一開始樓主以爲暴力會T,於是想了各種腦洞進行線性的搜索,後來發現upper_bound函數的時間複雜度爲logn,完全夠用,於是寫了一個暴力水過去了:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[200010];
int b[200010];//公式:b1-a1要最大化
int d[400020];
int main()
{
int n, m, cnt = 0, l, r;
int ans = -0x3f3f3f3f;
d[++cnt] = 0;//先把最前面的 d = 0 插進來
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]), d[++cnt] = a[i];
scanf("%d", &m);
for(int i = 1; i <= m; i++)
scanf("%d", &b[i]), d[++cnt] = b[i];
sort(a+1, a+n+1);
sort(b+1, b+m+1);
sort(d+1, d+cnt+1);
for(int i = 1; i <= cnt; i++)
{
int x = upper_bound(a+1, a+n+1, d[i])-a-1;
int y = upper_bound(b+1, b+m+1, d[i])-b-1;
int l1 = 2*x + 3*(n-x);
int r1 = 2*y + 3*(m-y);
if(l1-r1 > ans)
{
l = l1, r = r1;
ans = l1-r1;
}
}
//printf("%d %d\n", bi, bj);
printf("%d:%d\n", l, r);
return 0;
}