使用C语言实现某集合(假设为整数集合)上二元关系的输入、输出、运算(交、差、并、补、逆、复合、闭包)和判断(判断是否自反、对称和传递)。
时间: 2023-07-03 21:07:56 浏览: 53
以下是使用C语言实现整数集合上二元关系的输入、输出、运算和判断的示例代码:
```c
#include <stdio.h>
#include <stdlib.h>
#define MAXSIZE 100
typedef struct {
int data[MAXSIZE];
int length;
} Set;
void initSet(Set *s) {
s->length = 0;
}
void printSet(Set *s) {
int i;
printf("{");
for (i = 0; i < s->length; i++) {
printf("%d", s->data[i]);
if (i < s->length - 1) {
printf(", ");
}
}
printf("}\n");
}
void inputSet(Set *s) {
int i, n;
printf("请输入集合元素个数:");
scanf("%d", &n);
printf("请输入集合元素:");
for (i = 0; i < n; i++) {
scanf("%d", &(s->data[i]));
}
s->length = n;
}
void intersection(Set *a, Set *b, Set *c) {
int i, j;
c->length = 0;
for (i = 0; i < a->length; i++) {
for (j = 0; j < b->length; j++) {
if (a->data[i] == b->data[j]) {
c->data[c->length++] = a->data[i];
break;
}
}
}
}
void difference(Set *a, Set *b, Set *c) {
int i, j, flag;
c->length = 0;
for (i = 0; i < a->length; i++) {
flag = 1;
for (j = 0; j < b->length; j++) {
if (a->data[i] == b->data[j]) {
flag = 0;
break;
}
}
if (flag) {
c->data[c->length++] = a->data[i];
}
}
}
void unionSet(Set *a, Set *b, Set *c) {
int i, j;
c->length = 0;
for (i = 0; i < a->length; i++) {
c->data[c->length++] = a->data[i];
}
for (i = 0; i < b->length; i++) {
for (j = 0; j < c->length; j++) {
if (b->data[i] == c->data[j]) {
break;
}
}
if (j >= c->length) {
c->data[c->length++] = b->data[i];
}
}
}
void complement(Set *a, Set *b, Set *c) {
difference(b, a, c);
}
void inverse(Set *r, Set *i) {
int j, k;
i->length = 0;
for (j = 0; j < r->length; j++) {
for (k = 0; k < r->length; k++) {
if (r->data[k] == j && k != j) {
i->data[i->length++] = r->data[j];
break;
}
}
}
}
void composition(Set *r, Set *s, Set *c) {
int i, j, k;
c->length = 0;
for (i = 0; i < r->length; i++) {
for (j = 0; j < s->length; j++) {
if (r->data[i] == s->data[j]) {
for (k = 0; k < c->length; k++) {
if (c->data[k] == j) {
break;
}
}
if (k >= c->length) {
c->data[c->length++] = j;
}
}
}
}
}
int isReflexive(Set *r) {
int i, flag;
for (i = 0; i < r->length; i++) {
flag = 0;
for (j = 0; j < r->length; j++) {
if (r->data[j] == i && j != i) {
flag = 1;
break;
}
}
if (!flag) {
return 0;
}
}
return 1;
}
int isSymmetric(Set *r) {
int i, j, flag;
for (i = 0; i < r->length; i++) {
for (j = 0; j < r->length; j++) {
if (r->data[i] == j && r->data[j] == i && i != j) {
flag = 1;
break;
}
}
if (!flag) {
return 0;
}
}
return 1;
}
int isTransitive(Set *r) {
int i, j, k, flag;
for (i = 0; i < r->length; i++) {
for (j = 0; j < r->length; j++) {
if (r->data[i] == j) {
for (k = 0; k < r->length; k++) {
if (r->data[k] == i && r->data[j] == k && j != k) {
flag = 1;
break;
}
}
if (!flag) {
return 0;
}
}
}
}
return 1;
}
int main() {
Set a, b, c;
int choice;
do {
printf("1.输入集合\n");
printf("2.输出集合\n");
printf("3.求交集\n");
printf("4.求差集\n");
printf("5.求并集\n");
printf("6.求补集\n");
printf("7.求逆关系\n");
printf("8.求复合关系\n");
printf("9.判断是否自反\n");
printf("10.判断是否对称\n");
printf("11.判断是否传递\n");
printf("0.退出\n");
printf("请输入操作:");
scanf("%d", &choice);
switch (choice) {
case 1:
printf("输入集合A:\n");
inputSet(&a);
printf("输入集合B:\n");
inputSet(&b);
break;
case 2:
printf("集合A:");
printSet(&a);
printf("集合B:");
printSet(&b);
break;
case 3:
intersection(&a, &b, &c);
printf("A ∩ B = ");
printSet(&c);
break;
case 4:
difference(&a, &b, &c);
printf("A - B = ");
printSet(&c);
difference(&b, &a, &c);
printf("B - A = ");
printSet(&c);
break;
case 5:
unionSet(&a, &b, &c);
printf("A ∪ B = ");
printSet(&c);
break;
case 6:
complement(&a, &b, &c);
printf("B - A = ");
printSet(&c);
complement(&b, &a, &c);
printf("A - B = ");
printSet(&c);
break;
case 7:
printf("输入关系r:\n");
inputSet(&a);
inverse(&a, &c);
printf("r^-1 = ");
printSet(&c);
break;
case 8:
printf("输入关系r:\n");
inputSet(&a);
printf("输入关系s:\n");
inputSet(&b);
composition(&a, &b, &c);
printf("r o s = ");
printSet(&c);
break;
case 9:
printf("输入关系r:\n");
inputSet(&a);
if (isReflexive(&a)) {
printf("r是自反关系\n");
} else {
printf("r不是自反关系\n");
}
break;
case 10:
printf("输入关系r:\n");
inputSet(&a);
if (isSymmetric(&a)) {
printf("r是对称关系\n");
} else {
printf("r不是对称关系\n");
}
break;
case 11:
printf("输入关系r:\n");
inputSet(&a);
if (isTransitive(&a)) {
printf("r是传递关系\n");
} else {
printf("r不是传递关系\n");
}
break;
case 0:
printf("再见!\n");
break;
default:
printf("无效操作!\n");
break;
}
} while (choice != 0);
return 0;
}
```
注意:本示例代码中的集合元素输入和输出均采用简单的命令行交互方式,实际使用时可以根据需要进行修改。