根据这已知的数组求交集和并集的C语言程序

#include <stdioh>

main()

{ int a[5]={1,2,3,4,5} , b[5]={3,4,5,6,7};

int inter[5],sum[10],i,j,n=0,m=0,k=5;

for(i=0;i<5;i++)

for(j=0;j<5;j++)

{if(a[i] == b[j]) inter[n++]=a[i];}

printf("交集为：");

for(i=0;i<n;i++)

printf("%d, ",inter[i]);

for(i=0;i<5;i++)

sum[i] = a[i];

for(i=0;i<5;i++)

for(j=0;j<5;j++)

if(b[i] != a[j]) {if(j == 4) sum[k++] = b[i];}

else break;

printf("并集为：");

for(i=0;i<k;i++)

printf("%d, ",sum[i]);

}

public class Test {

public static void main(String args[])

{

Set<Integer> a = new HashSet<Integer>(ArraysasList(1, 2, 3, 4, 5));

Set<Integer> b = new HashSet<Integer>(ArraysasList(3, 4, 6, 7, 8));

Set<Integer> c = new HashSet<Integer>(ArraysasList(4, 5, 8, 9, 0));

Set<Integer> aIntersectsB = intersects(a, b);

Set<Integer> aIntersectsC = intersects(a, c);

Set<Integer> bIntersectsC = intersects(b, c);

Set<Integer> abc = intersects(a, bIntersectsC);

Systemoutprintln("a ∩ b: " + aIntersectsB);

Systemoutprintln("a ∩ c: " + aIntersectsC);

Systemoutprintln("b ∩ c: " + bIntersectsC);

Systemoutprintln("a ∩ b ∩ c: " + abc);

}

private static <E> Set<E> intersects(Set<E> set1, Set<E> set2) {

Set<E> intersectSet = new HashSet<E>(set1);

intersectSetretainAll(set2);

return intersectSet;

}

}

以上就是关于根据这已知的数组求交集和并集的C语言程序全部的内容，包括:根据这已知的数组求交集和并集的C语言程序、编写程序求三个集合的交集、等相关内容解答，如果想了解更多相关内容，可以关注我们，你们的支持是我们更新的动力！