Rating

const double eps = 1e-10;
double a[22 * 22][22 * 22], x[22 * 22]; //方程的左边的矩阵和等式右边的值，求解之后x存的就是结果
int equ, var;                            //方程数和未知数个数
int Gauss()
{
    int i, j, k, col, max_r;
    for (k = 0, col = 0; k < equ && col < var; k++, col++)
    {
        max_r = k;
        for (i = k + 1; i < equ; i++)
            if (fabs(a[i][col]) > fabs(a[max_r][col]))
            max_r = i;
        if (fabs(a[max_r][col]) < eps) return 0;
        if (k != max_r)
        {
            for (j = col; j < var; j++)
                swap(a[k][j], a[max_r][j]);
            swap(x[k], x[max_r]);
        }
        x[k] /= a[k][col];
        for (j = col + 1; j < var; j++) a[k][j] /= a[k][col];
        a[k][col] = 1;
        for (i = 0; i < equ; i++)
            if (i != k)
            {
                x[i] -= x[k] * a[i][k];
                for (j = col + 1; j < var; j++) a[i][j] -= a[k][j] * a[i][col];
                a[i][col] = 0;
            }
    }
    return 1;
}

double P;
int s[22 * 22][22 * 22];
void build()
{
    CLR(a, 0); CLR(x, 0);
    FE(i, 0, 20) FE(j, i, 20)
    {
        int cur = s[i][j];
        if (~cur)
        {
            a[cur][cur] = 1;
            if (i == 20 || j == 20)
                x[cur] = 0;
            else
            {
                int tx = min(i + 1, 20), ty = j;
                if (tx > ty) swap(tx, ty);
                int nxt = s[tx][ty];
                a[cur][nxt] -= P;

                tx = max(i - 2, 0); ty = j;
                if (tx > ty) swap(tx, ty);
                nxt = s[tx][ty];
                a[cur][nxt] -= 1 - P;

                x[cur] = 1;
            }
        }
    }
}
void bfs()
{
    CLR(s, -1);
    queue<int> qx, qy;
    qx.push(0); qy.push(0);
    int cnt = 0;
    s[0][0] = cnt++;
    while (!qx.empty())
    {
        int x = qx.front(); qx.pop();
        int y = qy.front(); qy.pop();

        int tx = min(x + 1, 20), ty = y;
        if (tx > ty) swap(tx, ty);
        if (!~s[tx][ty])
        {
            s[tx][ty] = cnt++;
            qx.push(tx); qy.push(ty);
        }

        tx = max(x - 2, 0), ty = y;
        if (!~s[tx][ty])
        {
            s[tx][ty] = cnt++;
            qx.push(tx); qy.push(ty);
        }
    }
    equ = var = cnt;
}

int main()
{
    bfs();
    while (cin >> P)
    {
        build();
        Gauss();
        printf("%.6lf\n", x[s[0][0]]);
    }
    return 0;
}