Description
设计一个求集合交的算法：输入集合A与集合B，求集合A与B之交。集合中的元素为整数（可以用c语言中的int表示），且互不相同。
Input
输入第一行为一个整数t（0<t<10），表示测试用例个数。
每个测试样例由3行构成。第1行是2个正整数a(1≤a≤1000000)，b(1≤a≤1000000)，分别表示集合A和B的元素个数；第2行为a个以空格分隔的整数，为集合A中元素；第3行为b个以空格分隔的整数，为集合B中元素。
Output
每个样例单独一行输出交集中元素的个数。
Sample Input
aaarticlea/jpeg;base64,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" alt="" /> Copy sample input to clipboard
2
7 5
0 1 2 4 7 8 9
1 2 5 6 7
7 8
1 2 3 4 5 6 8
1 2 4 5 6 7 8 9
Sample Output
3
6
#include<iostream>
using namespace std; struct Node
{
Node* next;
int data;
}; Node* Create(int n)
{
Node* head = new Node;
Node*p, *pre;
head->next = NULL;
pre = head;
int k;
for (int i = 1; i <= n; i++)
{
p = new Node;
cin >> k;
p->data = k;
p->next = NULL;
pre->next = p;
pre = p;
}
return head;
} void jiao(Node*A, Node*B)
{
Node*p1, *p2;
p1 = A->next;
int count = 0;
while (p1)
{
p2 = B->next;
while (p2)
{
if (p2->data == p1->data)
{
count++;
break;
}
p2 = p2->next;
}
p1 = p1->next;
}
cout << count << endl;
} int main()
{
int m;
cin >> m;
while (m-->0)
{
int n1, n2;
cin >> n1 >> n2;
Node*A = Create(n1);
Node*B = Create(n2);
jiao(A, B);
delete A, B;
} return 0;
}