hdu 1058 Humble Numbers

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

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.
 
 
思路：每个 humble number都只有2,3,5,7这些约数；所以，我们的思路是对于每个humble number从1开始，乘以2,3,5,7，然后从小到大排列就行了，存入数组中；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；此时，2已经乘过第一个 humble number数，故下一次2应乘以第二个 humble number数，即s[2]=2;此时，为3,4,5,7；以此类推。。
至于输出格式，只能悲催了。。。
 
</pre><pre name="code" class="cpp" style="font-size: 14px;">#include<stdio.h>#include<algorithm>using namespace std;#define N 5844int 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;}