1788

//2007-02-13 23:21:10    Accepted    1788    C++    00:00.18    3236K
//用时1小时15分，读题时间 =0，第二次做 ，一次AC 
/*
用数组存放4-树，Tree[0]为根
Tree[i]的四个儿子是Tree[4*i+1],Tree[4*i+2],Tree[4*i+3],Tree[4*i+4]
若i!=0，Tree[i]的父亲是Tree[(i-1)/4]
设N是输入的矩阵阶数
树的高度high=log(N)/log(2)
树的总占空间大小为4^0+4^1+4^2+...+4^high=(4^(high+1)-1)/3=(4*N*N-1)/3
当N=512时，上式=349525
*/
#include <iostream>
#include <vector>
#include <stack>
using namespace std;

int Tree[349525],high,tree_size;
int a[512][512],N;

int *treep;

void dfs(int size,int i,int j)
{
    if(size==1){
        *--treep=a[i][j];
    }
    else {
        dfs(size/2,i+size/2,j+size/2);
        dfs(size/2,i+size/2,j);
        dfs(size/2,i,j+size/2);
        dfs(size/2,i,j);
    }
}

bool eq(int a,int b,int c,int d,int e)
{
    //printf("%d %d %d %d %d",a,b,c,d,e);
    if(a!=b)return 0;
    if(b!=c)return 0;
    if(c!=d)return 0;
    if(d!=e)return 0;
    return 1;
}

int main()
{
    int k;
    scanf("%d",&k);
    while(k--){
        scanf("%d",&N);
        high=0;
        while((1<<high)!=N){
            high++;
        }
        tree_size=(4*N*N-1)/3;
        treep=Tree+tree_size;//定位在树的最后一个空间的后面一个位置
        for(int i=0;i<N;i++){
            for(int j=0;j<N;j++){
                scanf("%d",&a[i][j]);
            }
        }
        
        dfs(N,0,0);//初始化树的叶子结点
        
        /*利用叶子结点计算整个树
        结点和值的对应方法如下
        (0,0)->0,(0,1)->1,1->2,    null->-1*/
        int* treep2=Tree+tree_size;
        while(treep!=Tree){
            treep2-=4;
            if(eq(treep2[0],treep2[1],treep2[2],treep2[3],1)){
                *--treep=1;
                treep2[0]=treep2[1]=treep2[2]=treep2[3]=-1;
            }
            else if(eq(treep2[0],treep2[1],treep2[2],treep2[3],0)){
                *--treep=0;
                treep2[0]=treep2[1]=treep2[2]=treep2[3]=-1;
            }
            else {
                *--treep=2;
            }
        }
        
        //将二进制串压入vector，顺序遍历Tree[]即为BFS
        vector<int> V;
        for(int i=0;i<tree_size;i++){
            switch(Tree[i]){
            case 0:
                V.push_back(0);                
                V.push_back(0);
                break;
            case 1:
                V.push_back(0);
                V.push_back(1);
                break;
            case 2:
                V.push_back(1);
                break;
            }
        }
        
        //由二进制转换到16进制，压入栈
        stack<char> S;
        vector<int>::iterator p;
        for(p=V.end()-1;p>=V.begin();){
            int sum(0);
            if(p>=V.begin() && *p--)sum+=1;
            if(p>=V.begin() && *p--)sum+=2;
            if(p>=V.begin() && *p--)sum+=4;
            if(p>=V.begin() && *p--)sum+=8;
            if(0<=sum && sum<10){
                S.push(sum+'0');
            }
            else{
                S.push(sum-10+'A');
            }
        }

        //输出
        while(!S.empty()){
            putchar(S.top());
            S.pop();
        }
        putchar(' ');
    }
}