c语言实现集合的交、并、差和补运算

C语言中可以使用数组来实现集合的交、并、差和补运算。

交集运算：

#include <stdio.h>

void intersection(int a[], int b[], int n, int m) {
    int i, j;
    printf("Intersection of two sets:\n");
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            if (a[i] == b[j]) {
                printf("%d ", a[i]);
                break;
            }
        }
    }
}

int main() {
    int a[] = {1, 2, 3, 4, 5};
    int b[] = {3, 4, 5, 6, 7};
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);
    intersection(a, b, n, m);
    return 0;
}

并集运算：

#include <stdio.h>

void union_set(int a[], int b[], int n, int m) {
    int i, j, k = 0;
    int c[n + m];
    printf("Union of two sets:\n");
    for (i = 0; i < n; i++) {
        c[k++] = a[i];
    }
    for (i = 0; i < m; i++) {
        for (j = 0; j < n; j++) {
            if (b[i] == a[j]) {
                break;
            }
        }
        if (j == n) {
            c[k++] = b[i];
        }
    }
    for (i = 0; i < k; i++) {
        printf("%d ", c[i]);
    }
}

int main() {
    int a[] = {1, 2, 3, 4, 5};
    int b[] = {3, 4, 5, 6, 7};
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);
    union_set(a, b, n, m);
    return 0;
}

差集运算：

#include <stdio.h>

void difference(int a[], int b[], int n, int m) {
    int i, j, k = 0;
    int c[n];
    printf("Difference of two sets:\n");
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            if (a[i] == b[j]) {
                break;
            }
        }
        if (j == m) {
            c[k++] = a[i];
        }
    }
    for (i = 0; i < k; i++) {
        printf("%d ", c[i]);
    }
}

int main() {
    int a[] = {1, 2, 3, 4, 5};
    int b[] = {3, 4, 5, 6, 7};
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);
    difference(a, b, n, m);
    return 0;
}

补集运算：

#include <stdio.h>

void complement(int a[], int b[], int n, int m) {
    int i, j, k = 0;
    int c[n];
    printf("Complement of set B in set A:\n");
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            if (a[i] == b[j]) {
                break;
            }
        }
        if (j == m) {
            c[k++] = a[i];
        }
    }
    for (i = 0; i < k; i++) {
        printf("%d ", c[i]);
    }
}

int main() {
    int a[] = {1, 2, 3, 4, 5};
    int b[] = {3, 4, 5};
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);
    complement(a, b, n, m);
    return 0;
}