题目链接：https://codeforces.com/group/yz4qgUJSxI/contest/103115/problem/I
一道简单的二分查找题
Chino held a blind date event, there are n male guests and m female
guests attending.

The personality characteristic value of each person can be represented
by an integer, the personality characteristic value of the i-th male
guest is ai , and the personality characteristic value of the j-th
female guest is bj

.

Chino believes that when the calculation result of the personality
characteristic values of two people ai×bj

in a certain range [l,r], the personality of two person is suitable each of them.

Chino would like to know how many matching combinations possible satisfied personality suitability between male and female guests.
Input

The first line contain two positive integers n,m(1≤n,m≤105)

The next line contain n integers ai(−109≤ai≤109)

indicates the personality characteristic value of the male guests.

Then the next line contain m integersbi(−109≤bi≤109)

indicates the personality characteristic value of the female guests.

the final line only contain two integers l,r(−109≤l≤r≤109)

Output

Output an integer on a line to indicate the number of matching combinations. Examples
Input

5 5
1 2 3 4 5
1 2 3 4 5
1 20

Output

24

Input

3 4
-1 -3 -5
-7 -6 -8 -9
9 9

Output

1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Note

For the first example, except that the male guest 5 and the female guest 5 do not meet the conditions, the other conditions are all
legal, so the answer is 25-1=24

For the second example, there is only a legal matching combination for male guest 1 and female guest 4.
这个题本来很简单的二分，L<=ai*bi<=R，两边同时除以一个bi，然后二分查找ai，显然是左边向上取整右边向下取整的范围内找，然后就出现了问题，我当时觉得直接用int型的R除以bi截断小数点就本来是向下取整，不用额外处理，结果。。负数的时候截断就成了向上取整。。

#include <bits/stdc++.h>
typedef long long ll;
const int maxn=100000+10;
using namespace std;
ll a[maxn+10],b[maxn+10];
int main()
{
    ll N,M;
    cin>>N>>M;
    for(int i=0;i<N;i++){
        scanf("%lld",&a[i]);
    }
    for(int j=0;j<M;j++){
        scanf("%lld",&b[j]);
    }
    sort(a,a+N);
    sort(b,b+M);
    ll L,R;
    cin>>L>>R;
    ll ans=0;
    for(int i=0;i<N;i++){
        if(a[i]==0){
            if(L<=0&&0<=R){
                ans+=M;
            }
            continue;
        }
        ll l,r;
        if(a[i]>0){
            l=ceil((double)L/(double)a[i]),r=floor((double)R/a[i]);
        }
        if(a[i]<0){
            l=ceil((double)R/(double)a[i]),r=floor((double)L/a[i]);
        }
        ll left=lower_bound(b,b+M,l)-b;
        ll right=upper_bound(b,b+M,r)-b;
        ans+=right-left;
    }
    cout<<ans<<endl;
    return 0;
}