Codeforces 493C 枚举加二分

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#STL #二分查找 #枚举

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本文介绍了一道关于篮球比赛中划分2分球与3分球界限的问题。通过枚举不同的界限值，并利用二分查找确定各队得分，最终找到使一方优势最大的得分方案。

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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 a_i (1 ≤ a_i ≤ 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 b_i (1 ≤ b_i ≤ 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 - b is maximum. If there are several such scores, find the one in which number a is maximum.

Sample test(s)

Input

Output

9:6

Input

Output

15:10

题意：有两个篮球球队打比赛，分别记录下了他们投中时候距离框的距离（整数）。让你划定一个三分球的线，小于等于这个距离的算为2分球，大于这个距离的算为3分球。输出一个比分，要求第一队得分减第二队得分最大，当分差最大时，还要求第一队得分最高。

思路：从投球距离小到大枚举三分线，用二分查找得每个队到小于等于三分线距离的球的个数，从而算出在这个距离上的得分。二分可以使用STL里面的lower_bound()

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 200005
using namespace std;
int a[maxn],b[maxn],c[maxn<<1];
int main()
{
    int n,m;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        c[i]=a[i];
    }
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d",&b[i]);
        c[i+n]=b[i];
    }
    c[n+m+1]=0;
    sort(a+1,a+1+n);
    sort(b+1,b+1+m);
    sort(c+1,c+1+n+m+1);
    int len=unique(c+1,c+1+n+m+1)-c-1;
    int x,y;
    int ma=0x80000000;
    for(int i=1;i<=len;i++)
    {
        int n1=upper_bound(a+1,a+1+n,c[i])-a-1;
        int n2=upper_bound(b+1,b+1+m,c[i])-b-1;
        int xx=2*n1+3*(n-n1);
        int yy=2*n2+3*(m-n2);
        if(xx-yy>ma)
        {
            x=xx;
            y=yy;
            ma=xx-yy;
        }
    }
    printf("%d:%d\n",x,y);
    return 0;
}