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 | jiangly | 3640 |
2 | Benq | 3593 |
3 | tourist | 3572 |
4 | orzdevinwang | 3561 |
5 | cnnfls_csy | 3539 |
6 | ecnerwala | 3534 |
7 | Radewoosh | 3532 |
8 | gyh20 | 3447 |
9 | Rebelz | 3409 |
10 | Geothermal | 3408 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 173 |
2 | adamant | 164 |
3 | awoo | 161 |
4 | TheScrasse | 160 |
5 | nor | 159 |
6 | maroonrk | 156 |
7 | SecondThread | 152 |
8 | pajenegod | 146 |
9 | BledDest | 144 |
10 | Um_nik | 143 |
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...