模拟+思维(时间规划烧饼）

前言：这个题目就是我们小时候学的的活动规划烧饼，我们要先算出我们最大耗时是多少，然后再对我们的活动进行规划

题目地址

我们这个题目还要求算出k（执行次数）我的做法是写两遍代码，其实也可以存起来，因为最大操作次数为 2n

#include<bits/stdc++.h>
using namespace std;#define ll long long
#define endl "\n"const int N = (int)1e5 + 10;
int n, m;
ll a[N];int main() {cin >> n >> m;ll limi = 0;ll sum = 0;for (int i = 1; i <= n; i++) {cin >> a[i]; limi = max(limi, a[i]);sum += a[i];}// 限制好一个limi ll t = (sum % m) ? sum / m + 1 : sum / m;limi = max(limi, t);int l = 0, cnt = 1;ll k = 0;for (int i = 1; i <= n; i++) {if (l + a[i] >= limi) {//cout << i << " " << cnt << " " << l << " " << limi << endl;k++;l = (l + a[i]) % limi; cnt++;if (l) {//cout << i << " " << cnt << " " << 0 << " " << l << endl;k++;}}else {//cout << i << " " << cnt << " " << l << " " << l + a[i] << endl;l += a[i];k++;}}cout << k << endl;l = 0, cnt = 1;for (int i = 1; i <= n; i++) {if (l + a[i] >= limi) {cout << i << " " << cnt << " " << l << " " << limi << endl;l = (l + a[i]) % limi; cnt++;if (l) {cout << i << " " << cnt << " " << 0 << " " << l << endl;}}else {cout << i << " " << cnt << " " << l << " " << l+a[i] << endl;l += a[i];}}return 0;
}