HDU1058 Humble Numbers

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本文介绍了一种特殊的数列——谦逊数序列，并提供了一个高效的算法来找出序列中的特定元素。谦逊数是指仅由2、3、5或7作为质因数的数。通过预计算方法，文章展示了如何快速定位到第5842个谦逊数。

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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.

//水题了

#include<stdio.h>#define min(a,b) (a<b?a:b)#define min4(a,b,c,d) min(min(a,b),min(c,d))int a[5850];//存放丑数int main(){    int n=1;    int p2,p3,p5,p7;    p2=p3=p5=p7=1;//2,3,5,7的计数器    a[1]=1;    while(a[n]<2000000000)//从2开始递推计算，一共5842个丑数    {        a[++n] = min4(2*a[p2],3*a[p3],5*a[p5],7*a[p7]);//取最小值，相应的计数器加1        if(a[n]==2*a[p2]) p2++;        if(a[n]==3*a[p3]) p3++;        if(a[n]==5*a[p5]) p5++;        if(a[n]==7*a[p7]) p7++;    }    while(scanf("%d",&n) && n)    {        if(n%10 == 1&&n%100!=11)        printf("The %dst humble number is ",n);        else if(n%10 == 2&&n%100!=12)        printf("The %dnd humble number is ",n);        else if(n%10 == 3&&n%100!=13)        printf("The %drd humble number is ",n);        else        printf("The %dth humble number is ",n);        printf("%d.\n",a[n]);    }    return 0;}