POJ 2537 Tight Words 继续DP之路

最新推荐文章于 2017-12-14 20:26:18 发布

iteye_17312

最新推荐文章于 2017-12-14 20:26:18 发布

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文章标签：数据结构与算法

Tight words

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2079		Accepted: 1041

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.

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

Sample Output

100.00000
40.74074
17.38281
0.10130
题意就是给你一个从0到k的字母表，让你输出长度为n的由前面给你的字母表组成的符合Tight Words的字符串占左右字符串的百分比

动态规划的子结构划分还是有点不太清楚，看到题目，不知道从哪里下手，DP还得多练，多体会，DP无处不在，DP才是王道

Memory: 400K		Time: 0MS
Language: GCC		Result: Accepted

#include<stdio.h>

double opt[10][101];//opt[i][j]表示已i结尾的长度为j的tight字符串个数

int main()
{
    int k,n;
    int i,j;
    double sum;
	freopen("input","r",stdin);
    while(scanf("%d %d",&k,&n)!=EOF)
    {
        memset(opt,0,sizeof(opt));
        sum=0;
        for(i=1;i<=k+1;i++) opt[i][1]=1;//所有长度为1的字符串本身就是tight字符串

        for(j=2;j<=n;j++)//opt[i][j]=opt[i-1][j-1]+opt[i][j-1]+opt[i+1][j-1] 状态转移方程
        {
            for(i=1;i<=k+1;i++)
            {
                opt[i][j]=opt[i-1][j-1]+opt[i][j-1]+opt[i+1][j-1];
            }
        }
        for(i=0;i<=k+1;i++)//计算字符长度为n的字符串个数
        {
            sum+=opt[i][n];
        }
        printf("%.5f/n",sum/pow(k+1,n)*100);
    }
    return 0;
}