Cut the rope - offer to prove safety

Title Description

You give a string of length n, please cut the rope segment length integer m (m, n are integers, n> 1 and m> 1), the length of each rope is referred to as k [0], k [1], ..., k [m]. Will k [0] xk [1] x ... xk [m] maximum possible product is how much? For example, when the length of the rope is 8:00, we cut it into lengths of three sections 2,3,3 of the product obtained at this time is the maximum 18.

 

Idea: cut the length of the rope as many 3 2 then built a little note that if left, then built 4 2 3 * 2 * 1 than good

public class Solution {
    public int cutRope(int target) {
        if(target <= 1) return 0;
        if(target == 2) return 1;
        if(target == 3) return 2;
        int x=target/3;
        int y=target%3;
        if(y == 1) {
            return 4*(int)Math.pow(3,x-1);
        }
        else if(y == 2){
            return 2*(int)Math.pow(3,x);
        }else {
            return (int)Math.pow(3,x);
        }
    }
}

 Dynamic programming can also be

The length of the two halves is equal to the current product of the maximum length of each of the two half-product of the maximum - "optimal substructure and overlapping subproblems 

public class Solution {
    public int cutRope(int target) {
        if(target <= 1) return 0;
        if(target == 2) return 1;
        if(target == 3) return 2;
        int[] project=new int[target+1];
        project[0]=0;
        project[1]=1;
        project[2]=2;
        project[3]=3;
        for(int i = 4;i <= target;i ++){
            int max=0;
            for(int j = 1;j <= i/2;j ++){
                max=Math.max(max,project[j]*project[i-j]);
            }
            project[i] = max;
        }
        return project[target];
    }
}