蓝桥杯 工作时长_小蓝手里有一份 20222022 年度自己的上班打卡记录文件,文件包含若干条打卡记录,每

问题描述

小蓝手里有一份2022年度自己的上班打卡记录文件,文件包含若干条打
卡记录,每条记录的格式均为“yyyy-MM-dd HH:mm:ss”,即按照年-
月-日 时:分:秒的形式记录着一个时间点(采用24小时进制)。由于某些
原因,这份文件中的时间记录并不是按照打卡的时间顺序记录的,而是被
打乱了。但我们保证小蓝每次上班和下班时都会正常打卡,而且正好打卡
一次,其它时候不会打卡。每一对相邻的上-下班打卡之间的时间就是小蓝
本次的工作时长,例如文件内容如下的话:

2022-01-01 12:00:05
2022-01-02 00:20:05
2022-01-01 07:58:02
2022-01-01 16:01:35

表示文件中共包含了两段上下班记录,1)2022-01-01 07:58:02~2022
-01-01 12:00:05,工作时长为14523秒;2)2022-01-01 16:01:35~
2022-01-02 00:20:05,工作时长为29910秒;工作时长一共是
14523+29910=44433秒。现在小蓝想知道在2022年度自己的工作
时长一共是多少秒?

答案提交

这是一道结果填空的题,你只需要算出结果后提交即可。本题的结果为
个整数,在提交答案时只填写这个整数,填写多余的内容将无法得分。

#include<bits/stdc++.h>
#pragma GCC optimize(2)
#define int long long
#define PII pair<int, int>
 
using namespace std;
const int N = 6e2 + 10;
 
struct date {
    int mou, day;
    int h, mi, s;
}d[N];
 
int th, tmi, ts, ans;
 
bool cmp(date a, date b)
{
    if(a.mou != b.mou)
        return a.mou < b.mou;
    else
        if (a.day != b.day)
            return a.day < b.day;
        else
            if(a.h != b.h)
                return a.h < b.h;
            else
                if(a.mi != b.mi)
                    return a.mi < b.mi;
                else
                    return a.s < b.s;
}
 
void solve()
{
    for(int i = 0; i < 520; i += 2)
    {
        if(d[i].day == d[i + 1].day)
        {
            ans += 3600 * (d[i + 1].h - d[i].h) + 60 * (d[i + 1].mi - d[i].mi) + (d[i + 1].s - d[i].s);
        } else {
            ans += 3600 * (24 - d[i].h + d[i + 1].h) + 60 * (60 - d[i].mi + d[i + 1].mi) + (60 - d[i].s + d[i + 1].s);
        }
    }
    cout << ans << endl;
}
 
int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int _ = 1;
//    cin >> _;
 
    for (int i = 0; i < 520; ++i) {
        string s;
        getline(cin, s);
        
        d[i].mou = (s[6] - '0') + (s[5] - '0') * 10;
        d[i].day = (s[9] - '0') + (s[8] - '0') * 10;
        d[i].h = (s[12] - '0') + (s[11] - '0') * 10;
        d[i].mi = (s[15] - '0') + (s[14] - '0') * 10;
        d[i].s = (s[18] - '0') + (s[17] - '0') * 10;
    }
 
    sort(d, d + 520, cmp);
    
    while(_--) solve();
    
    return 0;
}

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2022-01-21 07:58:27
2022-02-07 22:41:33
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2022-01-07 13:36:54
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