URAL 1196. History Exam （二分）

1196. History Exam

Time limit: 1.5 second
Memory limit: 64 MB

Professor of history decided to simplify the examination process. At the exam, every student should write a list of historic dates she knows (she should write the years only and, of course, must be able to explain what event took place in this or that year). Professor has a list of dates that students must know. In order to decide upon the student's mark, Professor counts the number of dates in the student's list that are also present in his list. The student gets her mark according to the number of coincidences.

Your task is to automatize this process. Write a program that would count the number of dates in the student's list that also occur in Professor's list.

Input

The first line contains the number N of dates in Professor's list, 1 ≤ N ≤ 15000. The following Nlines contain this list, one number per line. Each date is a positive integer not exceeding 109. Professor's list is sorted in non-descending order. The following line contains the number M of dates in the student's list, 1 ≤ M ≤ 106. Then there is the list itself; it is unsorted. The dates here satisfy the same restriction. Both in Professor's and in the student's lists dates can appear more than once.

Output

Output the number of dates in the student's that are also contained in Professor's list.

Sample

input	output
2 1054 1492 4 1492 65536 1492 100	2

题意：找出第一和第二个序列中都出现（同意反复累加）过的字符个数。

解析：因为第一个有序的，所以我们遍历第二个序列的同一时候对第一个序列二分搜索答案。

PS：本题有个非常诡异的现象：G++跑了1.5s+。可是VC++居然才跑0.343s。

。。貌似仅仅有VC++才干过。

AC代码：

#include <cstdio>
using namespace std;

int a[15002];

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk

    int n, m, ans, foo;
    while(scanf("%d", &n)==1){
        ans = 0;
        for(int i=0; i<n; i++) scanf("%d", &a[i]);
        scanf("%d", &m);
        for(int i=0; i<m; i++){
            scanf("%d", &foo);
            int l = 0, r = n - 1, m;
            if(foo < a[0] || foo > a[n-1]) continue;
            else if(foo == a[0] || foo == a[n-1]){
                ans ++;
                continue;
            }
            while(l <= r){
                m = (r - l) / 2 + l;
                if(a[m] == foo){
                    ans ++;
                    break;
                }
                if(a[m] < foo) l = m + 1;
                else r = m - 1;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}