SPOJ BTCODE_H

dp[i][j]前一维是串的个数，后一维是串的长度

下面代码是队友写的，感觉好神

#include<cstdio>
#include<cstring>
#include<algorithm>
const int MAX = 300, MAXN = MAX+5;
int T, N, L;
double c[MAXN][MAXN], d[MAXN][MAXN];
using namespace std;
int main()
{
	c[0][0] = 1;
	for (int i = 1; i <= MAX; i++)
		c[i][0] = c[i-1][0]/2;
	for (int i = 1; i <= MAX; i++)
		for (int j = 1; j <= i; j++)
		{
			c[i][j] = (c[i-1][j]+c[i-1][j-1])/2;
//			printf("c %d %d = %f\n", i, j, c[i][j]);
		}
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d", &N, &L);
		for (int i = 1; i <= N; i++)
			d[i][0] = 1;
		for (int i = 1; i <= N; i++)
			for (int j = 1; j <= L; j++)
			{
				d[i][j] = 0;
				for (int k = 0; k <= i; k++)
				{
					d[i][j] += (1+d[k][j-1]+d[i-k][j-1])*c[i][k];
				}
			}
		printf("%.2lf\n", d[N][L]);
	}
	return 0;
}