Program3_D

我现在做的是第三专题编号为1004的试题，具体内容如下所示：

Problem D

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 71   Accepted Submission(s) : 25

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.

简单题意：

求出数一个。该数就是因子只含2，3，5，7；

解题思路：

建立一个数组，里面存放2,3,5,7这四个数据，求第i个f[i],第i个f[i]必定等于前i-1个数中一个数与这个数组中一个数据的乘积，结果就出来了

编写代码：

#include<iostream>
using namespace std;
long long a[5842];

int main()
{
    int p[4]={2,3,5,7};
    int i,j,k;
    int n;
    a[1]=1;
    for(i=2; i<=5842; i++)
    {
        a[i]=2000000001;
        for(j=0;j<4;j++)
        {
            for(k=i-1; k>=1; k--)
            {
                if(a[k] * p[j] <= a[i-1])
                    break;
                if(a[k] * p[j] < a[i])
                   a[i] = a[k] * p[j];
            }
        }
    }
    while(cin>>n,n)
    {
        cout<<"The "<<n;
        if(n%10==1 && n%100!=11)
          cout<<"st ";
        else if(n%10==2 && n%100!=12)
          cout<<"nd ";
        else if(n%10==3 && n%100!=13)
          cout<<"rd ";
        else
          cout<<"th ";
        cout<<"humble number is "<<a[n]<<"."<<endl;

    }
    return 0;
}