Codeforces Round #281 (Div. 2) C. Vasya and Basketball 排序

C. Vasya and Basketball

 

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 subtractiona - b is 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

题意：投篮比赛，给你两个人投进篮球的距离，让你找到一个d使得小于等于d的距离的投篮得分为2，大于d的距离得分为3，同时使得a-b得分差最大

        如果有a-b相同的情况，尽量使a最大

题解：按照分差排序同时a尽量大，就行了

///meek
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){
        if(ch=='-')f=-;ch=getchar();
    }
    while(ch>=''&&ch<=''){
        x=x*+ch-'';ch=getchar();
    }return x*f;
}
//****************************************
const int N=;
#define mod 10000007
#define inf 2000000007
#define maxn 10000

ll lna,lnm,n,a[N],b[N],m;

struct ss{
  ll x,in,index;
}s[N+N];
struct sss {
    ll x,aa;
    int in;
}W[N+N];
int cmp(ss s1,ss s2)  {
    if(s1.x==s2.x) return s1.in<s2.in;
    return s1.x<s2.x;
}
int cmp2(sss s1,sss s2)  {
    if(s1.x==s2.x) return s1.aa<s2.aa;
   else return s1.x<s2.x;
}
int main() {

     n=read();
    int k=;
    for(int i=;i<=n;i++) {
        scanf("%I64d",&a[i]);
        s[++k].x=a[i];
        s[k].in=;
        s[k].index=i;
    }
    m=read();
    for(int i=;i<=m;i++) {
        scanf("%I64d",&b[i]);
         s[++k].x=b[i];
        s[k].in=;
        s[k].index=i;
    }
    sort(s+,s+k+,cmp);
    ll l=,r=;
    for(int i=;i<=k;i++) {
        if(s[i].in==) {
            l++;
        }
        else r++;
        W[i].x=(l*+(n-l)*)-(r*+(m-r)*);
        W[i].aa=(l*+(n-l)*);
        W[i].in=i;
    }
    W[++k].x=n*-m*;
    W[k].aa=(n*);
    W[k].in=-;
     W[++k].x=n*-m*;
    W[k].aa=(n*);
    W[k].in=-;
    sort(W+,W+k+,cmp2);
    ll ansl=,ansr=;
    ansl=W[k].aa;ansr=W[k].aa-W[k].x;
    cout<<ansl<<":"<<ansr<<endl;
    return ;
}

代码