Sicily 11599. Tight words

11599. Tight words

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

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

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

Problem Source

2014年每周一赛第八场

// Problem#: 11599
// Submission#: 3525803
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include 
#include 

double dp[10][101];
double ans[10][101];
    
int main() {
    
    double sum = 0;
    
    for (int i = 0; i <= 9; i++) {
        
        for (int k = 0; k <= i; k++) dp[k][1] = ans[k][1] = 1;
        for (int j = 2; j <= 100; j++) {
            dp[0][j] = dp[0][j - 1] + (i > 0 ? dp[1][j - 1] : 0);
            for (int k = 1; k < i; k++) 
                dp[k][j] = dp[k - 1][j - 1] + dp[k][j - 1] + dp[k + 1][j - 1];  // recursion formula
            dp[i][j] = (i > 0 ? dp[i - 1][j - 1] : 0) + dp[i][j - 1];
            for (int k = 0; k <= i; k++) sum += dp[k][j];
            ans[i][j] = sum / pow(1.0 * i + 1, j);
            sum = 0;
            if (ans[i][j] < 0.0000001) break;  // ans[i][j] will decrease when j increase. If the ans[i][j] is small enough to ignore, it's unnecessary to calculate anymore
        }
    }

    int k, n;
    while (scanf("%d%d", &k, &n) != EOF) printf("%.5f\n", ans[k][n] * 100);

    return 0;
}