UVA 10081 - Tight Words （数论 dp）

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Problem B: Tight words

Given is an alphabet {0, 1, ... , k}, 0 <= k <= 9 .We say that a word of lengthn over this alphabet istight 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 andn, 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 12 53 58 7

Output for the sample input

100.0000040.7407417.382810.10130

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 <stdio.h>#include <string.h>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;}