多校训练6 F Hamburger Steak 贪心+构造

• 注意：输出时，如果同一块牛排在两个锅内完成烹饪，则需要先输出pos+1锅的起始和结束时间，再输出pos锅的起始和结束时间。

accode:

#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 +10;
inline ll max(ll a, ll b){return (a > b) ? a : b;}
int n, m, k, T, cnt, flag;
int ans, a[maxn];
ll sum = 0, maxx = 0, f, t;

int main()
{        
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        sum += a[i];
        maxx = max(maxx, 1ll * (a[i]));
    }
    f = sum / m;
    if(sum % m)
        f++;
    maxx = max(f, maxx);//最小总时间（每口锅的最长烹饪时间）
    
    int pos = 1; //锅的位置
    t = 0; //该锅消耗的时间
    for(int i = 1; i <= n; i++)
    {
        if(t + a[i] <= maxx)//满足煎完 a[i]的条件 
        {
            cout<<1<<" "<<pos<<" "<<t<<" "<<t + a[i]<<endl;
            t += a[i];
            if(t == maxx)
            {
                t = 0;
                pos++; 
            }
        }
        else//当前锅烹饪不完第i块肉饼
        {
            cout<<2<<" "<<pos + 1<<" "<<0<<" "<<a[i] - (maxx - t)<<" "; //注意pos+1排在pos前面 
            cout<<pos<<" "<<t<<" "<<maxx<<endl;
            t = a[i] - (maxx - t);
            pos++;
        }
    }
    
    return 0;
}