CodeForces 489B BerSU Ball （水题 双指针）

B. BerSU Ball

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.

We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.

For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.

Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.

Output

Print a single number — the required maximum possible number of pairs.

Sample test(s)

input

4
1 4 6 2
5
5 1 5 7 9

output

3

input

4
1 2 3 4
4
10 11 12 13

output

0

input

5
1 1 1 1 1
3
1 2 3

output

2

水题
双指针模拟一下就行了，用i和j分别指向两个数组，如果a[i]和b[j]相差小于等于1的话，就ans++ i++ j++，否则就较小那边向前移动一下

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int INF = 1e9;
const double eps = 1e-6;
const int N = 110;
int cas = 1;

int a[N],b[N];
int n,m;

bool ok(int x,int y)
{
    return x-y>=-1 && x-y<=1;
}

void run()
{
    for(int i=0;i<n;i++) scanf("%d",a+i);
    scanf("%d",&m);
    for(int i=0;i<m;i++) scanf("%d",b+i);
    sort(a,a+n);
    sort(b,b+m);
    int ans = 0;
    for(int i=0,j=0;i<n&&j<m;)
    {
//        cout<<a[i]<<' '<<b[j]<<'	';
        if(ok(a[i],b[j]))
            ans++,i++,j++;
        else if(a[i]<b[j])
            i++;
        else
            j++;
//        cout<<i<<' '<<j<<endl;
    }
    cout<<ans<<endl;
}

int main()
{
    #ifdef LOCAL
    freopen("case.txt","r",stdin);
    #endif
    while(scanf("%d",&n)!=EOF)
        run();
    return 0;
}