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;
}