11427 - Expect the Expected(概率期望)

11427 - Expect the Expected

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 definitions 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 definitions, 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 first 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

题解：题意是一个人玩牌，每次胜率是p，她每天晚上最多玩n局，如果胜的频率大于p就睡，明天继续，如果玩了n局还没大于p

就戒了，以后就不玩了；平均情况下，他可以玩几天；求期望，先求出每天哭着睡觉的概率，然后期望就是：s+=i*Q*pow(1-Q,i-1)；

s=Q+2Q(1-Q)+3Q*(1-Q)^2........;大神们通过一定的推算可以得到s=1/Q;

还可以假设期望是e天，情况分两类，第一天哭着睡觉：概率Q，期望1；第一天开心睡觉：期望1-Q,期望1+e；e=Q*1+(1-Q)*(1+e);

e=1/Q;

不过我还是不太理解，概率论没学好T_T；

两种代码：

代码1：直接套了1/Q

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
typedef long long LL;
const int MAXN=110;
double dp[MAXN][MAXN];
int main(){//dp[i][j]=dp[i-1][j-1]*p+dp[i-1][j]*(1-p)
int T,px,py,n,kase=0;
scanf("%d",&T);
while(T--){
scanf("%d/%d %d",&px,&py,&n);
//printf("%d/%d %d\n",px,py,n);
double p=1.0*px/py,Q=0;
mem(dp,0);dp[0][0]=1;
for(int i=1;i<=n;i++){
for(int j=0;j*py<=px*i;j++){
dp[i][j]=dp[i-1][j]*(1-p);
if(j)dp[i][j]+=dp[i-1][j-1]*p;
if(i==n)Q+=dp[i][j];
}
}
printf("Case #%d: %d\n",++kase,(int)(1/Q));
}
return 0;
}

　　代码2：暴力趋近o~o;到10w就抄了，1w就ac了，1e-15;

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
typedef long long LL;
const int MAXN=110;
double dp[MAXN][MAXN];
int main(){//dp[i][j]=dp[i-1][j-1]*p+dp[i-1][j]*(1-p)
int T,px,py,n,kase=0;
scanf("%d",&T);
while(T--){
scanf("%d/%d %d",&px,&py,&n);
//printf("%d/%d %d\n",px,py,n);
double p=1.0*px/py,Q=0;
mem(dp,0);dp[0][0]=1;
for(int i=1;i<=n;i++){
for(int j=0;j*py<=px*i;j++){
dp[i][j]=dp[i-1][j]*(1-p);
if(j)dp[i][j]+=dp[i-1][j-1]*p;
if(i==n)Q+=dp[i][j];
}
}
double s=1e-15;
//	printf("%lf\n",s);
for(int i=1;i<=10000;i++)s+=i*Q*pow(1-Q,i-1);
printf("Case #%d: %d\n",++kase,(int)s);
}
return 0;
}