Luo Gu P1167 brush title

Topic links: Luo Gu P1167 brush title

Algorithms 贪心Tags: ,模拟

topic

Title Description

noipNear, 小Aonly to find that he had not written the subject. Fortunately, now there is some time away from the competition, 小Adecided to brush day and night from now on topic. That 小Asleepless nights, 24 hours a day to brush the question.

Today's date (time) is min minutes and hh yyyy mm-dd-day, time of examination is yyyy 'in mm' Month dd min 'minutes and' day hh '. All the time between these 小Aare used to brush the question, then the maximum number of questions before the exam he can brush it? Note Well, consider a leap year.

Time constraints 小Ajust quantity regardless of quality. Of course, some topics easier, some harder topic. According to 小Aexperience, he can see at a glance the time to write a certain topic needed to record minutes.

Now given 洛谷Online Judgea list of topics, please pick up the topic 小Acan be written before the race.

We assume that from ancient times to the future, the calendar is represented as now.

Input Format

A first line integer N, the 洛谷Online Judgnumber of topics e, N≤5000.

Next N rows, each row represents an integer time of the issue requires a brush, in minutes and (≤10000). (This question itself is not what is important, is not it? 小AHas written several topics to 0).

The next two lines followed by the current time and race time. Format given time is: yyyy-mm-dd-hh: min, for example: 2007-06-23-02: 00, 24-hour, day, from 00: 00 to 23: 59 years from 0000 to 9999.

Output Format

Row, an integer noipnumber of the subject before the brush up.

Sample input and output

Input # 1

2
1
1
2007-06-23-11:59
2007-06-23-12:00

Output # 1

answer:

Most begin to see the solution to a problem is obvious greedy solution, to ensure that each time within a limited time you can take the minimum (use sort()), but the difficulty lies in this question of how simulation calculated the total number of minutes. There are some things that require special attention during the simulation:

leap year
The number of days each month
How to read? ? ? ( scanf("%d-%d-%d-%d:%d",……))

But then you need to brush up for judgment :( leap year every four years a leap, leap hundred years, four hundred years and then leap)

bool run(int x)
{
    if (x % 100 == 0)
    {
        if (x % 400 == 0)
            return 1;
    }
    else
        if (x % 4 == 0)
            return 1;
    return 0;
}

Then we simulated time, the calculation time processing "Year", "Day", "Hour", "Minute" are relatively simple, according to the band pass to the next, but in the "month" judgment is more complicated, simulation, considering the number of days per month, and in view of this year (12 months) if the case involves a leap year, the specific implementation, see AC codes .

AC Code

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct _TIME{
    int y, m, d, h, minn;
}_time[3];
int mon[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool run(int x)
{
    if (x % 100 == 0)
    {
        if (x % 400 == 0)
            return 1;
    }
    else
        if (x % 4 == 0)
            return 1;
    return 0;
}
int getminn(_TIME a, _TIME b)
{
    int ansy, ansm, ansd, ansh, ansminn;
    ansy = b.y - a.y;
    ansm = b.m - a.m;
    ansd = b.d - a.d;
    ansh = b.h - a.h;
    ansminn = b.minn - a.minn;
    if (ansy > 0)
        ansm += ansy * 12;
    if (ansm > 0)
    {
        int i = a.m, j = a.y;
        for (int k = 1; k <= ansm; k ++ )
        {
            if (i > 12)
            {
                i -= 12;
                j ++ ;
            }
            if (i == 1 && run(j))
            {
                ansd += mon[i - 1] + 1;
                i ++ ;
                continue ;
            }
            ansd += mon[i - 1];
            i ++ ;
        }
    }
    if (ansd > 0)
        ansh += ansd * 24;
    if (ansh > 0)
        ansminn += ansh * 60;
    return ansminn;
}
int num[5500];
int n;
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ )
        scanf("%d", &num[i]);
    sort(num + 1, num + 1 + n);
    scanf("%d-%d-%d-%d:%d", &_time[1].y, &_time[1].m, &_time[1].d, &_time[1].h, &_time[1].minn);
    scanf("%d-%d-%d-%d:%d", &_time[2].y, &_time[2].m, &_time[2].d, &_time[2].h, &_time[2].minn);
    int allminn = getminn(_time[1], _time[2]);
    int cnt = 0;
    for (int i = 1; i <= n; i ++ )
    {
        if (num[i] <= allminn)
        {
            cnt ++ ;
            allminn -= num[i];
        }
        else
            break ;
    }
    printf("%d\n", cnt);
    return 0;
}

Brush Valley P1167 Los problem - greedy, simulation