Hdu 1058 Humble Numbers

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最新推荐文章于 2024-08-25 08:45:00 发布

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分类专栏： ACM算法题目

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本文介绍了一种算法，用于找出特定序列中的谦逊数（仅包含2、3、5、7为质因数的数）。通过高效的数据结构和算法设计，程序能够快速找到序列中的第n个元素，并正确地输出带有适当序数后缀的结果。

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Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26020 Accepted Submission(s): 11474

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

100

1000

5842

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.

#include <iostream>  
#include <stdio.h>  using namespace std;  
int f[5843],n;  
int i,j,k,l;  
  
int min(int a,int b,int c,int d)  
{  
    int min=a;  
    if(b<min) min=b;  
    if(c<min) min=c;  
    if(d<min) min=d;  
  
    if(a==min) i++;  
    if(b==min) j++;  
    if(c==min) k++;  
    if(d==min) l++;  
  
    return min;  
}  
  
int main()  
{  
    i=j=k=l=1;  
    f[1]=1;  
    for(int t=2;t<=5842;t++)  
    {  
        f[t]=min(2*f[i],3*f[j],5*f[k],7*f[l]);  
    }  
    while(scanf("%d",&n)&&n!=0)  
    {  
        if(n%10==1&&n%100!=11)  
            printf("The %dst humble number is %d.\n",n,f[n]);  
        else if(n%10==2&&n%100!=12)  
            printf("The %dnd humble number is %d.\n",n,f[n]);  
        else if(n%10==3&&n%100!=13)  
            printf("The %drd humble number is %d.\n",n,f[n]);  
        else  
            printf("The %dth humble number is %d.\n",n,f[n]);  
    }  
  
    return 1;  
}