Garbage Heap

Time limit: ? seconds

Memory limit: 64 megabytes

Description
Farmer John has a heap of garbage formed in a rectangular parallelepiped.
It consists of 

 garbage pieces each of which has a value. The value of a piece may
be 0, if the piece is neither profitable nor harmful, and may be negative which means that the piece is not just unprofitable, but even harmful (for environment).



The farmer thinks that he has too much harmful garbage, so he wants to decrease the heap size, leaving a rectangular nonempty parallelepiped of smaller size cut of the original heap to maximize the sum of the values
of the garbage pieces in it. You have to find the optimal parallelepiped value. (Actually, if any smaller parallelepiped has value less than the original one, the farmer will leave the original parallelepiped).

Input

The first line of the input contains the number of the test cases, which is at most 15. The descriptions of the test cases follow. The first line of a test case description contains three integers A, B,
and C (1 ≤ A, B, C ≤ 20). The next lines contain 

 numbers,
which are the values of garbage pieces. Each number does not exceed 

 by absolute value. If we introduce
coordinates in the parallelepiped such that the cell in one corner is (1,1,1) and
the cell in the opposite corner is (A,B,C), then the values are listed in the order



The test cases are separated by blank lines.

Output

For each test case in the input, output a single integer denoting the maximal value of the new garbage heap.
Print a blank line between test cases.

Examples

Input	Output
1 2 2 2 -1 2 0 -3 -2 -1 1 5	6

这道题是Distant Galaxy的延伸，变成了三维的图形，同时在求废料块的价值时用到了Substance里的“前缀和”以及Open Credit System的维护最大值。代码如下：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define FOR(i,t,n) for(int i=(t);i<=(n);i++) //宏定义简化代码（Image Is Everything）
const long long int INF=1LL<<60; //注意极大的数如何表示
long long int s[25][25][25];

int main()
{
int T;
cin>>T;
while(T--)
{
memset(s,0,sizeof(s));
int A,B,C;
cin>>A>>B>>C;
FOR(i,1,A) FOR(j,1,B) FOR(k,1,C)
{
long long int x;
scanf("%lld",&x);
s[i][j][k]=x+s[i-1][j][k]+s[i][j-1][k]-s[i-1][j-1][k]
+s[i][j][k-1]-s[i-1][j][k-1]-s[i][j-1][k-1]+s[i-1][j-1][k-1];
}
long long int ans=-INF;
FOR(i,1,C) FOR(i1,i,C)
{
FOR(j,1,B) FOR(j1,j,B)
{
long long int l=0;
FOR(k,1,A)
{
long long int g;
g=s[k][j1][i1]-s[k][j1][i-1]-(s[k][j-1][i1]-s[k][j-1][i-1]);
ans=max(ans,g-l);
l=min(l,g);
}
}
}
cout<<ans<<endl;
if(T) cout<<endl; //特别注意
}
return 0;
}还是要特别注意以下输出数据的格式