Problem:
Search II

You are given a sequence of n integers S and a sequence of different q integers T. Write a program which outputs C, the number of integers in T which are also in the set S.

Input

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers are given.

Output

Print C in a line.

Constraints

Elements in S is sorted in ascending order

n ≤ 100000

q ≤ 50000

0 ≤ an element in S ≤ 109

0 ≤ an element in T ≤ 109

Sample Input 1

5
1 2 3 4 5
3
3 4 1

Sample Output 1

3

Sample Input 2

3
1 2 3
1
5

Sample Output 2

0

Sample Input 3

5
1 1 2 2 3
2
1 2

Sample Output 3

2

Notes

Solution in C (judge verdict: Accepted)

//2020 4.18 Binary Search

#include <stdio.h>
#define MAX1 100000
#define MAX2 50000
#define NOT_FOUND -1  //1にしていたが、そうするとbinarySearchの戻り値(mid)が1になる場合があり、if(res!=1)のところでカウントをしなくなってしまう。

int binarySearch(int *,int,int);

int main()
{
    int a,b,res=0,cnt=0;
    int i,j;
    int A[MAX1],B[MAX2];

    scanf("%d",&a);
    for(i=0;i<a;i++)
    {
        scanf("%d",&A[i]);
    }

    scanf("%d",&b);
    for(i=0;i<b;i++)
    {
        scanf("%d",&B[i]);

        res=binarySearch(A,a,B[i]);

        if(res!=-1) cnt++;
    }
    printf("%d\n",cnt);

    return 0;
}

int binarySearch(int *A,int n,int key)
{
    int left = 0;
    int right = n;
    int mid;
    while(left < right)
    {
        mid=(left + right)/2;
        if(A[mid] == key) {return mid;}
        else if(key<A[mid]) {right = mid;}
        else left = mid + 1;
    }
    return NOT_FOUND;    
}