求C语言编程

任意编一个求2个数组的交集和并集！

来自问问团队编程之道 的正解

原来还有这个数学模板的,受教了..orz

不过头上还要再加句话,不过编译有问题

#include <algorithm>
#include <stdio.h>
using namespace std;

#include<stdio.h> int main() { int a[5]={1,3,5,5,14}; int b[8]={1,3,4,4,7,8,9,14}; int t[8]; int u[13]; int x,y; int i,j,k=0,l,m=0; printf("a[5]原数组有:"); for (i=0;i<5;i++) printf("%d,",a[i]); printf("\nb[8]原数组有:"); for (i=0;i<8;i++) printf("%d,",b[i]); printf("\n"); for (i=0;i<5;i++) for (j=0;j<8;j++) if (a[i]==b[j]) { x=1; for (l=0;l<k;l++) if (a[i]==t[l]) {x=0;break;} if (x==1) { t[k]=b[j]; k++; } } printf("交集为:");for (i=0;i<k;i++) printf("%d,",t[i]); for (i=0;i<5;i++) { y=0; for (j=0;j<k;j++) if (a[i]!=t[j]) y++; if (y==k) { x=1; for (l=0;l<m;l++) if (u[l]==a[i]) {x=0;break;} if (x==1) {u[m]=a[i];m++;} } } for (i=0;i<8;i++) { y=0; for (j=0;j<k;j++) if (b[i]!=t[j]) y++; if (y==k) { x=1; for (l=0;l<m;l++) if (u[l]==b[i]) {x=0;break;} if (x==1) {u[m]=b[i];m++;} } }

printf("\n并集为:");for (i=0;i<k;i++) printf("%d,",t[i]); for (i=0;i<m;i++) printf("%d,",u[i]); printf("\n");

}

~~~纯循环编写的程序~其思路比较复杂~如果想了解下请联系QQ：9931097

#include <algorithm> #include <stdio.h> using namespace std; int main() { int a[] = { 1,2,3,4,5 }; int b[] = { 3,4,5,6,7 }; int c[10], *e, *p; e = set_intersection( a, a + 5, b, b + 5, c ); printf( "交集：" ); for ( p = c; p != e; ++p ) { printf( "%d ", *p ); } printf( "\n并集：" ); e = set_union( a, a + 5, b, b + 5, c ); for ( p = c; p != e; ++p ) { printf( "%d ", *p ); } } set_intersection和set_union等集合操作的前提的数组已经排序