hdu 1204糖果大战

题意：A有n个糖果，B有m个糖果，他们玩24点游戏，A能解答出来的概率是p，B能解答出来的概率是q，如果都解答出来或都没解答出来就平局，否则如果A解答出来了，B给A一颗糖，否则A给B一颗糖，某一方糖全部给光结束。求A赢的概率。

下面记pa = p(1-q)为A赢的概率，pb = q(1-p)为B赢的概率，(1-pa-pb)为平局的概率。

设f(i)为A有i个糖果时赢的概率，则

\[f(i) = pa * f(i+1) + pb * f(i-1) + (1 - pa - pb)*f(i) \]

\[pa * (f(i+1) - f(i)) = pb * (f(i) - f(i-1)) \]

如果n == 0，则f(n) = 0

否则如果m == 0，则f(n) = 1

(n == 0且m == 0题目没有定义，假装没有)

否则如果pa = 0，则f(n) = 0

设\(g(i) = f(i+1) - f(i)\)，则

\[g(i) / g(i-1) = pb / pa \]

记\(t = pb / pa\)

则\(g(i) = t g(i-1)\)

\[g(0) + g(1) + \cdots + g(n+m-1) = f(n+m) - f(0) = 1 \]

因为f(n+m) = 1, f(0) = 0

如果\(t\neq 1\)，

\[g(0) (1 + t + \cdots + t^{n+m-1}) = g(0) \frac{t^{n+m} - 1}{t - 1} = 1 \]

\[g(0) = \frac{t-1}{t^{n+m} - 1} \]

所求

\[f(n) = f(n) - f(0) = g(0) + g(1) + \cdots + g(n-1) = g(0) \frac{t^{n} - 1}{t - 1} = \frac{t^n-1}{t^{n+m} - 1} \]

如果\(t = 1\)，

\[g(0) (1 + t + \cdots + t^{n+m-1}) = g(0) (n+m) = 1 \]

\[g(0) = \frac{1}{n+m} \]

\[f(n) = f(n) - f(0) = g(0) + g(1) + \cdots + g(n-1) = n\ g(0) = \frac{n}{n+m} \]

代码：

#include <cstdio>
#include <cmath>

using namespace std;

double eps = 1e-8;
int sgn(double x) {
    return x < -eps ? -1 : (x > eps ? 1 : 0);
}

int main() {
    double n, m;
    double p, q;

    while (~scanf("%lf%lf%lf%lf", &n, &m, &p, &q)) {
        if (sgn(n) == 0) {
            puts("0.00");
        } else if (sgn(m) == 0) {
            puts("1.00");
        } else {
            double pb = q * (1 - p), pa = (p * (1 - q));
            if (sgn(pa) == 0) {
                puts("0.00");
            } else if (sgn(pb) == 0) {
                puts("1.00");
            } else {
                double t = pb / pa;
                if (sgn(t - 1)) {
                    printf("%.2f\n", (pow(t, n) - 1) / (pow(t, n + m) - 1));
                } else {
                    printf("%.2f\n", n / (n + m));
                }
            }
        }
    }
    return 0;
}