Sicily 11599. Tight words

11599. Tight words

    

    

Time Limit: 1sec    Memory Limit:256MB

Description

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. 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

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4 12 53 58 7

Sample Output

100.0000040.7407417.382810.10130

题意是用0到k这k+1个符号，组成一个n位的数，如果这个n位数相邻的位置上的数字差都不超过1，那么这个n位数就称为tight words，最后问tight words的个数占总数的百分比，因为总数有(k+1)^n之多，所以只能用double来存储了，我们用dp[i][j]表示i位数，最高位上是j的tight words的个数，初始化后，很容易就可以递推出来，因为个数比较大，long long也存不下，只能使用double了。

这道题目简单的数位dp，状态转移很简单，处理好边界就可以了。

// Problem#: 11599// Submission#: 3667449// 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 <cmath>#include <ctime>#include <cctype>#include <climits>#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <sstream>#include <iostream>#include <algorithm>double dp[111][111];int main(int argc, char *argv[]){    int k, n;    int N=111;     while (scanf("%d%d",&k,&n)!=EOF){        for (int i = 0; i < N; ++i){            for (int j = 0; j < N; ++j){                dp[i][j] = 1.0;            }        }        for (int i = 2; i<N; ++i){            for (int j = 0; j<= k;++j){                dp[i][j] = dp[i-1][j];                if (j > 0){                    dp[i][j] += dp[i-1][j-1];                }                if (j < k) {                    dp[i][j] += dp[i-1][j+1];                }            }        }        double ans = 0;        double base = pow(k + 1, n);        for (int i = 0; i <= k; ++i) {            ans += dp[n][i];        }        printf("%.5f\n", ans * 100.0 / base);    }    return 0;}