Time limit: 3.000 seconds

Given is an alphabet {0, 1, ... , k}, 0
<= k <= 9 . We say that a word of length n over this alphabet is tightif
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

题意：给定两个数k,n。

用 {0,
1, ... , k}的数组成一个n个数的序列。假设这个序

列每两个相邻的数相差<=1,就记为是tight，求这样的序列占总序列的比率。

思路： dp[i][j]表示第i为数字是j的概率 。

即   dp[i][j] = 1/(k+1) *  (dp[i-1][j-1]+dp[i-1][j] + dp[i+1][j] );

注意下边界就OK了。

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

const int maxn=105;

double dp[maxn][15],t,ans;

int n,k;

void initial()

{

    memset(dp,0,sizeof(dp));

    t=1.0/(k+1),ans=0.0;

    for(int j=0; j<=k; j++)   dp[1][j]=t;

}

void solve()

{

    for(int i=2; i<=n; i++)

        for(int j=0; j<=k; j++)

        {

            dp[i][j]+=t*dp[i-1][j];

            if(j!=0)  dp[i][j]+=t*dp[i-1][j-1];

            if(j!=k)  dp[i][j]+=t*dp[i-1][j+1];

        }

    for(int j=0; j<=k; j++) ans+=dp[n][j];

    printf("%.5lf\n",ans*100);

}

int main()

{

    while(scanf("%d %d",&k,&n)!=EOF)

    {

        initial();

        solve();

    }

    return 0;

}