Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.
For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).
Note: you may assume that n is not less than 2.
方法
归纳法，尽量分出3，用3算出的乘积是最大的
c代码
#include <assert.h>
int integerBreak(int n) {
if(n == 2)
return 1;
else if(n == 3)
return 2;
int product = 1;
while(n > 4) {
product *= 3;
n -= 3;
}
return product*n;
}
int main() {
assert(integerBreak(2) == 1);
assert(integerBreak(10) == 36);
assert(integerBreak(50) == 86093442);
return 0;
}
