Question
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
n ≤ 10000
q ≤ 500
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
3 1 2
1
5
Sample Output 2
0
Sample Input 3
5
1 1 2 2 3
2
1 2
Sample Output 3
2
Meaning
使用二分搜索，效率就会很大提高。但是需要注意，二分搜索的前提是数组已经排序好了。那么排序一般最高的效率希尔排序也是0(n1.25），岂不是效率还降低了？相对于O（n），这里题目限制说了
给出的数据就是按升序排列的，那么时间复杂度就是0（qlogn）
Coding
#include<iostream>
#include<cstdio>
using namespace std;
int A[100005];
int BinarySearch(int a[], int size, int key) {
int l = 0;
int r = size - 1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (key == a[mid])
return true;
else if (key > a[mid])
l = mid + 1;
else
r = mid - 1;
}
return false;//没有找到
}
int main() {
int n, q, key, sum = 0;
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d", &A[i]);
scanf("%d", &q);
for (int i = 0; i < q; i++) {
scanf("%d", &key);
if (BinarySearch(A,n,key))
sum++;
}
printf("%d\n", sum);
}Summary
二分搜索使用前需要对数组进行排序；
标准二分搜索算法：
int BinarySearch(int a[], int size ,int p){
int L=0; //查找区间的左端点
int R=size -1;//查找区间的的右端点
while(L<=R){
int mid =L+(R-L)/2;
if(p == a[mid])
return mid;
else if(p>a[mid])
L=mid +1;
else
R=mid-1;
}
return -1;
}
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