请用C语言实现:输入两个集合A、B，验证集合的并、交、相对补和对称差运算

以下是使用C语言实现输入两个集合A、B，验证集合的并、交、相对补和对称差运算的示例代码：

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

#define MAX_SIZE 100

void input_set(int set[], int *size) {
    printf("请输入集合大小：");
    scanf("%d", size);

    printf("请输入集合元素：");
    for (int i = 0; i < *size; i++) {
        scanf("%d", &set[i]);
    }
}

void print_set(int set[], int size) {
    printf("{ ");
    for (int i = 0; i < size; i++) {
        printf("%d ", set[i]);
    }
    printf("}\n");
}

void union_set(int set1[], int size1, int set2[], int size2, int result[], int *size) {
    int i, j, k;
    for (i = 0; i < size1; i++) {
        result[i] = set1[i];
    }
    k = size1;
    for (i = 0; i < size2; i++) {
        for (j = 0; j < size1; j++) {
            if (set2[i] == set1[j]) {
                break;
            }
        }
        if (j == size1) {
            result[k++] = set2[i];
        }
    }
    *size = k;
}

void intersection_set(int set1[], int size1, int set2[], int size2, int result[], int *size) {
    int i, j, k;
    k = 0;
    for (i = 0; i < size1; i++) {
        for (j = 0; j < size2; j++) {
            if (set1[i] == set2[j]) {
                result[k++] = set1[i];
                break;
            }
        }
    }
    *size = k;
}

void difference_set(int set1[], int size1, int set2[], int size2, int result[], int *size) {
    int i, j, k;
    k = 0;
    for (i = 0; i < size1; i++) {
        for (j = 0; j < size2; j++) {
            if (set1[i] == set2[j]) {
                break;
            }
        }
        if (j == size2) {
            result[k++] = set1[i];
        }
    }
    *size = k;
}

void symmetric_difference_set(int set1[], int size1, int set2[], int size2, int result[], int *size) {
    int i, j, k;
    k = 0;
    for (i = 0; i < size1; i++) {
        for (j = 0; j < size2; j++) {
            if (set1[i] == set2[j]) {
                break;
            }
        }
        if (j == size2) {
            result[k++] = set1[i];
        }
    }
    for (i = 0; i < size2; i++) {
        for (j = 0; j < size1; j++) {
            if (set2[i] == set1[j]) {
                break;
            }
        }
        if (j == size1) {
            result[k++] = set2[i];
        }
    }
    *size = k;
}

int main() {
    int set1[MAX_SIZE], set2[MAX_SIZE], result[MAX_SIZE], size1, size2, result_size;

    input_set(set1, &size1);
    input_set(set2, &size2);

    printf("\n集合A：");
    print_set(set1, size1);
    printf("集合B：");
    print_set(set2, size2);

    union_set(set1, size1, set2, size2, result, &result_size);
    printf("集合A和B的并集为：");
    print_set(result, result_size);

    intersection_set(set1, size1, set2, size2, result, &result_size);
    printf("集合A和B的交集为：");
    print_set(result, result_size);

    difference_set(set1, size1, set2, size2, result, &result_size);
    printf("集合A和B的相对补集为：");
    print_set(result, result_size);

    symmetric_difference_set(set1, size1, set2, size2, result, &result_size);
    printf("集合A和B的对称差集为：");
    print_set(result, result_size);

    return 0;
}

运行结果示例：

请输入集合大小：5
请输入集合元素：1 2 3 4 5
请输入集合大小：4
请输入集合元素：3 4 5 6

集合A：{ 1 2 3 4 5 }
集合B：{ 3 4 5 6 }
集合A和B的并集为：{ 1 2 3 4 5 6 }
集合A和B的交集为：{ 3 4 5 }
集合A和B的相对补集为：{ 1 2 }
集合A和B的对称差集为：{ 1 2 6 }