Batting Practice (LightOJ-1408, Probability & Expectation)

1. Title link:

LightOJ-1408

2. The main idea of ​​the topic:

There is a probability of p for each pitch. If you make k1 consecutive shots or consecutive k2 shots, the game is over.

Find the expectations for the number of pitches.

3. Analysis:

f[i]: I have made i consecutive shots, the expected number of shots before the end of the game.

g[i]: It has been pitched i times in a row, the expected number of pitches before the end of the game.

Easy to get transfer equation:

f[i]  = (1 - p) * f[i + 1] + p * g[1] + 1.

g[i] = p * g[i + 1] + (1 - p) * f[1] + 1.

After simple iteration:

Combine the two types to get:

Solve f[1] and back with it to get g[1].

The final answer ans = qf[1] + pg[1] + 1.

4. Code implementation:

#include <bits/stdc++.h>
using namespace std;

const double eps = 1e-8;

double quick(double a, int b)
{
    double sum = 1.0;
    while(b)
    {
        if(b & 1)   sum *= a;
        a *= a;
        b >>= 1;
    }
    return sum;
}

int main()
{
    int T;
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ++ca)
    {
        double p; int k1, k2;
        scanf("%lf %d %d", &p, &k1, &k2);
        if(fabs(p - 0.0) < eps)        printf("Case %d: %d\n", ca, k1);
        else if(fabs(p - 1.0) < eps)   printf("Case %d: %d\n", ca, k2);
        else
        {
            double q = 1.0 - p;
            double pk = quick(p, k2 - 1), qk = quick(q, k1 - 1);
            double f1 = (1.0 - qk) * ((1.0 - pk) / q + 1.0 / p) / (1.0 - (1.0 - qk) * (1.0 - pk));
            double g1 = (q * f1 + 1.0) * (1.0 - pk) / (1.0 - p);
            double ans = q * f1 + p * g1 + 1.0;
            printf("Case %d: %f\n", ca, ans);
        }
    }
    return 0;
}