hdu 1058 Humble Numbers

最新推荐文章于 2019-03-09 21:48:15 发布

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本文介绍了一种高效算法来寻找Humble Numbers，即仅包含2、3、5、7为质因数的数。通过逐步扩展这些数并记录下第n个这样的数，解决了特定的算法问题，并给出了完整的C++实现代码。

Humble Numbers

Problem Description

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers.

Write a program to find and print the nth element in this sequence

Input

The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line saying "The nth humble number is number.". Depending on the value of n, the correct suffix "st", "nd", "rd", or "th" for the ordinal number nth has to be used like it is shown in the sample output.

Sample Input

1 2 3 4 11 12 13 21 22 23 100 1000 5842 0

Sample Output

The 1st humble number is 1.
The 2nd humble number is 2.
The 3rd humble number is 3.
The 4th humble number is 4.
The 11th humble number is 12.
The 12th humble number is 14.
The 13th humble number is 15.
The 21st humble number is 28.
The 22nd humble number is 30.
The 23rd humble number is 32.
The 100th humble number is 450.
The 1000th humble number is 385875.
The 5842nd humble number is 2000000000.

思路：每个 humble number都只有2,3,5,7这些约数，每个humble number也只能由这些因子进行扩展，每次取出当前可以扩展到的最小数字，相应的因子向后挪一个位置。例如：2*1=2；3*1=3；5*1=5；7*1=7； 1是第一个数，然后从上面选出最小的数2；

#include<stdio.h>
#include<algorithm>
using namespace std;
#define N 5844
int s[N]= {0,1};
void inti()
{
    int a,b,c,d,i;
    a=b=c=d=1;
    for(i=2; i<N; i++)
    {
        s[i]=min(s[a]*2,min(s[b]*3,min(s[c]*5,s[d]*7)));    // 找出这四个中最小的 humble number数
        if(s[i]==s[a]*2)
            a++;
        if(s[i]==s[b]*3)
            b++;
        if(s[i]==s[c]*5)
            c++;
        if(s[i]==s[d]*7)
            d++;
    }
}
int main()
{
    int n,i;
    inti();
    while(scanf("%d",&n),n)
    {
        printf("The %d",n);
        int ten=n/10%10;
        if(n%10 == 1 && ten != 1)
            printf("st");
        else if(n%10 == 2 && ten != 1)
            printf("nd");
        else if(n%10 == 3 && ten != 1)
            printf("rd");
        else
            printf("th");
        printf(" humble number is %d.\n",s[n]);
    }
    return 0;
}