c语言实现集合的并交差

C语言可以通过数组来实现集合的并、交、差运算。以下是一个简单的示例程序：

#include <stdio.h>

// 定义集合数组
int set1[] = {1, 3, 5, 7, 9};
int set2[] = {2, 4, 6, 8, 10};
int set3[] = {1, 2, 3, 4, 5};
int set4[] = {4, 5, 6, 7, 8};

// 计算集合并
void set_union(int set1[], int size1, int set2[], int size2, int result[], int *result_size) {
    int i, j, k;
    i = j = k = 0;
    while (i < size1 && j < size2) {
        if (set1[i] < set2[j]) {
            result[k++] = set1[i++];
        } else if (set1[i] > set2[j]) {
            result[k++] = set2[j++];
        } else {
            result[k++] = set1[i++];
            j++;
        }
    }
    while (i < size1) {
        result[k++] = set1[i++];
    }
    while (j < size2) {
        result[k++] = set2[j++];
    }
    *result_size = k;
}

// 计算集合交
void set_intersection(int set1[], int size1, int set2[], int size2, int result[], int *result_size) {
    int i, j, k;
    i = j = k = 0;
    while (i < size1 && j < size2) {
        if (set1[i] < set2[j]) {
            i++;
        } else if (set1[i] > set2[j]) {
            j++;
        } else {
            result[k++] = set1[i++];
            j++;
        }
    }
    *result_size = k;
}

// 计算集合差
void set_difference(int set1[], int size1, int set2[], int size2, int result[], int *result_size) {
    int i, j, k;
    i = j = k = 0;
    while (i < size1 && j < size2) {
        if (set1[i] < set2[j]) {
            result[k++] = set1[i++];
        } else if (set1[i] > set2[j]) {
            j++;
        } else {
            i++;
            j++;
        }
    }
    while (i < size1) {
        result[k++] = set1[i++];
    }
    *result_size = k;
}

int main() {
    int result[10];
    int result_size;
    set_union(set1, 5, set2, 5, result, &result_size);
    printf("set1 union set2 = { ");
    for (int i = 0; i < result_size; i++) {
        printf("%d ", result[i]);
    }
    printf("}\n");
    set_intersection(set3, 5, set4, 5, result, &result_size);
    printf("set3 intersection set4 = { ");
    for (int i = 0; i < result_size; i++) {
        printf("%d ", result[i]);
    }
    printf("}\n");
    set_difference(set1, 5, set3, 5, result, &result_size);
    printf("set1 difference set3 = { ");
    for (int i = 0; i < result_size; i++) {
        printf("%d ", result[i]);
    }
    printf("}\n");
    return 0;
}

输出结果如下：

set1 union set2 = { 1 2 3 4 5 6 7 8 9 10 }
set3 intersection set4 = { 4 5 }
set1 difference set3 = { 7 9 }

注意，在这个示例程序中，我们假设集合元素都是整数，并且集合元素没有重复。如果集合元素可以重复，我们需要对算法做一些修改。