Sword finger offer 14- I. cut rope

The finger of the sword Offer 14- I. Cut the rope

I'll give you a length of n The rope of , Please cut the rope to the whole length m paragraph （m、n Are integers. ,n>1 also m>1）, The length of each rope is recorded as k[0],k[1]…k[m-1] . Excuse me, k[0]k[1]…*k[m-1] What's the maximum possible product ？ for example , When the length of the rope is 8 when , We cut it into lengths of 2、3、3 Three paragraphs of , The maximum product we get here is 18.

 Example  1：

 Input : 2
 Output : 1
 explain : 2 = 1 + 1, 1 × 1 = 1
 Example  2:

 Input : 10
 Output : 36
 explain : 10 = 3 + 3 + 4, 3 × 3 × 4 = 36

Tips ：

2 <= n <= 58

Ideas

Dynamic programming
Recurrence ：a[i]=max(a[i], max( (i-j),a[i-j] ) * max( j,a[j] ));

• max( Previous value , Current updated value )
• Current updated value =[i-j] The maximum value of the range * [0-j] Maximum range

Code

class Solution {
    
public:
    int cuttingRope(int n) {
    
        vector<int> a(60);
        a[0]=0;
        a[1]=1;
        for(int i=2;i<=n;i++){
    
            for(int j=1;j<i;j++){
    
                a[i]=max(a[i],max( (i-j),a[i-j] )*max( j,a[j] ));
            }
            // printf("%d--%d\n",i,a[i]);
        }
        return a[n];
    }
};