> #include<iostream.h>
> #include<stdio.h>
> #define Min	-10001 
> #define Max	10001
> int NEG_INF = 0x80000000;
> int POS_INF = 0x7fffffff;
> int a[5001];//数据
> int b[5001];//记录正负值
> int x[5001];//当前最优路径
> int r[5001];//当前路径
> 
> int MaxSumWithR(int * a, int n)
> {
> 	int i,j;
> 
> 	for(j=1;j<=n;j++)
> 	{
> 		r[j]=0;
> 		x[j]=0;
> 	}
> 	int sum=0;
> 	int s=0;
> 	for(i=1;i<=n;i++)
> 	{
> 		if(s>0)
> 		{
> 			s+=a[i];
> 			r[i]=1;
> 		}
> 		else//s<=0
> 		{
> 			s=a[i];
> 			for(j=1;j<=n;j++)
> 				r[j]=0;
> 			r[i]=1;
> 		}
> 		if(sum<s)
> 		{
> 			sum=s;
> 			for(j=1;j<=n;j++)
> 				x[j]=r[j];
> 		}
> 	}
> 	return sum;
> }
> 
> int MaxSum(int * a, int begin, int end)//求a[begin...end]的最大和
> {
> 	int n=end-begin+1;
> 	int i;
> 	int s=0;
> 	int sum=NEG_INF;
> 	for(i=begin;i<=end;i++)
> 	{
> 		if(s>0)
> 		{
> 			s+=a[i];
> 		}
> 		else//s<=0
> 		{
> 			s=a[i];
> 		}
> 		if(sum<s)
> 		{
> 			sum=s;
> 		}
> 	}
> 	return sum;
> }
> 
> int MaxNum(int s1,int s2, int s3)
> {
> 	int s4 = s1>s2 ? s1:s2;
> 	int s = s4>s3? s4:s3;
> 	return s;
> }
> 
> int main()
> {
> //	cout<<NEG_INF<<endl;
> 	int i,k,times;
> 	int T;
> 	scanf("%d",&T);
> 
> 	printf("\n");
> 	for(times=1;times<=T;times++)
> 	{
> 
> 		int sum=0;
> 		int N;
> 		cin>>N;
> 		for(i=1;i<=N;i++)
> 		{
> 			scanf("%d",&a[i]);
> 			if(a[i]>0)
> 			{
> 				b[i]=1;//正数表示为1,0或负数表示为0
> 			}
> 		}
> 		int IsAllPositive=0;
> 		int IsAllNegative=0;
> 		int pos_count=0;
> 		int neg_count=0;
> 		for(i=1;i<=N;i++)
> 		{
> 			if(b[i]==1)
> 				pos_count++;
> 			else
> 				neg_count++;
> 		}
> 		if(pos_count==N)//全部是正数
> 		{
> 			for(i=1;i<=N;i++)
> 				sum+=a[i];
> 			printf("%d\n\n", sum);
> 		}
> 
> 		else if(neg_count==N)//全部是负数
> 		{
> 			int s1=Min;
> 			int t1=0;
> 			int s2=Min;
> 			for(i=1;i<=N;i++)
> 			{
> 				if(a[i]>s1)
> 				{
> 					s1=a[i];
> 					t1=i;
> 				}
> 			}
> 			for(i=1;i<=N;i++)
> 			{
> 				if(a[i]>s2 && i!=t1)
> 				{
> 					s2=a[i];
> 				}
> 			}
> 			sum=s1+s2;
> 			printf("%d\n\n", sum);
> 		}
> 
> 		else//正负数都有
> 		{
> 			int sum1=MaxSumWithR(a,N);//x[i]==0的为未访问的
> 			k=0;
> 			int begin1;//最大子串的开始位置
> 			int end1;//最大子串的结串位置
> 			for(i=1;i<=N;i++)
> 			{
> 				if(x[i]==1)
> 				{
> 					if(k==0)
> 					{
> 						begin1=i;
> 					}
> 					k++;
> 				}
> 			}
> 			end1=begin1+k-1;
> //			cout<<"begin1:"<<begin1<<endl;
> //			cout<<"end1:"<<end1<<endl;
> 			if(begin1==end1)//只有一个正数的情况,即最大子串只有一个元素
> 			{
> 				int max=Min;
> 				for(i=1;i<=N;i++)
> 				{
> 					if(i!=begin1 && a[i]>max)
> 						max=a[i];
> 				}
> 				sum=sum1+max;
> 				printf("%d\n\n", sum);
> 			}
> 			else
> 			{
> 				int sum2=NEG_INF;
> 				int sum3=NEG_INF;
> 				int sum4=0;
> 				if(begin1!=1)
> 				{
> 					sum2=MaxSum(a,1,begin1-1);
> 				}
> 				if(end1!=N)
> 				{
> 					sum3=MaxSum(a,end1+1,N);
> 				}
> 				for(i=begin1;i<=end1;i++)//找出最小的负数
> 				{
> 					if(a[i]<sum4)
> 						sum4=a[i];
> 				}
> 				sum4=-sum4;
> //				cout<<"sum1:"<<sum1<<endl;
> //				cout<<"sum2:"<<sum2<<endl;
> //				cout<<"sum3:"<<sum3<<endl;
> //				cout<<"sum4:"<<sum4<<endl;
> 				int sum5=MaxNum(sum2, sum3, sum4);
> 				sum=sum1+sum5;
> 				printf("%d\n\n", sum);
> //				cout<<"sum5:"<<sum5<<endl;
> //				cout<<"sum :"<<sum<<endl;
> 			}
> 		}
> 
> 	}//for
> 	return 0;
> }

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