【杭电1058】Humble Numbers-动态规划

最新推荐文章于 2025-07-28 00:00:26 发布

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最新推荐文章于 2025-07-28 00:00:26 发布

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分类专栏： ==数学and思维== 杭电acm~bestcoder 编程语言动态规划DP

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本文介绍了杭电1058题目，该题目涉及动态规划的运用。内容包括问题描述、输入输出格式、样例输入输出以及相关提示，旨在帮助读者理解和解决这一编程挑战。

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Humble NumbersCrawling in process...

Crawling failed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

Input

Output

Sample Input

Sample Output

Hint

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.

题意：找出丑数：将一个数拆开，因数只有2，3，5，7；第一个数为1；找出所有的丑数，单调递增数列。

题解：累乘

           1*2 ..             2*2..           3*2..

           1*3 ..             2*3..           3*3..

      1    1*5 ..         2   2*5 ..       3  3*5..      5..        7..

           1*7...             2*7 ..          3*7..

code:

#include<cstdio>
#include<algorithm>
using namespace std;
long long f[5843];
int main()
{
	int n;
	f[1]=1;
	int g[4]={2,3,5,7};
	for(int i=2;i<=5843;i++)
	{
		f[i]=2000000001;//赋予最大值
	for(int j=0;j<4;j++)
	{
	for(int k=i-1;k>=1;k--)
	{
		if(f[k]*g[j]<=f[i-1])//比前一个还小跳出
		break;
		if(f[k]*g[j]<f[i])//插入使单增
		f[i]=f[k]*g[j];
		}	
	}
		
	}
	while(scanf("%d",&n) && n){
        printf("The %d",n);
        if(n%10==1 && n%100!=11)
            printf("st ");
        else if(n%10==2 && n%100!=12)
            printf("nd ");
        else if(n%10==3 && n%100!=13)
            printf("rd ");
        else
            printf("th ");
        printf("humble number is %I64d.\n",f[n]);
    }
	return 0;
}