poj1671

题意：一首n行诗，每一行有一种韵律，问这首诗总共可能有多少种韵律排列。

思路:dp or 递推

     f[i][j]表示前i个位置，用j种韵律的方案数

    f[i][j] = f[i-1][j]*j + f[i-1][j-1]

   {其实挺好理解的，如果没出现新的韵律，那么就用前面的j种任意一种韵律来填充，否则就新引入一种韵律}

/*
  State:Accpeted
  Time:2013-03-07 22:02:23
*/
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 100;
double f[maxn + 1][maxn + 1] , ans[maxn + 1];
int n;
int main(){
    freopen("poj1671.in","r",stdin);
    freopen("poj1671.out","w",stdout);
    for (int i = 1 ;  i <= maxn; ++i){
         f[i][1]  =  1;
         ans[i] = 1;
    }
    for (int i = 1 ; i <= maxn; ++i)
        for (int j = 2;  j <= i; ++j){
           f[i][j] = f[i - 1][j - 1] + f[i - 1][j] * j;
           ans[i] += f[i][j];
        }
    while (1){
           scanf("%d",&n);
           if (!n) break;
           printf("%d %.0f\n", n , ans[n]);
    }
    fclose(stdin); fclose(stdout);
}