I've tried all test cases from uDebug and some random test case but could not find out wrong answer.
Is there any corner cases, which i've been not taking?
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3985 |
2 | jiangly | 3814 |
3 | jqdai0815 | 3682 |
4 | Benq | 3529 |
5 | orzdevinwang | 3526 |
6 | ksun48 | 3517 |
7 | Radewoosh | 3410 |
8 | hos.lyric | 3399 |
9 | ecnerwala | 3392 |
9 | Um_nik | 3392 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 169 |
2 | maomao90 | 162 |
2 | Um_nik | 162 |
4 | atcoder_official | 161 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | Dominater069 | 154 |
8 | awoo | 154 |
10 | luogu_official | 150 |
I've tried all test cases from uDebug and some random test case but could not find out wrong answer.
Is there any corner cases, which i've been not taking?
Название |
---|
Not sure what your solution is or why are you taking max but i guess the solution should look something like this :
$$$dp[val][cnt] = \displaystyle\sum_{x=1}^{val} dp[val-x][cnt-1]$$$ with base case $$$dp[0][0] = 1$$$
Where $$$dp[i][j]$$$ means the number of ways to get sum $$$i$$$ with exactly $$$j$$$ coins. You can answer all the other types using these dp values.