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?
# | User | Rating |
---|---|---|
1 | jiangly | 3846 |
2 | tourist | 3799 |
3 | orzdevinwang | 3706 |
4 | jqdai0815 | 3682 |
5 | ksun48 | 3590 |
6 | Ormlis | 3533 |
7 | Benq | 3468 |
8 | Radewoosh | 3463 |
9 | ecnerwala | 3451 |
9 | Um_nik | 3451 |
# | User | Contrib. |
---|---|---|
1 | cry | 165 |
2 | -is-this-fft- | 161 |
3 | Qingyu | 160 |
4 | atcoder_official | 156 |
4 | Dominater069 | 156 |
6 | adamant | 154 |
7 | djm03178 | 151 |
8 | luogu_official | 149 |
9 | Um_nik | 148 |
10 | awoo | 147 |
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?
Name |
---|
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.