uva 10081 (Tight Words) (DP)

Problem B : Tight Words

From:UVA, 10081

Problem B: Tight words

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

Output for the sample input

100.00000
40.74074
17.38281
0.10130

#include 
#include 
#include 
using namespace std;

const int maxn = 110;
double dp[20][20][maxn];

void initial(){
	for(int i = 0;i < 20;i++){
		for(int j = 0;j < 20;j++){
			dp[i][j][1] = 1.0/(1.0*(1+i));
		}
	}
	for(int m = 1;m < 10;m++){
		for(int k = 2;k <= 100;k++){
			for(int i = 0;i <= m;i++){
				dp[m][i][k] = dp[m][i][k-1];
				if(i-1 >= 0 && i-1 <= m){
					dp[m][i][k] += dp[m][(i-1+m)%m][k-1];
				}
				if(i+1 >= 0 && i+1 <= m){
					dp[m][i][k] += dp[m][i+1][k-1];
				}
				dp[m][i][k] /= (1.0*(1+m));
			}
		}
	}
}


int main(){
	initial();
	int k , n;
	while(cin >> k >> n){
		if(k == 0 || k == 1 || n == 1){
			printf("100.00000\n");
		}else{
			double sum = 0.0;
			for(int i = 0;i <= k;i++){
				sum += dp[k][i][n];
			}
			printf("%.5lf\n" , sum*100.0);
		}
	}
	return 0;
}