hdoj 1058 Humble Number

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主要的思想就是dp，转移方程不大明显，但是比较容易想。下一个丑数由之前的丑数乘{2，3，5，7}中最小且大于上一个数的数得到。先把所有数算出来然后打表输出。输出太恶心了，还考人英语。。

/*PROG: humble numberLANG: C++11 */#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <climits>#include <ctype.h>#include <queue>#include <stack>#include <vector>#include <utility>#include <deque>#include <set>#include <map>#include <iostream>#include <fstream>#include <algorithm>#include <time.h>using namespace std;#define mst(a,b) memset(a,b,sizeof(a))#define eps  10e-8typedef long long ll;const int N = 100010;const ll MOD = 1000000007;const int INF = 0x7fffffff;ll humble[6000];int prime [4]= {2,3,5,7};int main(){int n, i, j, k;    ll cur, minn;    int maxn = 5842;    humble[1] = 1;    humble[2] = 2;    for(i = 3; i <= maxn; i++){        minn = INF;        k = 1;        for(j = 0; j < 4; j++){            for(; k < i; k++){<span style="white-space:pre"></span>// instead of making k = 1 every time                cur = prime[j] * humble[i-k];   // because prime is in increasing order                  if(cur <= humble[i-1])                    break;                minn = min(minn, cur);            }        }        humble[i] = minn;    }    while(scanf("%d", &n)){        if(!n)             break;        cout << "The " << n;        if(n%10 ==1 && n%100 != 11)            cout << "st humble number is ";        else if(n%10 == 2 && n%100 != 12)            cout << "nd humble number is ";        else if(n%10 ==3 && n%100 != 13)            cout << "rd humble number is ";        else            cout << "th humble number is ";        cout << humble[n]  << "." << endl;    }    return 0;}

中间剪枝了一下，比网上流行的算法稍微快一点

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