对功率大于50马力的机器、维修记录不全或已运行10以上的机器给予优先的维修处理的C语言程序代码及设计决策表和测试用例

以下是C语言程序代码及设计决策表和测试用例：

#include <stdio.h>
#include <stdlib.h>

#define MIN_POWER 50
#define MAX_AGE 10

struct Machine {
    int id;
    int power;
    int age;
};

int isEligible(struct Machine m) {
    return (m.power >= MIN_POWER && (m.age >= MAX_AGE || m.age < 0));
}

void repairMachine(struct Machine m) {
    printf("Machine with ID %d is being repaired.\n", m.id);
    // Add repair logic here
}

int main() {
    int numMachines;
    printf("Enter number of machines: ");
    scanf("%d", &numMachines);

    struct Machine *machines = malloc(numMachines * sizeof(struct Machine));

    // Read machine details
    for (int i = 0; i < numMachines; i++) {
        printf("Enter details of machine %d:\n", i + 1);
        printf("ID: ");
        scanf("%d", &machines[i].id);
        printf("Power: ");
        scanf("%d", &machines[i].power);
        printf("Age: ");
        scanf("%d", &machines[i].age);
    }

    // Repair machines
    for (int i = 0; i < numMachines; i++) {
        if (isEligible(machines[i])) {
            repairMachine(machines[i]);
        }
    }

    free(machines);
    return 0;
}

设计决策表：

|条件|动作|
|----|----|
|功率>=50且机器的年龄>=10或机器的年龄<0|维修机器|

测试用例：

|ID|功率|年龄|是否维修|
|--|----|----|--------|
|1|40|5|否|
|2|60|5|是|
|3|40|15|是|
|4|60|15|是|
|5|40|-5|是|
|6|60|-5|是|