as C_n^k = C_n^(k-1) * (n-k+1)/k;

the only thing is when computing C_n^(k-1) * (n-k+1)/k, we may suffer overflow.

```
int calc(int x, int y, int z){
return (x/z)*y + (x%z)*y/z;
}
vector<int> getRow(int rowIndex) {
std::vector<int> vec;
vec.push_back(1);
for(int k = 1; k <= rowIndex; ++k){
vec.push_back(calc(vec[k-1], (rowIndex-k+1), k));
}
return vec;
}
```