Humble Numbers找规律

Posted 2020-08-19 唐唐123

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 

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

1
2
3
4
11
12
13
21
22
23
100
1000
5842
0

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 }