int  dphe(int a[],int len)
{int i,j;int **he=new int*[len];for(i=0;i<len;i++)*(he+i)=new int[len]();for(i=0;i<len;i++)for(j=0;j<len;j++)he[i][j]=0;int length;for(length=0;length<len;length++)for(i=0;i<len-length;i++)for(j=i;j<=i+length;j++)he[i][i+length]+=a[j];int max=INT_MIN;for(i=0;i<len;i++)for(j=0;j<len;j++)if(he[i][j]>=max)max=he[i][j];for(i=0;i<len;i++)delete [] *(he+i);delete he;return max;
}

/*求子数组的和，现用DP做*/#include<iostream>
#include<climits>
#include<cstdlib>
#include<time.h>
using namespace std;int  dphe(int a[],int len)
{int i,j;int **he=new int*[len];for(i=0;i<len;i++)*(he+i)=new int[len]();for(i=0;i<len;i++)for(j=0;j<len;j++)he[i][j]=0;int length;for(length=0;length<len;length++)for(i=0;i<len-length;i++)for(j=i;j<=i+length;j++)he[i][i+length]+=a[j];int max=INT_MIN;for(i=0;i<len;i++)for(j=0;j<len;j++)if(he[i][j]>=max)max=he[i][j];for(i=0;i<len;i++)delete [] *(he+i);delete he;return max;
}/*下面用栈做，规则如下1.元素顺序入栈，保存2个变量sum和max,sum代表整个栈中元素的总和,sum代表到目前元素为止前面最大的子数组的和2.若sum<0,则清空整个栈3.若sum>max,则更新sum
*/typedef int DataType;
#define MAX 100
typedef struct
{DataType data[MAX];int top;
}Stack;Stack* init_stack(void)
{Stack* s=new Stack;if(s)s->top=-1;return s;
}bool empty_stack(Stack* s)
{if(s->top==-1)return true;return false;
}bool full_stack(Stack* s)
{if(s->top==MAX-1)return true;return false;
}void push_stack(Stack* s,DataType value)
{if(full_stack(s))    return ;s->top++;s->data[s->top]=value;
}void pop_stack(Stack* s,DataType *value)
{if(empty_stack(s))return ;*value=s->data[s->top];s->top--;
}void clear_stack(Stack* s)
{int value;while(!empty_stack(s))pop_stack(s,&value);
}void free_stack(Stack* s)
{delete s;
}int he(int a[],int length)
{int i;Stack* s;int sum=0,max=INT_MIN;s=init_stack();for(i=0;i<length;i++){push_stack(s,a[i]);sum=sum+a[i];if(sum<=0){clear_stack(s);sum=0;}else // sum>0
        {if(sum>max)max=sum;}}return max;free_stack(s);
}    int main(void)
{int b[]={3,-5,2,-4,7,-2,10,1};int c[]={-1,0,2,-1,0,4,2,-1};int a[8];int i,j;//比较10次for(j=1;j<=10;j++){srand(time(NULL));for(i=0;i<8;i++)a[i]=rand()%200-100; //随即产生-100-100的数for(i=0;i<8;i++)cout<<a[i]<<" ";cout<<"穷举和="<<dphe(a,8)<<" ";cout<<"栈和="<<he(a,8)<<endl;sleep(1);}return 0;
}

int he2(int a[],int length)
{int i,j;int sum=0,max=INT_MIN;for(i=0;i<length;i++){sum+=a[i];if(sum<=0)sum=0;else if(sum>max)max=sum;    }return max;
}