求任意两个集合的交集并集差集 用c语言求两个集合的交集，并集，差集
\u5982\u4f55\u6c42\u4e24\u4e2a\u96c6\u5408\u7684\u4ea4\u96c6\u5e76\u96c6\u8865\u96c6\u4ea4\u96c6\u662f\u4e24\u4e2a\u96c6\u5408\u7684\u516c\u5171\u5143\u7d20\uff0c\u5373\u4e24\u4e2a\u65b9\u7a0b\u7684\u516c\u5171\u89e3\uff1b
\u5e76\u96c6\u662f\u4e24\u4e2a\u96c6\u5408\u7684\u5143\u7d20\u7684\u603b\u4e2a\u6570\uff08\u76f8\u540c\u7684\u5143\u7d20\u53ea\u5199\u4e00\u6b21\uff09\uff1b
\u5dee\u96c6\uff1a\u5982\u679c\u4e24\u4e2a\u96c6\u5408\u6709\u4ea4\u96c6\uff0c\u5219\u5927\u96c6\u5143\u7d20\u4e2d\u6240\u6709\u4e0d\u5c5e\u4e8e\u5c0f\u96c6\u5408\u7684\u5143\u7d20\u7684\u96c6\u5408\u662f\u5dee\u96c6\uff0c\u5982\u679c\u6ca1\u6709\u4ea4\u96c6\uff08\u7a7a\u96c6\uff09\uff0c\u5219a-b=a,
b-a=b
#include #include #include #define ARR_LEN 255/*\u6570\u7ec4\u957f\u5ea6\u4e0a\u9650*/#define elemType char/* \u96c6\u5408\u5143\u7d20\u6570\u636e\u7c7b\u578b *//* \u96c6\u5408\u6570\u636e\u7ed3\u6784 */typedef struct set {elemType data[ARR_LEN];int length;} set;/* \u521d\u59cb\u5316\u96c6\u5408 */void initSet (set *S) {S->length = 0;} /* \u4ea4\u96c6 *//* A\u4e0eB\u7684\u4ea4\u96c6\uff08A\u2229B\uff09\uff1a\u65e2\u5c5e\u4e8eA\u53c8\u5c5e\u4e8eB\u7684\u5143\u7d20\u6784\u6210\u7684\u96c6\u5408 */int setIntersection (set A, set B, set *dest) {int i = 0, j = 0, k = 0;dest->length = 0;for (i=0; idata[k] = A.data[i];k++;}}}dest->length = k;if (dest->length)return 1;elsereturn 0;}/* \u5e76\u96c6 *//* A\u4e0eB\u7684\u5e76\u96c6\uff08A\u222aB\uff09\uff1aA\u4e0eB\u6240\u6709\u5143\u7d20\u6784\u6210\u7684\u96c6\u5408 */int setUnion (set A, set B, set *dest) {int i = 0, j = 0, k = 0;dest->length = 0;for (i=0; idata[k] = A.data[i];k++;}}for (j=0; jdata[k] = B.data[j];k++;}dest->length = k;if (dest->length)return 1;elsereturn 0;}/* \u8865\u96c6 *//* B\u5728A\u4e2d\u7684\u76f8\u5bf9\u8865\u96c6\uff08A\uff3cB\uff09\uff1a\u5c5e\u4e8eA\u4f46\u4e0d\u5c5e\u4e8eB\u7684\u5143\u7d20\u6784\u6210\u7684\u96c6\u5408 */int setComplement (set A, set B, set *dest) {int i = 0, j = 0, k = 0;dest->length = 0;for (i=0; idata[k] = A.data[i];k++;}}dest->length = k;if (dest->length)return 1;elsereturn 0;}/* \u6253\u5370\u96c6\u5408\u5185\u5bb9 */int printSet (set S) {int i;if (S.length == 0) {puts ("The set is empty! ");return 0;}for (i=0; i<S.length; i++)printf ("%c", S.data[i]);putchar ('\n');return 1;}int main (void) {set A, B;set AIB, AUB, ACB; /* \u4ea4\u96c6\u3001\u5e76\u96c6\u3001\u8865\u96c6 */initSet (&A); initSet (&B);initSet (&AIB); initSet (&AUB); initSet (&ACB);strcpy (A.data, "123");A.length = strlen (A.data);strcpy (B.data, "4532");B.length = strlen (B.data);printf ("A:\t");printSet (A);printf ("B:\t");printSet (B);putchar ('\n');printf ("A\u2229B:\t");setIntersection (A, B, &AIB);printSet (AIB);printf ("A\u222aB:\t");setUnion (A, B, &AUB);printSet (AUB);printf ("A\uff3cB:\t");setComplement (A, B, &ACB);printSet (ACB);getch (); /*\u5c4f\u5e55\u6682\u7559*/return 0;}
调试过了，成功。
想变换数组个数就改
#define M 10
#define N 10
后面的数字！
#include "stdio.h"
#define M 10
#define N 10
main() {
int i,j,k=-1,n=0;
int jj=-1,bb=-1;
int a[M],b[N],c[M*N],d[M+N],x[M*N],y[M+N];
for(i=0;i<M;i++)
scanf("%d",&a[i]);
for(i=0;i<M;i++)
scanf("%d",&b[i]);
for(i=0;i<M;i++){
for(j=0;j<N;j++)
if(a[i]==b[j]){
k++;
c[k]=a[i];
}
}
printf("
jiao ji :
");
for(i=0;i<=k;i++){
n=0;
for(j=i+1;j<=k;j++){
if(c[i]!=c[j]){
n++;}
}
if(n==k-i){
jj++;
x[jj]=c[i];
printf("%d " ,c[i]);
}
}
printf("
bing ji :
");
for(i=0;i<M;i++)
d[i]=a[i];
for(j=0;j<N;j++)
d[M+j]=b[j];
for(i=0;i<=M+N-1;i++){
n=0;
for(j=i+1;j<=M+N-1;j++){
if(d[i]!=d[j]){
n++;}
}
if(n==M+N-1-i){
bb++;
y[bb]=d[i];
printf("%d " ,d[i]);
}
}
printf("
cha ji :
");
for(i=0;i<=bb;i++){
n=0;
for(j=0;j<=jj;j++){
if(y[i]!=x[j]){
n++;
}
}
if(n-1==jj)
printf("%d ",y[i]);
}
}
不知道你想要什么= =
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