Jpeg（模拟）

Jpeg

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice Gym 100947K

Description

standard input/output  Announcement

 

• Statements

  In computing, JPEG is a commonly used method of lossy compression for digital images, particularly for those images produced by digital photography . The degree of compression can be adjusted, allowing a selectable tradeoff between storage size and image quality,and JPEG typically achieves 10:1 compression with little perceptible loss in image quality. Entropy coding is a special form of lossless data compression. It involves arranging the image components in a "zigzag" order employing run-length encoding (RLE) algorithm that groups similar frequencies together, inserting length coding zeros, and then using Huffman coding on what is left.

  

  Now i am so busy ,so i will give you a  square matrix that represents pixel intensities of the image.

  Your task is simple: reconstruct the image so that the value in the i^th row and j^th column of the resulting image is the value of the (i * N + j)^thpixel visited in the zigzag sequence .

Input

Your program will be tested on one or more test cases. The first line of the input contains a single integer T (1  ≤ T ≤  100) the number of test cases. Followed by T test cases.

Each test case consists of N+1 lines. The first line contains an integer N (2  ≤ N ≤  100). The next lines consists of an  squared pixel matrix.

Output

For each test case print the required transformed matrix.

Sample Input

Input

1 5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Output

1 2 6 11 7  3 4 8 12 16  21 17 13 9 5  10 14 18 22 23  19 15 20 24 25 
题解：水题，按照途中的划线访问，模拟下就好了；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 110;
int num[MAXN * MAXN];
int mp[MAXN][MAXN];
int main(){
    int T, N;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &N);
        for(int i = 1; i <= N; i++){
            for(int j = 1; j <= N; j++){
                scanf("%d", &mp[j][i]);
            }
        }
        int tp = 0;
        int x = 1, y = 1;
        num[tp++] = mp[x][y];
        while(tp < N*N){
            if(x + 1 <= N)
                num[tp++] = mp[++x][y];
            else if(y + 1 <= N)
                num[tp++] = mp[x][++y];
            while(x - 1 >= 1 && y + 1 <= N){
                num[tp++] = mp[--x][++y];
            }
            if(x == 1 && y + 1 <= N)
                num[tp++] = mp[x][++y];
            else if(y == N && x + 1 <= N)
                num[tp++] = mp[++x][y];
            while(x + 1 <= N && y - 1 >= 1){
                num[tp++] = mp[++x][--y];
            }
        }
        for(int i = 0; i < N; i++){
            for(int j = 0; j < N; j++){
                mp[i][j] = num[i * N + j];
            }
        }
        for(int i = 0; i < N; i++){
            for(int j = 0; j < N; j++){
                if(j)printf(" ");
                printf("%d", mp[i][j]);
            }
            puts("");
        }
    }
    return 0;
}