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 lengthn 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 andn, 1 <= n <= 100. For each line 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

4、代码：

#include<stdio.h>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;}