HDU - 1058（Humble Numbers）动态规划

原创于 2019-01-28 23:56:03 发布 · 262 阅读

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本文介绍了一种使用动态规划算法来解决求第n个Humble Number的问题，Humble Number是指其质因数仅包含2、3、5或7的整数。通过维护四个指针并迭代更新动态规划数组，可以高效地找到给定位置的Humble Number。

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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.
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1058

这道题得意思是求第n个humble number，即它得因数只能为2，3，5，7。
这道题用到得动态规划，humble number肯定是前面得某一个humble number乘以2，3，5，7中的其中一个，然后取最小的。



#include "pch.h"

#include <iostream>

#include<algorithm>

using namespace std;

long long dp[6000];

 

int main()
{
    int n;
    int a, b, c, d;
         dp[1]= a = b = c = d = 1;
         for (int i = 2; i <=5842; i++)
         {
             dp[i]= min(dp[a] * 2, min(dp[b] * 3, min(dp[c] * 5, dp[d] * 7)));
             if (dp[i] == dp[a] *2) a++;
             if (dp[i] == dp[b] *3) b++;
             if (dp[i] == dp[c] *5) c++;
             if (dp[i] == dp[d] *7) d++;
         } 

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

}