1006. Sign In and Sign Out (25)
2018-01-23 22:11
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简析:
(1)求第一个登录的人和最后一个登出的人; (2)m:记录的个数; (3)记录:id 登录的时间 登出的时间(hh:mm:ss; id为string,少于15字符)(4)关键在于如何比较时间,化为整数比 h*3600 + m*60 + s; (5)采用结构体, id, t1, t2;题目很简单,一次ac.#include <iostream> using namespace std; struct node{ string id; int t1; int t2; }; int main(int argc, const char * argv[]) { int m; cin >> m; struct node n[m]; int hh, mm, ss; for (int i = 0; i < m; i++) { cin >> n[i].id; scanf("%d:%d:%d", &hh, &mm, &ss); n[i].t1 = hh * 3600 + mm * 60 + ss; scanf("%d:%d:%d", &hh, &mm, &ss); n[i].t2 = hh * 3600 + mm * 60 + ss; } int maxi = 0, max = n[0].t2, mini = 0, min = n[0].t1; for (int i = 1; i < m; i++) { if (max < n[i].t2) { max = n[i].t2; maxi = i; } if (min > n[i].t1) { min = n[i].t1; mini = i; } } cout << n[mini].id << ' ' << n[maxi].id << endl; return 0; }
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