Humble Numbers（hdu1058）

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14047    Accepted Submission(s): 6108

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
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.
 
 
 
带大一的不容易啊！（感慨一下。）这道题目就是刚刚看的时候没看懂什么意思，
o(︶︿︶)o 唉！硬伤啊！
下面我来详细解释一下吧。。。。。
 
Humble Numbers
    如果一个数的质因子只有2、3、5或7，那么这个数被称为Humble Numbers（差数）。
将正整数正序排列（1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... ）
你会发现前20个Humble Numbers（差数）。
现在，你的任务是编写一个程序将第n个Humble Numbers（差数）找出并打印出来。  
输入规范：
输入的数有许多组，其中的n均满足1 <= n <= 5842。如果输入0，则表示终止。
输出规范：
对于每一组数，分列几行输出“The nth humble number is number.”，其中nth必须根据英语词法规范写
为“1st、2nd、3rd、4th、5th、6th、7th、8th、9th、10th... ”。
 
 
 
详见代码：
#include<stdio.h>
#define min(a,b) ((a)<(b)?(a):(b))//一定要打括号。。。
int num[5850];
int main()
{
    int n,p2,p3,p5,p7;
    p2=p3=p5=p7=1;
    int m=1;
    num[1]=1;
    while(m<=5842)
    {
        num[++m]=min(min(2*num[p2],3*num[p3]),min(5*num[p5],7*num[p7]));
        if(num[m]==2*num[p2])
            p2++;
        if(num[m]==3*num[p3])
            p3++;
        if(num[m]==5*num[p5])
            p5++;
        if(num[m]==7*num[p7])
            p7++;    
    }
    while(scanf("%d",&n),n)
    {
        printf("The %d",n);
        m=n/10%10;//十位不能为1，eg：11 12 13 111 112 113，，，
        if(n%10==1&&m!=1)
            printf("st ");
        else if(n%10==2&&m!=1)
            printf("nd ");
        else if(n%10==3&&m!=1)
            printf("rd ");
        else
            printf("th ");
        printf("humble number is %d.\n",num[n]);
    }
    return 0;
}

 

 我想想这句num[++m]=min(min(2*num[p2],3*num[p3]),min(5*num[p5],7*num[p7]));其实，现在想想用快排也可以古，哈哈！    
 
　　　　　　　　关键是题意！嗯嗯。
 
 
posted @ 2013-11-13 21:01  寻找&星空の孩子  阅读(943)  评论(0)  编辑  收藏  举报