c语言实现集合的交并差运算

集合的交、并、差运算可以通过数组来实现，具体实现如下：

#include <stdio.h>

void intersection(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];
            i++;
            j++;
            k++;
        }
    }
}

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];
            i++;
            k++;
        } else if (b[j] < a[i]) {
            c[k] = b[j];
            j++;
            k++;
        } else {
            c[k] = a[i];
            i++;
            j++;
            k++;
        }
    }
    while (i < m) {
        c[k] = a[i];
        i++;
        k++;
    }
    while (j < n) {
        c[k] = b[j];
        j++;
        k++;
    }
}

void difference(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];
            i++;
            k++;
        } else if (b[j] < a[i]) {
            j++;
        } else {
            i++;
            j++;
        }
    }
    while (i < m) {
        c[k] = a[i];
        i++;
        k++;
    }
}

int main() {
    int a[] = {1, 2, 3, 4, 5};
    int b[] = {3, 4, 5, 6, 7, 8};
    int m = sizeof(a) / sizeof(a[0]);
    int n = sizeof(b) / sizeof(b[0]);
    int c[m + n], d[m + n], e[m + n];
    intersection(a, b, c, m, n);
    union_set(a, b, d, m, n);
    difference(a, b, e, m, n);
    printf("A ∩ B = { ");
    for (int i = 0; i < sizeof(c) / sizeof(c[0]); i++) {
        if (c[i] != 0) {
            printf("%d ", c[i]);
        }
    }
    printf("}\n");
    printf("A ∪ B = { ");
    for (int i = 0; i < sizeof(d) / sizeof(d[0]); i++) {
        if (d[i] != 0) {
            printf("%d ", d[i]);
        }
    }
    printf("}\n");
    printf("A - B = { ");
    for (int i = 0; i < sizeof(e) / sizeof(e[0]); i++) {
        if (e[i] != 0) {
            printf("%d ", e[i]);
        }
    }
    printf("}\n");
    return 0;
}

输出结果为：

A ∩ B = { 3 4 5 }
A ∪ B = { 1 2 3 4 5 6 7 8 }
A - B = { 1 2 }