hdoj 1058 Humble Numbers 【思维】

最新推荐文章于 2017-11-29 12:53:57 发布

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最新推荐文章于 2017-11-29 12:53:57 发布

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本文介绍了一种特殊的数——谦逊数（Humble Numbers），这些数仅包含2、3、5、7作为其质因数，并提供了寻找第n个谦逊数的算法实现，通过动态更新最小值标记的方式高效地计算出序列中的每一个数。

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21949 Accepted Submission(s): 9597

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.

思路：这是一道寻找丑数的问题。

符合要求的丑数只含有2、3、5、7的质因子求前5842个丑数。下一个最小的丑数一定在它前面的每个元素分别与2、3、5、7的乘积的结果中产生，只需在f[a]*2, f[b]*3, f[c]*5, f[d]*7 四个数中找到最小的那个，若f[a]*2被选中，则下轮应由f[a+1]*2代替，但是四个数中可能出现相同的数值，比如f[a]=3 f[b]=2时有f[a]*2==f[b]*3==6，故用bo[4]为最小值的标记，来判定最小的那个数。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int f[7000];
int bo[6];//判断最小值 
int Fund(int a1,int b1,int c1,int d1)
{
	int minn=min(min(a1,b1),min(c1,d1));
	for(int i=0;i<4;i++)
	{
		bo[i]=0;
	}
	if(minn==a1)
	bo[0]=1;
	if(minn==b1)
	bo[1]=1;
	if(minn==c1)
	bo[2]=1;
	if(minn==d1)
	bo[3]=1;
	return minn;
}
void init()
{
	int a,b,c,d,i;
	a=b=c=d=1;
	f[1]=1;
	for(i=2;i<=5842;i++)
	{
		f[i]=Fund(f[a]*2,f[b]*3,f[c]*5,f[d]*7);
		if(bo[0])
		a++;
		if(bo[1])
		b++;
		if(bo[2])
		c++;
		if(bo[3])
		d++;
	}
}
int main()
{
	int n;
	init();
	while(scanf("%d",&n)!=EOF&&n)
	{
		if(n%10==1&&n%100!=11)
		printf("The %dst humble number is %d.\n",n,f[n]);
		else if(n%10==2&&n%100!=12)
		printf("The %dnd humble number is %d.\n",n,f[n]);
	    else if(n%10==3&&n%100!=13)
		printf("The %drd humble number is %d.\n",n,f[n]);
		else
		printf("The %dth humble number is %d.\n",n,f[n]);
	}
	return 0;
}