upc 6621: HSI（数学期望，数学推导能力）

6621: HSI

时间限制: 1 Sec  内存限制: 128 MB
提交: 544  解决: 112
[提交] [状态] [讨论版] [命题人:admin]

题目描述

Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is YES or NO.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1⁄2 probability in 1900 milliseconds, and correctly solve each of the other N−M cases without fail in 100 milliseconds.
Now, he goes through the following process:
Submit the code.
Wait until the code finishes execution on all the cases.
If the code fails to correctly solve some of the M cases, submit it again.
Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).

Constraints
All input values are integers.
1≤N≤100
1≤M≤min(N,5)

输入

Input is given from Standard Input in the following format:
N M

输出

Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 109.

样例输入

1 1

样例输出

3800

提示

In this input, there is only one case. Takahashi will repeatedly submit the code that correctly solves this case with 1⁄2 probability in 1900 milliseconds.
The code will succeed in one attempt with 1⁄2 probability, in two attempts with 1⁄4 probability, and in three attempts with 1⁄8 probability, and so on.
Thus, the answer is 1900×1⁄2+(2×1900)×1⁄4+(3×1900)×1⁄8+…=3800.

来源/分类

ABC078&ARC085 

[提交] [状态]

【总结】

我已经第n次读不懂atcoder的题是啥意思了，他就不能说明白一点，细节强调一下。md

【题意】

Takahashi是打acm的，他提交了一个题，后台有n组测试数据，他提交后，有m组超时了。

那么这个意思呢就是 我只要进行一次提交，不管过没过，这个题目中总是有m组测试数据耗时1900ms，n-m组耗时100ms。

每次提交，对于每一组超时的数据，有1/2的概率能通过。

问：当他过了这个题的时候，总的耗时的数学期望是多少。总的耗时是指：每次提交每组数据用的时间，而且每进行一次提交，耗费的时间是固定的=1900m+100(n-m)

【分析】

令x=1900m+100(n-m)，即每次提交需要花费的时间

令p=1/(2^m)，意思是 某次提交，m组超时的数据全部通过的概率

期望公式

,k是提交次数，下面我们求出提交次数的数学期望来，再乘上每次的耗时，就是耗时期望了。

第一次提交的贡献：1*p

第二次提交的贡献：2*(1-p)*p   （ (1-p)*p表示第一次提交没通过，并且第二次通过的概率）

第三次提交的贡献：3*(1-p)^2*p  （ (1-p)^2*p 表示前两次提交都没通过，并且第三次通过的概率）

第四次.....一直到第正无穷项。。。

将上面的式子加起来得：

令           

             ①式则

   ②式

①-②得：

根据等比数列前n项和公式可得：

化简可得：

由于k趋向于正无穷，其中被减数有一项k作为指数，那被减数的指定服从这个值。因为1-p<1，所以被减数趋近于0

故

所以

这就是提交次数的期望，再乘上1900m+100(n-m)即为答案；

【代码】

/****
***author: winter2121
****/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    int n,m;
    cin>>n>>m;
    //神仙题的化简
    cout<<(1900*m+100*(n-m))*(1<<m)<<endl;
}