Uva 10081 - Tight Words 解题报告（递推）

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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 12 53 58 7

Output for the sample input

100.0000040.7407417.382810.10130

解题报告：直接求概率。在第一个数字的时候，0到k每个数字的概率都是1/k，都符合条件。第二个数字时，除了0和k，其他数字x可以由x-1,x,x+1转移，概率为1/k*(p(x)+p(x-1)+p(x+1))，0和k个少一个。按照此方式转移n-1次，最终求出概率和即可。复杂度为n*k，大约1000。如果n很大，达到100W级别那种，可以使用矩阵+二分快速幂，用矩阵表示每次转移时概率的变化。复杂度为k^3*log n，这题的话就别了。本人代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;double may[111][11];void work(int n, int k){    if(k==0 || k==1)    {        puts("100.00000");        return;    }    for(int i=0;i<=k;i++)        may[1][i]=1.0/(k+1);    for(int i=2;i<=n;i++)    {        for(int j=1;j<k;j++)            may[i][j]=(may[i-1][j-1]+may[i-1][j]+may[i-1][j+1])/(k+1);        may[i][0]=(may[i-1][0]+may[i-1][1])/(k+1);        may[i][k]=(may[i-1][k]+may[i-1][k-1])/(k+1);    }    double ans=0;    for(int i=0;i<=k;i++)        ans+=may[n][i];    printf("%.5lf\n", ans*100);}int main(){    int k, n;    while(~scanf("%d%d", &k, &n))        work(n, k);}

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