最长子序列和（分治法实现）

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<stack>
#include<set>
#include<queue>
#include<algorithm>
using namespace std;
const long long mod=1000000007;
const long long min_=-1000000000;
int maxsub(const int a[],int left,int right){
       int maxleftsum,maxrightsum;
       int maxleftbordersum,maxrightbordersum;
       int leftbordersum,rightbordersum;
       int center,i;
       if(left==right)
           if(a[left]>0)
           return a[left];
       else return 0;
       center=(left+right)/2;
       maxleftsum=maxsub(a,left,center);
       maxrightsum=maxsub(a,center+1,right);
       maxleftbordersum=leftbordersum=0;
       for(i=center;i>=left;i--)
       {
           leftbordersum+=a[i];
           if(leftbordersum>maxleftbordersum)
               maxleftbordersum=leftbordersum;
       }
       maxrightbordersum=rightbordersum=0;
       for(i=center+1;i<=right;i++)
       {
           rightbordersum+=a[i];
           if(rightbordersum>maxrightbordersum)
            maxrightbordersum=rightbordersum;
       }
       int t=max(maxleftsum,maxrightsum);
       return max(t,maxleftbordersum+maxrightbordersum);
}
int main()
{
    int a[55];
    int n;
    cin>>n;
    int sum=0;
    for(int i=0;i<n;i++){
        cin>>a[i];
        sum+=a[i];
    }
        cout<<maxsub(a,0,n-1)<<" "<<sum<<endl;
    return 0;
}

最长子序列和，产生情况，用分治法就是左边或者右边，或者横跨俩边，

建立base case就是left=right则返回a[left]的值。

分治关键在于后面的和，俩边可以递归实现。