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最新推荐文章于 2019-10-25 21:45:02 发布

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最新推荐文章于 2019-10-25 21:45:02 发布

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分类专栏： C++语言 ACM 算法与数据结构

本文链接：https://blog.youkuaiyun.com/doufei_ccst/article/details/12111659

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本博客探讨了一个特定序列中第n个谦数的查找算法，该序列由仅包含2,3,5,7这四个质因数的整数组成。通过高效的算法实现，用户可以快速找到序列中的任意位置的谦数。

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Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13689 Accepted Submission(s): 5956

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

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.

代码如下：

#include <iostream>
#include <algorithm>
using namespace  std;
#define  N 5842
int values[N + 1] ;

inline int minValue(int n1, int n2, int n3, int n4)
{
	return min(min(n1, n2), min(n3, n4)) ;
}

int main(int argc, char ** argv)
{
	memset(values, 0, ( N + 1 ) * sizeof(int)) ;
	values[1] = 1 ;
	int p1, p2, p3, p4 ;
	p1 = p2 = p3 = p4 = 1 ;
	
	for (int i = 2; i <= N; ++ i)
	{
		values[i] = minValue(values[p1] * 2, values[p2] * 3, values[p3] * 5, values[p4] * 7) ;
		if (values[p1] * 2 == values[i]) p1 ++ ;
		if (values[p2] * 3 == values[i]) p2 ++ ;
		if (values[p3] * 5 == values[i]) p3 ++ ;
		if (values[p4] * 7 == values[i]) p4 ++ ;
	}

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