UVA 10081 Tight Words

UVA_10081

    我们可以用f[i][j]表示第i个字符为j时的tight words的比率，那么有f[i][j]=1.0/(K+1)*(f[i-1][j-1]+f[i-1][j]+f[i-1][j+1])，当然括号里面的三项不一定都存在，分情况讨论一下即可。

#include<stdio.h>
#include<string.h>
#define MAXK 15
#define MAXN 110
int N, K;
double f[MAXN][MAXK];
void solve()
{
    int i, j;
    double res = 0;
    for(i = 0; i <= K; i ++)
        f[1][i] = 100.0 / (K + 1);
    for(i = 2; i <= N; i ++)
    {
        f[i][0] = 1.0 / (K + 1) * (f[i - 1][0] + f[i - 1][1]);
        for(j = 1; j < K; j ++)
            f[i][j] = 1.0 / (K + 1) * (f[i - 1][j - 1] + f[i - 1][j] + f[i - 1][j + 1]);
        f[i][K] = 1.0 / (K + 1) * (f[i - 1][K - 1] + f[i - 1][K]);
    }
    for(i = 0; i <= K; i ++)
        res += f[N][i];
    printf("%.5lf\n", res);
}
int main()
{
    while(scanf("%d%d", &K, &N) == 2)
    {
        if(K <= 1)
            printf("100.00000\n");
        else
            solve();
    }
    return 0;
}