Humble Numbers-类DP

Humble Numbers
Source:hdu-1058

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
1
2
3
4
11
12
13
21
22
23
100
1000
5842
0

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.

源代码:

#include<iostream>#define MAX 2000000000using namespace std;int dp[6000];int min( int a , int b ){ return a <= b ? a : b ; }int main( ){    int i , n , r , num2 , num3 , num5 , num7;    dp[1] = i = 1;    num2 = num3 = num5 = num7 = 1;    do {        dp[++i] = min( min( 2*dp[num2] , 3*dp[num3] ) , min( 5*dp[num5] , 7*dp[num7] ) );        if( dp[i]==2*dp[num2] ) num2++;        if( dp[i]==3*dp[num3] ) num3++;        if( dp[i]==5*dp[num5] ) num5++;        if( dp[i]==7*dp[num7] ) num7++;    }while( i<=5842 );    while( cin>>n && n ){        r = n / 10 % 10;        cout<<"The "<<n;        if( n%10==1 && r!=1 ) cout<<"st ";        else if( n%10==2 && r!=1 ) cout<<"nd ";        else if( n%10==3 && r!=1 ) cout<<"rd ";        else cout<<"th ";        cout<<"humble number is ";        cout<<dp[n]<<"."<<endl;    }    return 0;}

代码分析:其实该算法是以O(n^2)的算法演变而来的,原算法是每求一个dp[n]的值时候,i遍历1->n-1,寻找dp[i]*2/3/5/7中大于dp[n-1]的最小值.由于范围只限制在5842,所以O(n^2)应该也能在1000ms内解决.
O(n)的算法如上,比较简洁,不过确实琢磨了很久才弄懂.优化的地方在于不用遍历每一个dp[1->n-1],而是有四个位置记录标志num2,num3,num5,num7来标志准备下一个更新的数值值可能是这几个位置的对应数乘2/3/5/7得来的,没有具体证明,不过草稿纸写一下应该就能大致得出该结论,本算法渣认为想到这个算法的那人确实不容易啊,点赞.