AT_abc302_d [ABC302D] Impartial Gift

容易发现我们可以先对两个序列排序。

依次枚举 $a_i$，那么另一个数应该 $x$ 应该满足 $x \in [a_i - D, a_i + D]$。因此我们可以二分在这个区间的最大值 $p$，那么这个贡献就是 $a_i+p$，最后将所有 $a_i+p$ 取最大值即可。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <cstdlib>
#include <string>
using namespace std;

#define ll long long

const int N = 2e5 + 5, INF = 2e9, MOD = 1e9 + 7;

inline int read()
{
	int op = 1, x = 0;
	char ch = getchar();
	while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
	while (ch == '-')
	{
		op = -op;
		ch = getchar();
	}
	while (ch >= '0' and ch <= '9')
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * op;
}

inline void write(int x)
{
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}

int n, m, t;
long long a[N], b[N], k;
long long ans = -1;

#define mabs(x) ((x) < 0 ? (-(x)) : (x))

int main()
{
	// freopen("*.in", "r", stdin);
	// freopen("*.out", "w", stdout);
	scanf("%d%d%lld", &n, &m, &k);
	for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
	for (int i = 1; i <= m; i++) scanf("%lld", &b[i]);
	sort(a + 1, a + n + 1);
	sort(b + 1, b + m + 1);
	for (int i = 1; i <= n; i++)
	{
		if (mabs(b[m] - a[i]) <= k)
		{
			ans = max(ans, b[m] + a[i]);
			continue;
		}
		int x = upper_bound(b + 1, b + m + 1, a[i] + k) - b;
		for (int j = max(1, x - 10); j <= min(m, x + 10); j++)
		{
			if (mabs(b[j] - a[i]) <= k)
			{
				ans = max(ans, b[j] + a[i]);
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}