Codeforces 1132D（二分模拟）

要点

• 二分显然，关键在于怎么判断
• 题解方法：开k个队列代表每个时间有哪些电脑会挂掉，这部分O（n）预处理一下，之后扫一遍时间，每个时间点贪心选取最靠前的会挂的电脑未雨绸缪给它充电，然后看看充电以后要不要换队列，这样就把n * k的变成了n + k的

const int maxn = 2e5 + 5;
int n;
ll k;
ll a[maxn], b[maxn], cur[maxn];
queue<int>	Q[maxn];

bool ok(ll mid) {
	rep(i, 0, k)	while(Q[i].size())	Q[i].pop();
	rep(i, 1, n) {
		cur[i] = a[i];
		ll t = cur[i] / b[i] + 1;
		if (t <= k)	Q[t].push(i);
		cur[i] %= b[i];
	}
	int p = 0;
	rep(i, 0, k) {
		while (p <= k && Q[p].empty())	p++;
		if (p > k)	return true;
		if (p <= i)	return false;

		int t = Q[p].front();
		if (cur[t] + mid < b[t]) {
			cur[t] += mid;
			continue;
		}
		Q[p].pop();
		ll d = (cur[t] + mid) / b[t];
		if (p + d <= k)	Q[p + d].push(t), cur[t] = (cur[t] + mid) % b[t];
	}
	return true;
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n >> k;
	k--;
	rep(i, 1, n)	cin >> a[i];
	rep(i, 1, n)	cin >> b[i];

	ll l = 0, r = INF, ans = -1;
	while (l <= r) {
		ll mid = (l + r) >> 1;
		if (ok(mid)) {
			r = mid - 1;
			ans = mid;
		} else l = mid + 1;
	}
	cout << ans << endl;
	return 0;
}