[UVA] 11427 玩纸牌 期望dp

Some mathematical background. This problem asks you to compute the expected value of a random
variable. If you haven’t seen those before, the simple de nitions are as follows. A random variable is a
variable that can have one of several values, each with a certain probability. The probabilities of each
possible value are positive and add up to one. The expected value of a random variable is simply the
sum of all its possible values, each multiplied by the corresponding probability. (There are some more
complicated, more general de nitions, but you won’t need them now.) For example, the value of a fair,
6-sided die is a random variable that has 6 possible values (from 1 to 6), each with a probability of 1/6.
Its expected value is 1=6 + 2=6 + : : : + 6=6 = 3:5. Now the problem.
I like to play solitaire. Each time I play a game, I have probability p of solving it and probability
(1 p) of failing. The game keeps statistics of all my games { what percentage of games I have won.
If I simply keep playing for a long time, this percentage will always hover somewhere around p 100%.
But I want more.
Here is my plan. Every day, I will play a game of solitaire. If I win, I’ll go to sleep happy until
the next day. If I lose, I’ll keep playing until the fraction of games I have won today becomes larger
than p. At this point, I’ll declare victory and go to sleep. As you can see, at the end of each day, I’m
guaranteed to always keep my statistics above the expected p 100%. I will have beaten mathematics!
If your intuition is telling you that something here must break, then you are right. I can’t keep
doing this forever because there is a limit on the number of games I can play in one day. Let’s say that
I can play at most n games in one day. How many days can I expect to be able to continue with my
clever plan before it fails? Note that the answer is always at least 1 because it takes me a whole day
of playing to reach a failure.
Input
The rst line of input gives the number of cases, N. N test cases follow. Each one is a line containing
p (as a fraction) and n.
1 N 3000, 0 p < 1,
The denominator of p will be at most 1000,
1 n 100.
Output
For each test case, print a line of the form `Case #x: y’, where y is the expected number of days,
rounded down to the nearest integer. The answer will always be at most 1000 and will never be within
0.001 of a round-off error case.
Sample Input
4
1/2 1
1/2 2
0/1 10
1/2 3
Sample Output
Case #1: 2
Case #2: 2
Case #3: 1
Case #4: 2

蓝书上面的题， 讲得很经典， 这里就不写题解了.

好吧还是大概说一下. 每天晚上都是独立的， 求出每天再也不玩纸牌的概率， 发现数学期望是一个无穷几何级数， 但是通过部分数学技巧就可以化解(注意那个是在无穷意义下)

没有蓝书的， 这里有近似刘汝佳的详解: Click Here

#include<stdio.h>
#include<cstring>
#define clear(a) memset(a, 0, sizeof(a))
int a, b, n, T;
double lose, p, dp[101][101];
int main(){
    scanf("%d", &T);
    for(int cas = 1; cas <= T; ++cas){
        scanf("%d/%d%d", &a, &b, &n);
        p = (double) a / b;
        clear(dp), dp[0][0] = 1;
        for(int i = 1; i <= n; ++i)
            for(int j = 0; j * b <= a * i; ++j){
                dp[i][j] = dp[i-1][j] * (1 - p);
                if(j) dp[i][j] += dp[i-1][j-1] * p;
            }
        lose = 0.0;
        for(int j = 0; j * b <= a * n; ++j) lose += dp[n][j];
        printf("Case #%d: %d\n", cas, (int) (1 / lose)); 
    }
}