hdu 1058 Humble Numbers(dp)

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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;}

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