void solve(int N,int P){
if(N==0)
{
cout<<"yes\n";
return;
}
if(N-2>=0){
solve(N-2,P);
}
if(N-3>=0){
solve(N-3,P);
}
}
It's showing runtime error for N = 200000(Ignore P) . Why ??
# | User | Rating |
---|---|---|
1 | tourist | 3857 |
2 | jiangly | 3747 |
3 | orzdevinwang | 3706 |
4 | jqdai0815 | 3682 |
5 | ksun48 | 3591 |
6 | gamegame | 3477 |
7 | Benq | 3468 |
8 | Radewoosh | 3463 |
9 | ecnerwala | 3451 |
10 | heuristica | 3431 |
# | User | Contrib. |
---|---|---|
1 | cry | 165 |
2 | -is-this-fft- | 161 |
3 | Qingyu | 160 |
4 | Dominater069 | 158 |
5 | atcoder_official | 157 |
6 | adamant | 154 |
7 | Um_nik | 151 |
8 | djm03178 | 150 |
9 | luogu_official | 149 |
10 | awoo | 147 |
void solve(int N,int P){
if(N==0)
{
cout<<"yes\n";
return;
}
if(N-2>=0){
solve(N-2,P);
}
if(N-3>=0){
solve(N-3,P);
}
}
It's showing runtime error for N = 200000(Ignore P) . Why ??
Name |
---|
Even if you write simple fibonacci recursive function, you will get RE at big tests. It is because program did a lot of recursive calls.
To know more and unterstand how to avoid it, learn basics of DP (dynamic programming).
wont work for n>=3 coz it will end up at n=1(which doesnt have any return statement), also there will be plenty of yes printed for every number and i dont think you would want that...