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}
1 /* 2 State:Accpeted 3 Time:2013-03-07 22:02:23 4 */ 5 #include <iostream> 6 #include <cstring> 7 #include <string> 8 #include <cstdlib> 9 #include <cstdio> 10 #include <algorithm> 11 #include <cmath> 12 using namespace std; 13 const int maxn = 100; 14 double f[maxn + 1][maxn + 1] , ans[maxn + 1]; 15 int n; 16 int main(){ 17 freopen("poj1671.in","r",stdin); 18 freopen("poj1671.out","w",stdout); 19 for (int i = 1 ; i <= maxn; ++i){ 20 f[i][1] = 1; 21 ans[i] = 1; 22 } 23 for (int i = 1 ; i <= maxn; ++i) 24 for (int j = 2; j <= i; ++j){ 25 f[i][j] = f[i - 1][j - 1] + f[i - 1][j] * j; 26 ans[i] += f[i][j]; 27 } 28 while (1){ 29 scanf("%d",&n); 30 if (!n) break; 31 printf("%d %.0f\n", n , ans[n]); 32 } 33 fclose(stdin); fclose(stdout); 34 }
