# I think it

I think it

Xiao Ming is only seven years old, Now I give him some numbers, and ask him what is the second largest sum if he can choose a part of them. For example, if I give him 1 、 2 、 3 ， then he should tell me 5 as 6 is the largest and 5 is the second. I think it is too hard for him, isn ’ t it?

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <=10) which is the number of test cases. And it will be followed by T consecutive test cases.
Each test case starts with a line containing an integer N (1<N<10) , the number I give Xiao Ming . The second line contains N Integer numbers ai (-10<ai<10),

For each test case, output the answer.

``````2
3
1 2 3
4
0 1 2 3``````

``````5
5``````

``````#include <iostream>
using namespace std;
int main()
{
int t=20,n,a[19]= {0};
int i,j,k;
while(t>10||t<1)
{
cin>>t;
}
for(k=0; k<t; k++)
{
cin>>n;
while(n>9||n<2)
{
cin>>n;
}
for(i=0; i<n; i++)
{
cin>>j;
while(j>=10||j<=-10)
{
cin>>j;
}
if(j>0)
a[j+9]=j;
if(j<0)
a[-j]=j;
if(j==0)
a[j]=j;
}
j=0;
for(i=18; i>=0&&j<10; i--)
{

if(a[i]!=0)
j=j+a[i];
}
cout << j << endl;
}
return 0;
}
``````

``````#include <iostream>
#include <stdlib.h>
using namespace std;

int main()
{
int T,N;
int i,j,k,t,b;
int a[10];
int sum,second;//sum为所有正数的和，second为要求的数
cin>>T;
for(i=1;i<=T;i++)
{
b=0;
sum=0;
cin>>N;
for(j=0;j<N;j++)
cin>>a[j];
for(j=0;j<N;j++)
{
if(a[j]>0)
{
b++;
break;
}
}
if(b>0)
{
for(j=0;j<N;j++)
if(a[j]!=0)
{
k=abs(a[j]);
break;
}                    //让k不为0，如果k一开始就为0，那之后就找不到一个比k小的正数；
for(j=0;j<N;j++)
{
t=abs(a[j]);
if(a[j]>0)
sum+=a[j];       //求所有正数的和
if(t<=k&&t!=0)
k=t;             //使k等于最小的正数
}
second=sum-k;
}
else
{
for(b=0;b<N;b++)
{
k=b;
for(j=b+1;j<N;j++)
{
if(a[j]>a[k])
k=j;
}
t=a[b];
a[b]=a[k];
a[k]=t;
}
if(a[0]==0&&a[0]==a[1])
{
for(j=1;j<N;j++)
{
if(a[j]<a[0])
{
second=a[j];
break;
}
}
}
else if(a[0]!=a[1])
{
second=a[1];
}
else
{
for(j=1;j<N;j++)
{
if(a[j]<a[0])
{
break;
}
}
sum=a[0]+a[1];
if(sum>=a[j])
second=sum;
else
second=a[j];
}
}
cout<<second<<endl;
}
return 0;
}

``````