求两个整型数组的异集，即A+B-(A与B的交集)

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/*求两个整型数组的异集，即A+B-(A与B的交集)。*/
#include <iostream>
using namespace std;

int search_one(int num, int *b_set_p, int b_set_size);

int
main(int argc, char *argv[])
{
    int array_a[] = {2, 3, 4, 8, 23, 21, 68, 79, 9, 15, 19};
    int array_b[] = {23, 68, 94, 74, 89, 70, 20, 15};
    int array_a_size, array_b_size;
    array_a_size = sizeof(array_a)/sizeof(int);
    array_b_size = sizeof(array_b)/sizeof(int);

    int i;
    cout << "different at array_a:" << endl;
    for(i = 0; i < array_a_size; i++){
        if(!search_one(array_a[i], array_b, array_b_size)){
            /*打印出该元素*/
            cout << array_a[i] << " ";
        }
    }

    cout << endl << "different at array_b:" << endl;
    for(i = 0; i < array_b_size; i++){
        if(!search_one(array_b[i], array_a, array_a_size)){
            /*打印出该元素*/
            cout << array_b[i] << " ";
        }
    }

    cout << endl << "array_a and array_b have the same elements:" <<endl;
    for(i = 0; i < array_a_size; i++){
        if(search_one(array_a[i], array_b, array_b_size)){
            /*打印出该元素*/
            cout << array_a[i] << " ";
        }
    }
    return 0;
}

/*在B集合里寻找一个元素，找到返回非零，否则返回零。*/
int
search_one(int num, int *b_set_p, int b_set_size)
{
    int i;
    for(i = 0; i < b_set_size; i++){
        if(num == *(b_set_p+i))
            return 1;
    }
    return 0;
}