Problem:
Matrix-like Computation

Your task is to develop a tiny little part of spreadsheet software.

Write a program which adds up columns and rows of given table as shown in the following figure:

Input

The input consists of several datasets. Each dataset consists of:

n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table

The input ends with a line consisting of a single 0.

Output

For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.

Sample Input

4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0

Output for the Sample Input

52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815

Solution in JAVA (judge verdict: Accepted)

import java.util.*;

public class Main {	
	public static void main(String[] args){
		new Main();
	}
	public Main(){
		new AOJ0102();
	}
	
	class AOJ0102{
		
		public AOJ0102() {
			Scanner in = new Scanner(System.in);
			while(in.hasNext()){
				int n = in.nextInt();
				if(n==0)break;
				int[][] hyou = new int[n][n];
				int tate[] = new int[n+1];
				for(int i=0;i<n*n;i++)hyou[i/n][i%n]=in.nextInt();
				for(int s=0;s<n;s++){
					int sum=0;
					for(int i=0;i<n;i++){
						System.out.printf("%5d",hyou[s][i]);
						sum+=hyou[s][i];
						tate[i]+=hyou[s][i];
					}
					System.out.printf("%5d\n",sum);
					tate[n]+=sum;
				}
				for(int i=0;i<n+1;i++){
					System.out.printf("%5d",tate[i]);
				}
				System.out.println();
			}
			
			
			
			
			
		}
	}

	class AOJ2503{
		int MAX = 0;
		public AOJ2503() {
			Scanner in = new Scanner(System.in);
			int n = in.nextInt();
			int m = in.nextInt();

			int[][] cost = new int[n][n];

			int[] dp = new int[n];
			for(int i=0;i<n*n;i++){
				cost[i/n][i%n]=MAX;
			}
			for(int i=0;i<m;i++){
				int a = in.nextInt();
				int b = in.nextInt();
				cost[a][b]=in.nextInt();
			}	
			//			TODO 一番大きいパスを0のポイントからn-1のポイントまでの

			for(int s=1;s<n;s++){
				for(int i=0;i<=s;i++){
					if(cost[i][s]==MAX)continue;
					dp[s]=Math.max(dp[s],dp[i]+cost[i][s]);
				}
			}
			for(int s=0;s<n;s++)for(int i=0;i<n;i++){
				if(cost[s][i]==MAX)continue;
				dp[i]=Math.max(dp[s]+cost[s][i],dp[i]);
			}
			for(int i=0;i<n;i++)System.out.print(dp[i]+" ");
			System.out.println();
			System.out.println(dp[n-1]);
		}
	}



	class  AOJ0155{
		int d[];
		int n;
		int x[];
		int y[];
		int ban[];
		boolean sw;
		ArrayList<Integer> result;
		int back[];
		int[][] cost;
		void dijkstra(int s,int g){
			d = new int[n+1];d[s]=0;
			Arrays.fill(d,0);
			boolean used[] = new boolean[n+1];
			int sumn[] = new int[n+1];
			int v=s;
			int save = -1;
			back = new int[n+1];
			while(true){
				save = v;
				v=-1;
				for(int u=1;u<n+1;u++)if(!used[u]&&(v==-1||d[u]<d[v])){
					v=u;
				}
				System.out.println(v);
				if(v==-1)break;
				else back[v]=save;
				used[v]=true;
				for(int u=1;u<n+1;u++)if(cost[v][u]!=100000)d[u]=Math.min(1+d[v],d[u]);
			}
			for(int i=1;i<d.length;i++){
				System.out.println(i+" "+d[i]);
			}
			int now=g;
			result = new ArrayList<Integer>();
			int a=0;
			//			for(int i=0;i<back.length;i++){
			//				System.out.println(i+" "+back[i]);
			//			}
			while(true){
				if(cost[a][now]==1000000)sw = true;
				result.add(now);
				if(now==s)break;
				a = now;
				now=back[now];
			}
			return;
		}
		public AOJ0155(){
			Scanner in = new Scanner(System.in);
			while(true){
				n = in.nextInt();
				if(n==0)break;
				cost = new int[n+1][n+1];
				ban = new int[n+1];
				x = new int[n+1];
				y = new int[n+1];
				for(int i=1;i<n+1;i++){
					ban[i] = in.nextInt();
					x[ban[i]] = in.nextInt();
					y[ban[i]] = in.nextInt();
				}
				for(int i=1;i<n+1;i++){
					for(int s=i+1;s<n+1;s++){
						int x1 = x[i],y1= y[i],x2=x[s],y2=y[s];
						int costs = (int)Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));		
						cost[i][s] = costs > 50 ? 1000000:costs;
						cost[s][i] = cost[i][s];
					}
				}
				int kai = in.nextInt();
				for(int i=0;i<kai;i++){
					int s = in.nextInt();
					int g = in.nextInt();
					dijkstra(s, g);
//					if(sw){
//						System.out.println("NA");
//						break;
//					}
					for(int a=result.size()-1;a>=0;a--){
						System.out.print(result.get(a)+" ");
					}
					System.out.println();
				}
			}
		}
	}
}