c - 如何在C中检查矩阵的行或列是否具有相同的值

Question

我的家庭作业有问题。它要求我们根据用户的输入制作一个矩阵。例如:如果用户输入 4，则矩阵将为 4 X 4。之后，程序将检查矩阵的行或列是否具有相同的值。它会给出 yes 或 no 输出。

例如:

输入:

21 22 1

Output :

Yes

(because that matrix doesnt has a same value in a row or a column.)

Input 2 :

34 5 67 8 97 3 3

Output :

No 

(Because that matrix have same values in a row or column (3 & 3 and 7 & 7)

Input 3:

21 23 2

Output :

No 

(because that matrix have same value on column 1.)

Input 4

21 13 4

Output :

No

(because that matrix has same value in first row(1 1)

I have tried to do that, but some 'cases' still doesnt work. For example, i tried to include a count in my code but some of the count is not true.

example :input :

43 4 5 62 3 4 56 5 6 35 4 6 3

OUTPUT :

Nocount : 2

(it supposed to be 3 because it has the same value which are 6 (on row 3), 6 on column 3, and 3 on column 4.)

#include "stdio.h"

int main()
{

    int matrix[500][500];
    int testcase;
    int count = 0;
    scanf("%d",&testcase); getchar();

    for(unsigned i = 0; i < testcase; i++) {
        for(unsigned j = 0; j < testcase; j++) {
            scanf("%d",&matrix[i][j]); getchar();
        }
    }
    // printf("[0,0] = %c",matrix[0][0]);
    // printf("\n[0,1] = %c",matrix[0][1]);
    // printf("\n[1,0] = %c",matrix[1][0]);
    // printf("\n[1,1] = %c",matrix[1][1]);

    for(unsigned i = 0; i < testcase; i++) {
        for(unsigned j = 0; j < testcase; j++) {
            if(matrix[i][j] == matrix[i][j+1]) {
                count = count + 1;
            }
            else if(matrix[i][j] == matrix[i+1][j]) {
                count = count + 1;
            }

        }
    }
    if(count > 0) {
        printf("No\n");
    } else{
        printf("Yes\n");
    }

    printf("Count : %d\n",count );
    getchar();
    return 0;
}

Answer 1

正如我所见，您检查两个具有相同值的数字是否仅一列或一行不同:

if(matrix[i][j] == matrix[i][j+1]) {
      count = count + 1;
}
else if(matrix[i][j] == matrix[i+1][j]) {
      count = count + 1;
}

我认为您可能需要一个临时变量，以便您可以分别扫描每一行，然后扫描每一列，例如:

temp = matrix[i][j];
if(checkRow(temp, i, j, matrix, testcase) == true) count++;
if(checkColumn(temp, i, j, matrix, testcase) == true) count++;

checkRow 会是这样的:

bool checkRow(int temp, int row, int col, int matrix[][500], int size)
{
   for(int i=col; i < size;)
   {
      if(temp == matrix[row][i]) return true;
   }

    return false;
}

您将分别构建 checkColumn 函数。

编辑:既然你告诉我你还没有学会如何使用函数，这将是你的最终程序。它有效，我可能建议最终的测试用例应该输出“count = 4”，因为您可能会错过一个案例。这是代码:

#include "stdio.h"
int main()
{

    int matrix[500][500];
    int testcase;
    int count = 0;
    scanf("%d",&testcase); getchar();
    int temp;
    for(unsigned i = 0; i < testcase; i++) {
        for(unsigned j = 0; j < testcase; j++) {
            scanf("%d",&matrix[i][j]); getchar();
        }
    }
    // printf("[0,0] = %c",matrix[0][0]);
    // printf("\n[0,1] = %c",matrix[0][1]);
    // printf("\n[1,0] = %c",matrix[1][0]);
    // printf("\n[1,1] = %c",matrix[1][1]);

    for(unsigned i = 0; i < testcase; i++) {
        for(unsigned j = 0; j < testcase; j++) {
            temp = matrix[i][j];
            //Scan current row
            for(unsigned k = j+1; k < testcase; k++)
            {
                if(temp == matrix[i][k])
                {
                    count++;
                    break;
                }
            }
            //Scan current column
            for(unsigned k = i+1; k < testcase; k++)
            {
                if(temp == matrix[k][j])
                {
                    count ++;
                    break;
                }
            }
        }
    }
    if(count > 0) {
        printf("No\n");
    } else{
        printf("Yes\n");
    }

    printf("Count : %d\n",count );
    getchar();
    return 0;
}

我建议您在复制代码之前必须了解其背后的算法。这是简单而暴力的思维。