cf493C Vasya and Basketball

C. Vasya and Basketball

time limit per test 2 seconds

memory limit per test 256 megabytes

input standard input

output standard output

Vasya follows a basketball game and marks the distances from which each team makes a throw. He knows that each successful throw has value of either 2 or 3 points. A throw is worth 2 points if the distance it was made from doesn't exceed some value of d meters, and a throw is worth 3 points if the distance is larger than d meters, where d is some non-negative integer.

Vasya would like the advantage of the points scored by the first team (the points of the first team minus the points of the second team) to be maximum. For that he can mentally choose the value of d. Help him to do that.

Input

The first line contains integer n (1 ≤ n ≤ 2·105) — the number of throws of the first team. Then follow n integer numbers — the distances of throws ai (1 ≤ ai ≤ 2·109).

Then follows number m (1 ≤ m ≤ 2·105) — the number of the throws of the second team. Then follow m integer numbers — the distances of throws of bi (1 ≤ bi ≤ 2·109).

Output

Print two numbers in the format a:b — the score that is possible considering the problem conditions where the result of subtraction a - bis maximum. If there are several such scores, find the one in which number a is maximum.

Sample test(s)

input

3
1 2 3
2
5 6

output

9:6

input

5
6 7 8 9 10
5
1 2 3 4 5

output

15:10

ab排序

for一遍在a数组枚举d=a[i]-1，这样可以保证最优（因为d=a[i]-1一定比d=a[i]-2优）

然后二分d在b数组中的位置就可以知道有多少个2分多少个3分了

还要考虑d=inf的情况

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m;
int a[200010];
int b[200010];
int now2;
int ans1,ans2,mx,mx2;
int main()
{
	n=read();
	for(int i=1;i<=n;i++)a[i]=read();
	m=read();
	for(int j=1;j<=m;j++)b[j]=read();
	sort(a+1,a+n+1);
	sort(b+1,b+m+1);
	int kk=a[1]-1,s1=3*n,s2=0;
	for (int i=1;i<=m;i++)if (b[i]>kk)s2+=3;else s2+=2;
	mx=s1-s2;mx2=s1;
	ans1=s1;ans2=s2;
	if (2*n-2*m>mx||2*n-2*m==mx&&2*n>mx2)
	{
		mx=2*n-2*m;mx2=2*n;
		ans1=2*n;ans2=2*m;
	}
	for (int now=1;now<=n;now++)
	{
		if (a[now]==a[now-1])continue;
		now2=lower_bound(b+1,b+m+1,a[now])-b;
		int res1=3*(n-now+1)+2*(now-1),res2=3*(m-now2+1)+2*(now2-1);
		if (res1-res2>mx||res1-res2==mx&&res1>mx2)
		{
			mx=res1-res2;
			mx2=res1;
			ans1=res1;
			ans2=res2;
		}
	}
	printf("%d:%d
",ans1,ans2);
}

 其实因为排序完b数组是递增的，所以可以直接用一个指针来模拟二分