Uva 10081 Tight words （概率DP）

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;
}