練習

請用二維陣列輸出九九乘法表

a015: 矩陣的翻轉 https://zerojudge.tw/ShowProblem?problemid=a015

b367: 翻轉世界 https://zerojudge.tw/ShowProblem?problemid=b367

a694: 吞食天地二 https://zerojudge.tw/ShowProblem?problemid=a694

d625: 踩地雷真好玩(較困難，本堂課可以不做) https://zerojudge.tw/ShowProblem?problemid=d625

解答

a015: 矩陣的翻轉

從對角線對稱翻轉，所以把arr[i][j]填入arr[j][i]即可

#include <iostream>
using namespace std;

int main() {

    int arr[100][100];
    int x, y;
    while (cin >> x >> y) {
        for (int i= 0; i < x;i++){
            for (int j= 0; j < y;j++){
                cin >> arr[i][j];
            }
        }
        for (int i= 0; i < y;i++){
            for (int j= 0; j < x;j++){
                cout << arr[j][i] << " ";
            }
            cout << endl;
        }
    }

    return 0;
}

b367: 翻轉世界

旋轉180度，等於把整個陣列給reverse

測資解釋

範例輸入

2      <= 代表2個二維陣列

3 3    <= 第1個二維陣列是 3 X 3
1 1 0  <= 3 X 3 的二維陣列
0 1 0
0 1 1

1 5    <= 第2個二維陣列是 1 X 5
0 0 1 0 1 <= 1 X 5 的二維陣列

範例輸出

go forward     <= 第1個二維陣列轉180度之後，還是長得一樣
keep defending <= 第2個二維陣列，長得不一樣

#include <iostream>
using namespace std;

int main()
{

    int dataCount, row, col, arr[11][11], arr180[11][11];

    cin >> dataCount;

    while (dataCount--) {

        cin >> row >> col;

            for (int i = 0; i < row; i++)
            {
                for (int j = 0; j < col; j++)
                {
                    cin >> arr[i][j];
                    arr180[row - i - 1][col - j - 1] = arr[i][j];
                }
            }
            bool flag = false;
            for (int i = 0; i < row; i++)
            {
                for (int j = 0; j < col; j++)
                {
                    if (arr[i][j] != arr180[i][j]) {
                        flag = true;
                        break;
                    }
                }
            }
            cout << (flag ? "keep defending" : "go forward") << endl;
        }

    return 0;
}

a694: 吞食天地二

測資解釋

3 3 <= 第一個3代表下面會有3*3的正方形，後面3那個代表正方形下面會有3行

1 2 3 <= 3*3正方形
4 5 6
7 8 9

1 1 3 3 <= 代表座標(1,1)與(3,3)圍成的正方形
1 1 1 3 <= 代表座標(1,1)與(1,3)圍成的長方形
1 1 3 1 <= 代表座標(1,1)與(3,1)圍成的長方形

#include <iostream>
    using namespace std;

int main() {

    int arr[500][500];
    int n, m, x1[10000], y1[10000], x2[10000], y2[10000];
    int sum = 0;

    cin >> n >> m;

    for (int i= 0; i < n;i++){
        for (int j= 0; j < n;j++){
            cin >> arr[i][j];
        }
    }


    for (int i= 0; i < m;i++){
        cin >> x1[i];
        cin >> y1[i];
        cin >> x2[i];
        cin >> y2[i];
    }


    int eatTimes = 0;
    while (eatTimes<m){
        sum = 0;
        for (int i= x1[eatTimes]-1; i <= x2[eatTimes]-1;i++){
            for(int j= y1[eatTimes]-1; j <= y2[eatTimes]-1;j++){
                sum += arr[i][j];
            }
        }
        eatTimes++;
        cout<<sum<<endl;
    }

}

d625: 踩地雷真好玩

二維地圖走 八方位

#include <bits/stdc++.h>
using namespace std;

int N;
char arr[105][105];
int cnt[105][105];

int dx[]={ -1, 0, 1, -1, 1, -1, 0, 1 };
int dy[]={ -1, -1, -1 , 0 , 0 , 1 , 1 , 1 };
int main(){
    cin >> N;

    for ( int i = 0 ; i < N; i++ )
        cin >> arr[i];
    for ( int i = 0 ; i < N; i++ )
        for ( int j = 0 ; j < N; j++ )
            cnt[i][j] = 0;
    for ( int i = 0 ; i < N; i++ )
        for ( int j = 0 ; j < N; j++ )
            if ( arr[i][j] == '-' ){
                for ( int d = 0 ; d < 8 ; d++ ){
                    int nx = j + dx[d];
                    int ny = i + dy[d];    
                    if ( nx < 0 || ny < 0 || nx >= N || ny >= N )
                        continue;
                    else{
                        if ( arr[ny][nx] == '*' )    
                            cnt[i][j]++;

                    }
                }
            }

    for ( int i = 0 ; i < N ; i++ ){
        for ( int j = 0 ; j < N ; j++ ){

            if ( arr[i][j] == '*' )
                cout << '*';
            else if ( cnt[i][j] == 0 )
                cout << '-';
            else
                cout << cnt[i][j];
        }
        cout << endl;
    }


    return 0;    
}