洛谷：P1167 刷题(模拟 / 贪心)

传送门

题意： 就有个初始时间和结束时间，给出n个题的AC时间，试问这段时间24小时不停歇最多能AC多少题。

思路： 其实也没大好说的，就是模拟求时间段总共有多少分钟，然后再将n个问题sort一下求答案即可。

代码实现：

#include
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9 + 7;
const int  N = 2e5 + 5;

int n, a[N];
int year = 365, yearr = 366;
int month[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
int monthr[13] = {0,31,29,31,30,31,30,31,31,30,31,30,31};
int year1, month1, day1, hour1, min1;
int year2, month2, day2, hour2, min2;
ll time1, time2;

int check(int x){
	if(((x%4 == 0) && (x%100 != 0)) || ((x%100 == 0) && (x%400 == 0))) return 1;
	else return 0;
}

ll work(ll year3, ll month3, ll day3, ll hour3, ll min3){
	int flag = check(year3), time = 0;
	for(int i = 0; i < year3; i ++){
		if(check(i)) time += yearr*24*60;
		else time += year*24*60;
	}
	ll monthtime = 0;
	for(int i = month3 - 1; i >= 0; i --){
		if(flag) monthtime += monthr[i]*24*60;
		else monthtime += month[i]*24*60;
	}
	time += min3 + hour3*60 + (day3-1)*24*60 + monthtime;
	return time;
}

signed main()
{
	scanf("%d",&n);
	for(int i = 1; i <= n; i ++) cin >> a[i];
	sort(a + 1, a + n + 1);
	scanf("%lld-%lld-%lld-%lld:%lld", &year1, &month1, &day1, &hour1, &min1);
    scanf("%lld-%lld-%lld-%lld:%lld", &year2, &month2, &day2, &hour2, &min2);
	ll ans = work(year2, month2, day2, hour2, min2) - work(year1, month1, day1, hour1, min1);
	int tmp = 0;
	for(int i = 1; i <= n; i ++){
		if(ans-a[i] >= 0){
			tmp ++;
			ans -= a[i];
		}
	}
	cout << tmp << endl;

	return 0;
}