Humble Numbers(HDU-1058)

最新推荐文章于 2019-05-19 10:24:49 发布

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最新推荐文章于 2019-05-19 10:24:49 发布

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本文介绍了一种特殊的数列——谦逊数序列，并提供了一个程序来找出该序列中的特定元素。谦逊数是指仅由2、3、5或7作为质因数的整数。文章中的代码实现了生成这些数并按顺序排列的功能。

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

1234111213212223100100058420

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.

代码如下：

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
long long n[5842];
int main()
{
    long long i,j,k,l,m,t,c=0;
    for(i=1;c<=5842&&i<=2000000000;i=i*2)
        for(j=1;c<=5842&&i*j<=2000000000;j=j*3)
            for(k=1;c<=5842&&i*j*k<=2000000000;k=k*5)
               for(l=1;c<=5842&&i*j*k*l<=2000000000;l=l*7)
               {
                   t=i*j*k*l;
                    n[c++]=t;
               }
    sort(n,n+5842);
    while(~scanf("%lld",&m)&&m!=0)
    {
        if(m%10==1&&m%100!=11)
        printf("The %lldst humble number is %lld.\n",m,n[m-1]);
        else if(m%10==2&&m%100!=12)
        printf("The %lldnd humble number is %lld.\n",m,n[m-1]);
        else if(m%10==3&&m%100!=13)
        printf("The %lldrd humble number is %lld.\n",m,n[m-1]);
        else
        printf("The %lldth humble number is %lld.\n",m,n[m-1]);
    }
    return 0;
}