Educational Codeforces Round 61 (Rated for Div. 2) D. Stressful Training 详解

Educational Codeforces Round 61 (Rated for Div. 2) D. Stressful Training

D. Stressful Training
题目大意：
比赛开始每个同学都带了电脑，但没带充电器，已知有n个同学，每个同学电脑剩余电量为a[i]，每秒耗电为b[i]，比赛时长为k秒。教练买了个充电器，每秒钟可以给一个电脑冲x的电；求最小的x使全体同学完成比赛。ps：第一秒不耗电，最后一秒可以小于0，电量可以为0
大佬解答：
二分答案后发现可以O(nlogn)O(nlogn)判断，想办法卡卡常即可，时间复杂度O(nlognlogs)O(nlognlogs)，其中s为答案的范围。
菜鸟思维:
计算x的最小值，可以很容易想到二分；
如何通过mid验证x是否成立呢？
大佬告诉我们可以在电脑电量低于0的时刻，加一，表示此刻到此刻以前电脑至少需要充电一次。维护cnt数组，cnt[i]的前缀和即表示i以前至少要充电n次，如果n>i则表示无法满足充电需求；
那么代码为

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
#pragma warning (disable:4996)
typedef long long LL;
LL a[210000], need[210000];
int b[210000], n, k, cnt[210000];
bool check(LL mid)
{for (int i = 0; i <= k; i++)cnt[i] = 0;int tot = k - 1;for (int i = 1; i <= n; i++){if (!b[i])continue;LL hadtime = a[i];while (hadtime < need[i]){    cnt[hadtime / b[i] + 1]++;hadtime += mid;if (!tot)return false;tot--;}}for (int i = 1; i <= k; i++){cnt[i] = cnt[i] + cnt[i - 1];if (cnt[i] > i)return false;}return true;
}
int main()
{scanf("%d%d", &n, &k);for (int i = 1; i <= n; i++)cin >> a[i];for (int i = 1; i <= n; i++){scanf("%d", &b[i]);need[i] = LL(k - 1)*b[i];}LL left, right, mid, ans = -1;left = 0; right = 2e12;while (left <= right){mid = (left + right) / 2;if (check(mid)){ans = mid;right = mid - 1;}elseleft = mid + 1;}cout << ans << '\n';}

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Educational Codeforces Round 61 (Rated for Div. 2) D. Stressful Training 详解

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