Hello everyone, i just wanted to know why this solution cannot be solved using greedy approach https://codeforces.net/contest/1005/submission/76980469
# | User | Rating |
---|---|---|
1 | tourist | 3803 |
2 | jiangly | 3707 |
3 | Benq | 3627 |
4 | ecnerwala | 3584 |
5 | orzdevinwang | 3573 |
6 | Geothermal | 3569 |
6 | cnnfls_csy | 3569 |
8 | Radewoosh | 3542 |
9 | jqdai0815 | 3532 |
10 | gyh20 | 3447 |
# | User | Contrib. |
---|---|---|
1 | awoo | 163 |
2 | maomao90 | 160 |
2 | adamant | 160 |
4 | maroonrk | 152 |
5 | -is-this-fft- | 151 |
6 | atcoder_official | 148 |
6 | nor | 148 |
8 | SecondThread | 147 |
9 | TheScrasse | 146 |
10 | Petr | 144 |
Hello everyone, i just wanted to know why this solution cannot be solved using greedy approach https://codeforces.net/contest/1005/submission/76980469
Name |
---|
I guess, this is not the correct greedy approach. For number $$$20302$$$, your code gives answer $$$0$$$ whereas it should be $$$3$$$.
In my opinion, the correct greedy would be what you are doing, and additionally if you encounter $$$0$$$, $$$3$$$, $$$6$$$ or $$$9$$$ make a separator there.
Exactly, the better greedy aprroach is using DP, then $$$dp_i$$$ means the answer when we consider first $$$i$$$ elements, it can be updated in two ways, the last block which has $$$ith$$$ position is divisible by 3, or not, if not then $$$dp_i$$$ will be equal to $$$dp_{i-1}$$$, and otherwise cut the smallest block which is divisible by 3, then $$$dp_i$$$ will be $$$dp_j+1$$$(finding the smallest block can be done whit a precalculation with $$$O(n)$$$ time).
The whole problem will be $$$O(n)$$$.
This might help you.
Thanks a lot poduingi