
    sort(a,a+cnt);
}
int main()
{
    f();
    int n;
    while(scanf("%d",&n)&&n)
        printf("%d\n",a[n-1]);
    return 0;
}