#include <stdio.h>int main() { int set1[100], set2[100], intersection[100], unionSet[200], complement[100], i, j, k, n1, n2, n3 = 0; // 输入第一个集合 printf("Enter number of elements in set1: "); scanf("%d", &n1); printf("Enter elements of set1: "); for (i = 0; i < n1; i++) { scanf("%d", &set1[i]); } // 输入第二个集合 printf("Enter number of elements in set2: "); scanf("%d", &n2); printf("Enter elements of set2: "); for (i = 0; i < n2; i++) { scanf("%d", &set2[i]); } // 计算交集 k = 0; for (i = 0; i < n1; i++) { for (j = 0; j < n2; j++) { if (set1[i] == set2[j]) { intersection[k++] = set1[i]; break; } } } printf("Intersection of set1 and set2: "); for (i = 0; i < k; i++) { printf("%d ", intersection[i]); } printf("\n"); // 计算并集 k = 0; for (i = 0; i < n1; i++) { unionSet[k++] = set1[i]; } for (i = 0; i < n2; i++) { for (j = 0; j < n1; j++) { if (set2[i] == set1[j]) { break; } } if (j == n1) { unionSet[k++] = set2[i]; } } printf("Union of set1 and set2: "); for (i = 0; i < k; i++) { printf("%d ", unionSet[i]); } printf("\n"); // 计算补集 k = 0; for (i = 0; i < n1; i++) { for (j = 0; j < n2; j++) { if (set1[i] == set2[j]) { break; } } if (j == n2) { complement[k++] = set1[i]; } } printf("Complement of set1 with respect to set2: "); for (i = 0; i < k; i++) { printf("%d ", complement[i]); } printf("\n"); return 0;}
时间: 2024-04-07 11:28:52 浏览: 76
这段代码实现了离散数学中集合的交集、并集和补集的计算,输入两个集合的元素个数和元素值,然后输出它们的交集、并集和补集。
具体实现方法是用数组来存储集合中的元素,分别用两个数组 set1 和 set2 存储输入的两个集合,用数组 intersection 存储它们的交集,用数组 unionSet 存储它们的并集,用数组 complement 存储 set1 相对于 set2 的补集。
程序首先输入两个集合,然后计算它们的交集、并集和补集,并输出结果。
相关问题
#include <stdio.h>#define MAX_SIZE 10int main() { int set1[MAX_SIZE], set2[MAX_SIZE], union_set[MAX_SIZE], intersection_set[MAX_SIZE], complement_set[MAX_SIZE]; int n1, n2, i, j, k, flag; // 输入第一个集合 printf("Enter the size of set 1: "); scanf("%d", &n1); printf("Enter the elements of set 1: "); for (i = 0; i < n1; i++) { scanf("%d", &set1[i]); } // 输入第二个集合 printf("Enter the size of set 2: "); scanf("%d", &n2); printf("Enter the elements of set 2: "); for (i = 0; i < n2; i++) { scanf("%d", &set2[i]); } // 计算并集 k = 0; for (i = 0; i < n1; i++) { union_set[k++] = set1[i]; } for (i = 0; i < n2; i++) { flag = 1; for (j = 0; j < n1; j++) { if (set2[i] == set1[j]) { flag = 0; break; } } if (flag) { union_set[k++] = set2[i]; } } printf("The union of set 1 and set 2 is: "); for (i = 0; i < k; i++) { printf("%d ", union_set[i]); } printf("\n"); // 计算交集 k = 0; for (i = 0; i < n1; i++) { for (j = 0; j < n2; j++) { if (set1[i] == set2[j]) { intersection_set[k++] = set1[i]; } } } printf("The intersection of set 1 and set 2 is: "); for (i = 0; i < k; i++) { printf("%d ", intersection_set[i]); } printf("\n"); // 计算补集 k = 0; for (i = 0; i < n1; i++) { flag = 1; for (j = 0; j < n2; j++) { if (set1[i] == set2[j]) { flag = 0; break; } } if (flag) { complement_set[k++] = set1[i]; } } printf("The complement of set 1 with respect to set 2 is: "); for (i = 0; i < k; i++) { printf("%d ", complement_set[i]); } printf("\n"); return 0;}解释一下这个程序
