7 = 2+2+3

8 = 2+3+3

9 = 3+3+3

10 = 3+3+4

we can see that when numbers on the right are as close as possible,their product is the largest.

```
class Solution {
public:
int integerBreak(int n) {
if (n <= 1) {
return 0;
}
int result = 0;
int positive;
int num_larger = 0;
for (int num = 2; num <= n; num++) {
positive = n / num;
if (n % num == 0) {
result = max(result, (int)pow(positive, num));
}
else {
num_larger = n - num * positive;
result = max(result, (int)pow(positive, num - num_larger) * (int)pow(positive + 1, num_larger));
}
}
return result;
}
};
```