数学期望，I - Beating the Dataset，Light OJ 1274

You are in a contest, and unfortunately you don't have much time. You have one problem in hand; you just glanced at the sample output and found that it just wants 'YES' or 'NO'. So, you have made another plan instead of solving the problem as you know the system very well.

For this problem, every test case is stored in a separate file. When a submission is found, the system successively runs the solution on all tests of a problem, and for each test the checking process goes as follows. The input is copied to the file input.txt. Then the solution is launched. It reads the input from the file input.txt and writes the result to the file output.txt. When it finishes, the correct answer is copied to the file answer.txt. If the contents of the files answer.txt and output.txt match, the test is assumed to be passed; otherwise, the test is not passed.

So, you decided to write a program that would operate as follows. If the folder containing the program doesn't contain the file answer.txt (i.e. the program is run on the first test), then the program outputs "YES". Otherwise, the program outputs the contents of the file answer.txt. And before the contest, the sizes of the data files are given to you.

And it's clear that the size of the file with the answer "YES" is 3 bytes, the size of the file with the answer "NO" is 2 bytes, and all the variants of the order of tests are equally probable. Now you want to calculate the average number of tests that your solution won't pass.

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 5000) and s (2n ≤ s ≤ 3n) where n denotes the number of data sets and s denotes the total size of the answer files.

For each case, print the case number and the average number of tests your solution won't pass. Error less than 10^-6 will be ignored.

4

3 7

1 2

1 3

4 10

Case 1: 2

Case 2: 1

Case 3: 0

Case 4: 2.5000000000

For the first case, one of the three answers is "YES" and two answers are "NO". If the order of tests is "YES-NO-NO", then your solution won't pass the second test only; if the order is "NO-YES-NO", then it will pass none of the tests; if the order is "NO-NO-YES", the solution won't pass the first and the third tests.

这题深深的打击了我，真tm智障。。。。

开始看题目都没看懂，还是看的题解才看懂题目。

然后想的方向又错了，看了将近一个小时。。。

这题是求全排列中在首字母前加个yes然后与原来的字符串匹配，看有几个不同的。

这题肯定是有规律的，因为yes和no的个数被确定的。并且我们只要求出第i个字符与i+1个字符不同的情况数；

如果是NO-YES或者YES-NO那么就能确定有一种不同，所以期望是x*y/(n*(n-1))*（n-1)，乘以n-1是因为n个字符有n-1种22配对的情况，因为YES—NO和NO-YES是2中情况，所以还有乘以2；

再加上如果是NO为首字符的话也是不能匹配成功的，所以还要加上y/n；

然后合并得出ans=2*x*y+y/(x+y)

得出结论，数学期望题只要推得出公式代码贼短。。。。

ac代码：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
#define debug(x) cout<<#x<<'='<<x<<endl;
#define ya(x) scanf("%d",&x)
#define Cas(x) printf("Case %d: ",x)
#define wen(x) printf("%d\n",x)
#define FOR(T) for(int cas=1;cas<=T;cas++)
int main()
{
    int T;
    ya(T);
    FOR(T)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        x=y-2*x;
        y=(y-3*x)/2;
        double ans=(2.0*x*y+y)/(x+y);
        Cas(cas);
        printf("%lf\n",ans);
    }
    return 0;
}