这是一个 C 语言程序,用于计算两个集合的并集、交集和补集。程序首先定义了三个数组,用于存储并集、交集和补集。然后程序要求用户输入两个集合的元素,分别存储在 set1 和 set2 数组中。程序使用 for 循环计算并集,并将结果存储在 union_set 数组中。计算交集的方式也是使用 for 循环,将结果存储在 intersection_set 数组中。最后,程序计算 set1 相对于 set2 的补集,并将结果存储在 complement_set 数组中。程序最后输出计算结果。
用c写一段代码,实现“A = set(map(int, input().strip().split())) B = set(map(int, input().strip().split())) # 计算并集和交集 union = A.union(B) intersection = A.intersection(B) # 输出结果 print("{" + ",".join(map(str, union)) + "}" if union else "{}") print("{" + ",".join(map(str, intersection)) + "}" if intersection else "{}")”
在 C 语言中,没有现成的集合类型,但可以使用数组和指针等基本数据类型来模拟集合的实现。下面是一个示例代码:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_LEN 100 // 数组的最大长度
// 定义集合类型
typedef struct Set {
int *data; // 数据指针
int size; // 集合大小
} Set;
// 创建一个新集合
Set *new_set() {
Set *s = (Set *)malloc(sizeof(Set));
s->data = (int *)calloc(MAX_LEN, sizeof(int));
s->size = 0;
return s;
}
// 向集合中加入元素
void add_element(Set *s, int e) {
for (int i = 0; i < s->size; i++) {
if (s->data[i] == e) {
return; // 如果元素已经在集合中了,则不加入
}
}
s->data[s->size++] = e;
}
// 计算两个集合的并集
Set *set_union(Set *A, Set *B) {
Set *s = new_set();
for (int i = 0; i < A->size; i++) {
add_element(s, A->data[i]);
}
for (int i = 0; i < B->size; i++) {
add_element(s, B->data[i]);
}
return s;
}
// 计算两个集合的交集
Set *set_intersection(Set *A, Set *B) {
Set *s = new_set();
for (int i = 0; i < A->size; i++) {
for (int j = 0; j < B->size; j++) {
if (A->data[i] == B->data[j]) {
add_element(s, A->data[i]);
break;
}
}
}
return s;
}
// 打印集合
void print_set(Set *s) {
printf("{");
for (int i = 0; i < s->size; i++) {
printf("%d", s->data[i]);
if (i != s->size - 1) {
printf(",");
}
}
printf("}");
}
int main() {
Set *A = new_set();
Set *B = new_set();
// 读取输入集合A和B
char input[MAX_LEN];
fgets(input, MAX_LEN, stdin);
char *token = strtok(input, " \n");
while (token != NULL) {
add_element(A, atoi(token));
token = strtok(NULL, " \n");
}
fgets(input, MAX_LEN, stdin);
token = strtok(input, " \n");
while (token != NULL) {
add_element(B, atoi(token));
token = strtok(NULL, " \n");
}
// 计算并集和交集
Set *union_set = set_union(A, B);
Set *intersection_set = set_intersection(A, B);
// 输出结果
printf(" ");
print_set(union_set);
printf("\n ");
print_set(intersection_set);
printf("\n");
// 释放内存
free(A->data);
free(A);
free(B->data);
free(B);
free(union_set->data);
free(union_set);
free(intersection_set->data);
free(intersection_set);
return 0;
}
```
需要注意的是,C 语言中没有现成的字符串类型,需要使用字符数组来存储和处理输入输出。另外,由于没有动态数组类型,需要手动申请和释放内存。
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