Does anyone has a proof of the formula for problem C, the time T that the two flames meet is expressed as follows:
Thanks
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 166 |
2 | maomao90 | 163 |
2 | Um_nik | 163 |
4 | atcoder_official | 161 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | nor | 153 |
9 | Dominater069 | 153 |
Does anyone has a proof of the formula for problem C, the time T that the two flames meet is expressed as follows:
Thanks
Name |
---|
Assume that left and right flame meet at point x , and the time in which left flame reaches the point x is the same time right flame takes . T[l] = T[r] = t , and total time it will take to burn complete system is T* which is summation of A[i] / B[i] , and now 2*t = T = > t = T/2
The main idea is that we can change the burning speed of all the fuses equal to 1. By doing this the new length of any ith fuse becomes (Ai/Bi) that is its length/speed.
suppose we call new length as upgraded length.
Now the problem reduces to a simpler problem that all the fuses has burning speed 1 and length equal to (Ai/Bi). Due to equal burning speed the two flames will meet at half of the total upgraded distance.
Hence , the given equation is proved.
I have tried my best to explain it :)
If u want u can cheak my submission too https://atcoder.jp/contests/abc223/submissions/26645120