判断一个矩形矩阵是否有两行正元素

Question

I need to determine if a rectangular matrix has two rows of positive elements in C. I write part code for the set matrix and output its elements. I don't know how to check the positive elements in the row. Please help me

#include <stdio.h>
#include <conio.h>
#include <locale.h>

#define M 3
#define N 4

int main() {
    setlocale(LC_ALL, "Rus");
    float a[M][N]; //set matrix with 3 row and 4 column
    int i, j;     // row and column index
    for (i = 0; i < M; i++) {
        for (j = 0; j < N; j++)
            scanf_s("%f", &a[i][j]);
    }
    for (i = 0; i < M; i++) {
        printf("%d line:", i + 1);
        for (j = 0; j < N; j++)
            printf("%f", a[i][j]);
        printf("\n");
    }
    _getch();
    return 0;
}

So, I make some changes in my code after reading comments. Thanks a lot. But it's not working.

#include <stdio.h>
#include <conio.h>
#include <locale.h>

#define M 3
#define N 4

int main(){
    setlocale(LC_ALL, "Rus");
    float a[M][N]; //обьявление матрицы 3 строки и 4 столбца
    int i, j;     // индексы строки и столбца
    int count;
    for (i = 0; i < M; i++){
        for (j = 0; j < N; j++)
            scanf_s("%f", &a[i][j]);
    }
    for (i = 0; i < M; i++){
        printf("%d-я строка:", i + 1);
        for (j = 0; j < N; j++)
            printf("%f", a[i][j]);
        printf("\n");
    }
    count = 0;
    for (i = 0; i < M; i++){
        for (j = 0; j < N; j++)
            if (a[i][j] > 0){
                count++;
                printf("%d", count);
        }
    }
    _getch();
    return 0;
}

Answer 1

What you need to do is to set a counter to 0. Every time all the values are positive in row add 1 your counter. You can do that by a loop. If a value of your matrix a[i][j] is less than 0 just continue and move on to then next row.