Humble Numbers（dp）

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Humble Numbers

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Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 30 Accepted Submission(s) : 16

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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.

Source

University of Ulm Local Contest 1996
题目大意：
许多含有唯一的质因数是2,3,5或7被称为humble numbers。序列1，2，3，4，5，6，7，8，9，10，12，14，15，16，18，20，21，24，25，27，...示出了第一20humble numbers。写一个程序来查找和打印此序列中的第n个元素。
我们可以发现第二个数是2,3,5,7中最小的数，第三个数是2^2,3,5,7中最小的数，第四个数是2^2,2*3,5,7中最小的数。。。。。。。。。。。。。。。。。。。(每一次都是前面某一个humble numbers乘以2，乘以3，乘以7，乘以5中最小的数)
可以得出第n个数HN[n]=min(HN[f2]*2,HN[f3]*3,HN[f5]*5,HN[f7]*7);(状态转移方程感觉不是太好想)；

#include <stdio.h>#define N 5843int HN[5850],f2=1,f3=1,f5=1,f7=1;int min(int a,int b,int c,int d){return a>b?b>c?c>d?d:c:d>b?b:d:c<a?d<c?d:c:d<a?d:a;//求出最小值}void count(int HN[]){    int i;    HN[1]=1;    for(i=2; i<=N; i++)    {        HN[i]=min(HN[f2]*2,HN[f3]*3,HN[f5]*5,HN[f7]*7);//状态转移方程        if(HN[i]==HN[f2]*2) f2++;        if(HN[i]==HN[f3]*3) f3++;        if(HN[i]==HN[f5]*5) f5++;        if(HN[i]==HN[f7]*7) f7++;    }}int main(){    int t;    count(HN);    while(~scanf("%d",&t)&&t!=0)    {        if(t%10==1&&t%100!=11)            printf("The %dst humble number is %d.\n",t,HN[t]);        else if(t%10==2&&t%100!=12)            printf("The %dnd humble number is %d.\n",t,HN[t]);        else if(t%10==3&&t%100!=13)            printf("The %drd humble number is %d.\n",t,HN[t]);        else            printf("The %dth humble number is %d.\n",t,HN[t]);    }    return 0;}

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