### Using CF's Feature to pass CF1705F in practise↵
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#### Sill useful till 2022/7/21 9:27, the post time.↵
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It's found that [A well known blog](https://codeforces.com/blog/entry/78817) has talked about that trick two years ago, you can read (and upvote) that because that's really interesting (and **maybe** useful). ↵
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Let's think of randomize.↵
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For two adjacent problems ${a_i,a_{i+1}}$, randomly question once.↵
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It can be easily calculated that the posibility of instantly getting the correct answer is $\frac{1}{2}$.↵
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If we get the correct answer, go on querying ${a_{i+2},a_{i+3}}$.↵
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For other possiblilties, we can know the relationship between ${a_i,a_{i+1}}$, so we can go on querying ${a_{i+1},a_{i+2}}$.↵
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Finally, we query once to get the answer of the last problem. By using the relationships, we can get the overall answer.↵
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The expected query number is ${\frac{2}{3}n+2}$.↵
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The expected failure chance per testcase is ${\frac{1}{3}}$.↵
↵
But there is a feature that CF rejudges the TLE submissions per testcase $3$ times.↵
↵
So, every time we exceed query limit, ```while(1)``` instantly to get TLE instead of WA for rejudging.↵
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This makes the failure chance per testcase $\frac{1}{27}$, making the overall pass chance $\frac{1}{10}$.↵
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In practise, we just need to resubmit several times to get AC.↵
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In contests, this method can be used in randomized solutions to make the AC possibility larger.↵
↵
Accepted link: ↵
[164438874](https://codeforces.com/contest/1705/submission/164438874) and [164437895](https://codeforces.com/contest/1705/submission/164437895)↵
↵
(Mike, plz don't fix this.It's so useful that... qwq)
↵
#### Sill useful till 2022/7/21 9:27, the post time.↵
↵
It's found that [A well known blog](https://codeforces.com/blog/entry/78817) has talked about that trick two years ago, you can read (and upvote) that because that's really interesting (and **maybe** useful). ↵
↵
Let's think of randomize.↵
↵
For two adjacent problems ${a_i,a_{i+1}}$, randomly question once.↵
↵
It can be easily calculated that the posibility of instantly getting the correct answer is $\frac{1}{2}$.↵
↵
If we get the correct answer, go on querying ${a_{i+2},a_{i+3}}$.↵
↵
For other possiblilties, we can know the relationship between ${a_i,a_{i+1}}$, so we can go on querying ${a_{i+1},a_{i+2}}$.↵
↵
Finally, we query once to get the answer of the last problem. By using the relationships, we can get the overall answer.↵
↵
The expected query number is ${\frac{2}{3}n+2}$.↵
↵
The expected failure chance per testcase is ${\frac{1}{3}}$.↵
↵
But there is a feature that CF rejudges the TLE submissions per testcase $3$ times.↵
↵
So, every time we exceed query limit, ```while(1)``` instantly to get TLE instead of WA for rejudging.↵
↵
This makes the failure chance per testcase $\frac{1}{27}$, making the overall pass chance $\frac{1}{10}$.↵
↵
In practise, we just need to resubmit several times to get AC.↵
↵
In contests, this method can be used in randomized solutions to make the AC possibility larger.↵
↵
Accepted link: ↵
[164438874](https://codeforces.com/contest/1705/submission/164438874) and [164437895](https://codeforces.com/contest/1705/submission/164437895)↵
↵
(Mike, plz don't fix this.It's so useful that... qwq)
