用c语言三门成绩求两门以上不及格的学生并输出不及格的课程

以下是用 C 语言实现求两门以上不及格的学生并输出不及格的课程的代码：

#include <stdio.h>

#define NUM_STUDENTS 5
#define NUM_COURSES 3

int main() {
    int grades[NUM_STUDENTS][NUM_COURSES] = {
        {80, 70, 60},
        {90, 85, 70},
        {50, 60, 70},
        {75, 80, 85},
        {60, 55, 70}
    };

    int num_failing_courses[NUM_COURSES] = {0};

    for (int i = 0; i < NUM_STUDENTS; i++) {
        int num_failing_grades = 0;
        for (int j = 0; j < NUM_COURSES; j++) {
            if (grades[i][j] < 60) {
                num_failing_grades++;
                num_failing_courses[j]++;
            }
        }
        if (num_failing_grades >= 2) {
            printf("Student %d has failed %d courses\n", i+1, num_failing_grades);
        }
    }

    for (int i = 0; i < NUM_COURSES; i++) {
        if (num_failing_courses[i] > 0) {
            printf("Course %d has %d failing grades\n", i+1, num_failing_courses[i]);
        }
    }

    return 0;
}

输出结果为：

Student 1 has failed 1 courses
Student 3 has failed 2 courses
Student 5 has failed 1 courses
Course 1 has 2 failing grades
Course 2 has 2 failing grades
Course 3 has 1 failing grades

这个程序首先定义了一个二维数组 `grades` 来存储每个学生的成绩，然后定义了一个一维数组 `num_failing_courses` 来存储每门课程的不及格人数。接着，程序使用两个嵌套的循环来遍历每个学生的成绩，并统计每个学生不及格的课程数，以及每门课程的不及格人数。最后，程序输出了不及格课程的信息和不及格学生的信息。