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Humble Numbers

2024-07-04 12:31:34   228人阅读

                                                                              Humble Numbers

题目描述
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
输入
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.
输出
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.
样例输入
1
2
3
4
11
12
13
21
22
23
100
1000
5842
0
样例输出
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.


#include<iostream>
usingnamespace std;
longint x[5900];
intnum2=0,num3=0,num5=0,num7=0;
voidt()
{
    x[0]=1;
    for(inti=1;i<5900;i++)
    {
      intmin=x[num2]*2,b=x[num3]*3,c=x[num5]*5,d=x[num7]*7;
        if(min>b)
        {
            min=b;
        }
        if(min>c)
        {
            min=c;
        }
        if(min>d)
        {
            min=d;
        }
        x[i]=min;
        if(x[i]==x[num2]*2)
            num2++;
        if(x[i]==x[num3]*3)
            num3++;
        if(x[i]==x[num5]*5)
            num5++;
        if(x[i]==x[num7]*7)
            num7++;
        }
}
intmain()
{
    intn;
    t();
    while(cin>>n&&n!=0)
    {
        if(n%10==1&&n!=11)
        cout<<"The"<<" "<<n<<"st"<<" "<<"humble number is "<<x[n-1]<<"."<<endl;
        elseif(n%10==2&&n!=12)
            cout<<"The"<<" "<<n<<"nd"<<" "<<"humble number is "<<x[n-1]<<"."<<endl;
        elseif(n%10==3&&n!=13)
            cout<<"The"<<" "<<n<<"rd"<<" "<<"humble number is "<<x[n-1]<<"."<<endl;
        elsecout<<"The"<<" "<<n<<"th"<<" "<<"humble number is "<<x[n-1]<<"."<<endl;
 
    }
    return0;
}
此题思路同ugly number 一样，也就是要缩小穷举范围，

一直因子只能是2,3,5,7，所以分别用四个量来表示，通过将最小值放入数组中构造成功

另外需要考虑输出形式，在11，12,13处分别有小陷阱需注意


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