顺序表应用7：最大子段和之分治递归法(SDUT 3664)

#include <bits/stdc++.h>

using namespace std;
const int maxn = 50005;
int num = 0;
struct node
{
    int *elem;
    int len;
};
void Creatlist(struct node &list, int n)
{
    list.elem=new int[maxn];
    if(!list.elem)
        exit(OVERFLOW);
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&list.elem[i]);
        list.len++;
    }
}
int get_ans(struct node &list, int l, int r)
{
    num ++;
    int sum1, sum2, sum3, sum4, mav3, mav4;
    if(l == r)
    {
        if(list.elem[l] < 0)
            return 0;
        else
            return list.elem[l];
    }
    int m = (l + r) / 2;
    sum1 = get_ans(list, l, m);
    sum2 = get_ans(list,m + 1, r);
    sum3 = sum4 = 0;
    mav3 = mav4 = 0;
    for(int i = m; i >= l; i --)
    {
        sum3 += list.elem[i];
        mav3 = max(mav3,sum3);
    }
    for(int i = m + 1; i <= r; i ++)
    {
        sum4 += list.elem[i];
        mav4 = max(mav4,sum4);
    }
    int ans = 0;
    ans = max(sum1, sum2);
    ans = max(ans, mav3 + mav4);
    return ans;
}

int main()
{
    int n;
    struct node list;
   scanf("%d",&n);
    Creatlist(list, n);
    num = 0;
    int ans = get_ans(list,1,n);
    printf("%d %d\n", ans, num);
    return 0;
}