poj 2247 Humble Numbers

最新推荐文章于 2017-07-30 21:58:22 发布

转载最新推荐文章于 2017-07-30 21:58:22 发布 · 170 阅读

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原文链接：http://www.cnblogs.com/w0w0/archive/2011/11/22/2258285.html

本文介绍了一种算法，用于计算Humble Numbers序列中的特定元素。Humble Numbers是指仅包含2、3、5或7作为质因数的数。文章提供了一个C++程序示例，该程序能够高效地找到序列中的第n个Humble Number，并正确地输出带有适当序数后缀的结果。

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Humble Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7754		Accepted: 3666

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

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.

#include<iostream>  
using namespace std;  
int Min(int a ,int b)  
{  
return a<b?a:b;  
}  
int main()  
{    
int i,j,k,l;  
    i=j=k=l=1;
int data[5843];  
    memset(data,0,sizeof(data));  
    data[0]=1;  
for(int m=1;m<=5842;m++)  
    {  
int num1=data[i-1]*2;  
int num2=data[j-1]*3;  
int num3=data[k-1]*5;  
int num4=data[l-1]*7;  
int min=Min(Min(num1,num2),Min(num3,num4));  
        data[m]=min;  
if(num1==min) i++;  
if(num2==min) j++;  
if(num3==min) k++;  
if(num4==min) l++;  

    }  
int n;  
while(cin>>n,n)  
    {
if(n%10==1&&n%100!=11)   
            cout<<"The "<<n<<"st humble number is "<<data[n-1]<<'.'<<endl;  
else if(n%10==2&&n%100!=12)   
            cout<<"The "<<n<<"nd humble number is "<<data[n-1]<<'.'<<endl;  
else if(n%10==3&&n%100!=13)   
            cout<<"The "<<n<<"rd humble number is "<<data[n-1]<<'.'<<endl;  
else   
            cout<<"The "<<n<<"th humble number is "<<data[n-1]<<'.'<<endl;    
    }  
return 0;  
}

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/22/2258285.html