PAT (Advanced Level) Practise 1006. Sign In and Sign Out (25)

2019-11-06 09:14:55

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1006. Sign In and Sign Out (25) 时间限制: 400 ms内存限制: 65536 kB代码长度限制: 16000 B判题程序：Standard At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in’s and out’s, you are supposed to find the ones who have unlocked and locked the door on that day. Input Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format: ID_number Sign_in_time Sign_out_time where times are given in the format HH:MM:SS, and ID number is a string with no more than 15 characters. Output For each test case, output in one line the ID numbers of the persons who have unlocked and locked the door on that day. The two ID numbers must be separated by one space. Note: It is guaranteed that the records are consistent. That is, the sign in time must be earlier than the sign out time for each person, and there are no two persons sign in or out at the same moment. Examples

Input
3CS301111 15:30:28 17:00:10SC3021234 08:00:00 11:25:25CS301133 21:45:00 21:58:40
Output
SC3021234 CS301133

Notes 作者 CHEN, Yue

　　利用好scanf格式化输入。

#include <iostream>#include <algorithm>#include <map>#include <vector>#include <functional>#include <string>#include <cstring>#include <queue>#include <set>#include <stack>#include <cmath>#include <cstdio>#include <sstream>#include <iomanip>using namespace std;#define IOS ios_base::sync_with_stdio(false)#define TIE std::cin.tie(0)#define MIN2(a,b) (a<b?a:b)#define MIN3(a,b) (a<b?(a<c?a:c):(b<c?b:c))#define MAX2(a,b) (a>b?a:b)#define MAX3(a,b,c) (a>b?(a>c?a:c):(b>c?b:c))typedef long long LL;typedef unsigned long long ULL;const int INF = 0x3f3f3f3f;const double PI = 4.0*atan(1.0);const double eps = 1e-6;int M, h, m, s, t, t1, t2;char str[20],up[20],lp[20];int main(){ scanf("%d", &M); t1 = INF; t2 = 0; for (int i = 0; i < M; i++){ scanf(" %s", str); scanf("%d:%d:%d", &h, &m, &s); t = h * 3600 + m * 60 + s; if (t < t1){ t1 = t; strcpy(up, str); } scanf("%d:%d:%d", &h, &m, &s); t = h * 3600 + m * 60 + s; if (t > t2){ t2 = t; strcpy(lp, str); } } PRintf("%s %s/n", up, lp); //system("pause");}

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