UVA 10081 - Tight Words （数论 dp）

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 differby more than 1.

Input is a sequence of lines, each line contains two integer numbersk and n, 1 <= n <= 100. For eachline of input, output the percentage of tight words of lengthn 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

题意：给定k，n。要求用0-k的数字组成长度为n的序列中，相邻两两差小于等于1的概率。

思路：dp，dp[i][j]表示组成长度为i，最后一个数字为j的符合条件的序列的概率。那么每次多一个数字的时候，只要加上dp[i][j - 1]，dp[i][j + 1],dp[i][j]，然后在除上多取一个数字的种数（k + 1）即可。

代码：

#include 
#include 

int k, n, i, j;
double dp[105][10];

int main() {
	while (~scanf("%d%d", &k, &n)) {
		memset(dp, 0, sizeof(dp));
		for (j = 0; j <= k; j ++)
			dp[1][j] = 1.0 / (k + 1);
		for (i = 2; i <= n; i ++) {
			for (j = 0; j <= k; j ++) {
				dp[i][j] += dp[i - 1][j];
				if (j) dp[i][j] += dp[i - 1][j - 1];
				if (j != k) dp[i][j] += dp[i - 1][j + 1];
				dp[i][j] /= (k + 1);
			}
		}
		double ans = 0;
		for (j = 0; j <= k; j ++)
			ans += dp[n][j];
		printf("%.5lf\n", ans * 100);
	}
	return 0;
}