Humble Numbers
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
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 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】找规律，你会发现这个数列的每个数都是2， 3， 5， 7的倍数，那么f[n]=f[i]*2 f[j]*3 f[k]*5 f[h]*7中最小的一个，然后对应的i,j,k,h加一
此外需注意第1到第一千的表示方法，11th, 12th, 13th,别的 *1st，*2nd，*3rd, 剩余的数为**th。
AC代码：
1 #include<stdio.h>
2 int f[6000];
3 int min(int a,int b)
4 {
5 if(a<b)return a;
6 else return b;
7 }
8 int main()
9 {
10 int i,a,b,c,d,n;
11 f[1]=1;
12 a=b=c=d=1;
13 for(i=2;i<=5842;i++)
14 {
15 f[i]=min(f[a]*2,min(f[b]*3,min(f[c]*5,f[d]*7)));
16 if(f[i]==f[a]*2)a++;
17 if(f[i]==f[b]*3)b++;
18 if(f[i]==f[c]*5)c++;
19 if(f[i]==f[d]*7)d++;
20 }
21 while(scanf("%d",&n),n)
22 {
23 if(n%10==1&&n % 100!=11)printf("The %dst humble number is %d.\n",n,f[n]);
24 else
25 if(n%10==2&&n % 100!=12)printf("The %dnd humble number is %d.\n",n,f[n]);
26 else
27 if(n%10==3&&n % 100!=13)printf("The %drd humble number is %d.\n",n,f[n]);
28 else
29 printf("The %dth humble number is %d.\n",n,f[n]);
30 }
31 return 0;
32 }
