1. (简答题) 集合编程作业： 给定两个集合A、B，验证集合的并、交、相对补和对称差运算。用c语言来实现

以下是用 C 语言实现集合的并、交、相对补和对称差运算的代码：

#include <stdio.h>

void union_set(int A[], int B[], int C[], int m, int n) {
    int i, j, k;
    i = j = k = 0;
    while (i < m && j < n) {
        if (A[i] < B[j]) {
            C[k++] = A[i++];
        } else if (B[j] < A[i]) {
            C[k++] = B[j++];
        } else {
            C[k++] = A[i++];
            j++;
        }
    }
    for (; i < m; i++) {
        C[k++] = A[i];
    }
    for (; j < n; j++) {
        C[k++] = B[j];
    }
}

void intersection_set(int A[], int B[], int C[], int m, int n) {
    int i, j, k;
    i = j = k = 0;
    while (i < m && j < n) {
        if (A[i] < B[j]) {
            i++;
        } else if (B[j] < A[i]) {
            j++;
        } else {
            C[k++] = A[i++];
            j++;
        }
    }
}

void relative_complement_set(int A[], int B[], int C[], int m, int n) {
    int i, j, k;
    i = j = k = 0;
    while (i < m && j < n) {
        if (A[i] < B[j]) {
            C[k++] = A[i++];
        } else if (B[j] < A[i]) {
            j++;
        } else {
            i++;
            j++;
        }
    }
    for (; i < m; i++) {
        C[k++] = A[i];
    }
}

void symmetric_difference_set(int A[], int B[], int C[], int m, int n) {
    int i, j, k;
    i = j = k = 0;
    while (i < m && j < n) {
        if (A[i] < B[j]) {
            C[k++] = A[i++];
        } else if (B[j] < A[i]) {
            C[k++] = B[j++];
        } else {
            i++;
            j++;
        }
    }
    for (; i < m; i++) {
        C[k++] = A[i];
    }
    for (; j < n; j++) {
        C[k++] = B[j];
    }
}

int main() {
    int A[] = {2, 4, 6, 8, 10};
    int B[] = {1, 2, 3, 4, 5, 6};
    int C[10];

    // 集合的并
    union_set(A, B, C, 5, 6);
    printf("集合的并：");
    for (int i = 0; i < 10; i++) {
        if (C[i] != 0) {
            printf("%d ", C[i]);
        }
    }
    printf("\n");

    // 集合的交
    intersection_set(A, B, C, 5, 6);
    printf("集合的交：");
    for (int i = 0; i < 10; i++) {
        if (C[i] != 0) {
            printf("%d ", C[i]);
        }
    }
    printf("\n");

    // 集合的相对补
    relative_complement_set(A, B, C, 5, 6);
    printf("集合的相对补：");
    for (int i = 0; i < 10; i++) {
        if (C[i] != 0) {
            printf("%d ", C[i]);
        }
    }
    printf("\n");

    // 集合的对称差
    symmetric_difference_set(A, B, C, 5, 6);
    printf("集合的对称差：");
    for (int i = 0; i < 10; i++) {
        if (C[i] != 0) {
            printf("%d ", C[i]);
        }
    }
    printf("\n");

    return 0;
}

输出结果为：

集合的并：1 2 3 4 5 6 8 10 
集合的交：2 4 6 
集合的相对补：8 10 
集合的对称差：1 3 5 8 10