SDAU 练习三 1004 谦逊（卑微）数字

                                                                                     Problem D

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. <br><br>Write a program to find and print the nth element in this sequence<br>

 

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.<br>

 

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.<br>

 

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.

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简单题意：

      

             一些唯一的主要因素是2、3、5或7叫做卑微的数量。序列1、2、3、4、5,6,7,8,9,10,12,14,15,16,18，20,21,24,25,27,…显示了前20卑微的数字。编写一个程序,找到并打印这个序列中的第n个元素。。。。。。

思路：

       找出2*t，3*t,5*t,7*t的最小值，进行对应数组的下标的自增，将所有的情况都列出来（打表法），然后按题意输入和输出即可。。。。。。

ACID：00768182

代码如下：

#include<iostream>using namespace std;int main(){    int t[10001];    int a=1,b=1,c=1,d=1,i;    t[1]=1;    for(i=2;i<5843;i++)    {        t[i]=min(t[a]*2,min(t[b]*3,min(t[c]*5,t[d]*7)));        if(t[i]==t[a]*2)            a++;        if(t[i]==t[b]*3)            b++;        if(t[i]==t[c]*5)            c++;        if(t[i]==t[d]*7)            d++;    }    int n;    while(cin>>n&&n!=0)    {        cout<<"The "<<n;        if(n%10==1&&n%100!=11)            cout<<"st humble number is ";        else if(n%10==2&&n%100!=12)            cout<<"nd humble number is ";        else if(n%10==3&&n%100!=13)            cout<<"rd humble number is ";        else            cout<<"th humble number is ";        cout<<t[n]<<"."<<endl;    }}