hdu 1058 Humble Numbers(dp)

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

题意：求出前5842个质因数只在（2，3，5，7）中的数.

思路：感觉有点像递推。若一个数是Humble Numbers，那么它的2,3,5,7倍也是Humble Numbers。但是要求数组是按下标递增的，所以 f[n] = min(2*f[i], 3*f[j], 5*f[k], 7*f[l])，i, j,k,l在选择后进行右移。做了这道题还学会了英语计数，st,nd,rd,th终于搞明白了.....

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int f[5845];


int main()
{
	f[1] = 1;
	int i = 1,j = 1,k = 1,l = 1;

	for(int n = 2; n <= 5842; n++)
	{
		int ans = min( min(2*f[i], 3*f[j]),min(5*f[k], 7*f[l]) );
		if(ans == 2*f[i]) i++;
		if(ans == 3*f[j]) j++;
		if(ans == 5*f[k]) k++;
		if(ans == 7*f[l]) l++;
		f[n] = ans;
	}

	int m;
	while(~scanf("%d",&m)&& m)
	{
		if(m%10 == 1 && m%100 != 11)
			printf("The %dst humble number is %d.\n",m,f[m]);


		else if(m%10 == 2 && m%100 != 12)
			printf("The %dnd humble number is %d.\n",m,f[m]);


		else if(m%10 == 3 && m%100 != 13)
			printf("The %drd humble number is %d.\n",m,f[m]);

		else printf("The %dth humble number is %d.\n",m,f[m]);
	}
	return 0;
}