uva 10081 - Tight Words（dp）

1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1022

2、题目大意：

给定两个数k n

用1-k的数组成一个n个数的序列，如果这个序列每两个相邻的数相差<=1,就记为是tight，求这种序列占总序列的比率

3、题目：

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

4、代码：

#include
double dp[110][12];//第i个位置填充j的概率
int main()
{
    int k,n;
    while(scanf("%d%d",&k,&n)!=EOF)
    {
        if(k<=1)
        printf("100.00000\n");
        else
        {
            for(int i=0;i<=k;i++)
            dp[1][i]=100.0/(k+1);
            for(int i=2;i<=n;i++)
            {
                for(int j=0;j<=k;j++)
                {
                    if(j==0)
                    {
                        dp[i][j]=1.0/(k+1)*(dp[i-1][j]+dp[i-1][j+1]);
                    }
                    else if(j==k)
                    {
                        dp[i][j]=1.0/(k+1)*(dp[i-1][j-1]+dp[i-1][j]);
                    }
                    else
                    {
                        dp[i][j]=1.0/(k+1)*(dp[i-1][j-1]+dp[i-1][j]+dp[i-1][j+1]);
                    }
                }
            }
            double ans=0;
            for(int i=0;i<=k;i++)
            {
                ans+=dp[n][i];
            }
            printf("%.5lf\n",ans);
        }
    }
    return 0;
}