ZOJ1883(Tight Words)

Tight Words

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Given is an alphabet {0, 1, ... , k}, 0 <= k <= 9 . We say that a word of length n over this alphabet is tight if any two neighbour digits in the word do not differ by more than 1.

Input

Input is a sequence of lines, each line contains two integer numbers k and n, 1 <= n <= 100.

Output

For each line of input, output the percentage of tight words of length n over the alphabet {0, 1, ... , k} with 5 fractional digits.

Sample Input

4 1
2 5
3 5
8 7

Sample Output

100.00000
40.74074
17.38281
0.10130

/**//*
    dp[x][y] 表示以x结尾的长度为y的有多少种
        by Xredman
*/
#include <iostream>
#include <cmath>
using namespace std;

const int N = 105;

double dp[N][N];
int k, n;
int dir[3]= {-1, 0, 1};

double solve()
{
    int i, j;
    double ans = 0.0;

    for(i = 0; i <= k; i++)
        dp[i][1] = 1.0;

    for(i = 2; i <= n; i++)
    {
        for(j = 0; j <= k; j++)
        {    
            dp[j][i] = dp[j][i - 1];
            if(j - 1 >= 0)
                dp[j][i] += dp[j - 1][i - 1];
            if(j + 1 <= k)
                dp[j][i] += dp[j + 1][i - 1];
        }
    }

    for(i = 0; i <= k; i++)
        ans += dp[i][n];
    return ans;

}
int main()
{

    while(scanf("%d%d", &k, &n) != EOF)
    {
        printf("%.5lf\n", solve() / (pow((double)k + 1, (double) n)) * 100);
    }
    return 0;
}