编程竞赛TLE问题解决策略-优快云博客

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

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.

解：Repeat until the code correctly solve all the cases in one submission.，一定要注意这句话，他的意思是重复这句话“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.”再重写这m个问题

理解1：

首先感谢dalao写的，转一下。

https://blog.youkuaiyun.com/winter2121/article/details/81489377

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 $\sum k*p(k)$ ,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。

k的解释：第一次提交耗时x，第二次2x，第n次就是nx

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：

$ans=1p + 2(1-p)^{1}p+3(1-p)^{2}p+...+k(1-p)^{k-1}p$

$=p(1 + 2(1-p)^{1}+3(1-p)^{2}+...+k(1-p)^{k-1})$

令 $res=1 + 2(1-p)^{1}+3(1-p)^{2}+...+k(1-p)^{k-1}$ ①式

则 $(1-p)res=1(1-p)^{1} + 2(1-p)^{2}+3(1-p)^{3}+...+k(1-p)^{k}$ ②式

其实这个1.2式就是等差数列与等比数列想乘的前n项和，写出Sn，q*Sn，相减即可。

①-②得：

$p*res=1+(1-p)^{1}+(1-p)^{2}+....+(1-p)^{k-1}-k(1-p)^{k}$

根据等比数列前n项和公式可得：

$p*res=\frac{1-(1-p)^{k}}{p} -k(1-p)^{k}$

化简可得：

$res= \frac{1}{p^{2}}-\frac{(1-p)^k(1+p^{2}k)}{p^{2}}$

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 $res= \frac{1}{p^{2}}-0$

所以 $ans=p*res= \frac{1}{p}$

这就是提交次数的期望，再乘上1900m+100(n-m)即为答案；

理解2：

虽说比较简单，但是不容易想到这个。。。

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll n,m;
    cin>>n>>m;
    cout<<(100*(n-m)+1900*m)*(1<<m)<<endl;
    return 0; 
}