【TopCoder SRM 157 DIV 1】 Salary 解题报告_html/css_WEB-ITnose

本来想把原题给贴过来的，后来看到人家有版权保护，在网上搜了一下，好像也没有人在博客中贴过原题，为避免麻烦，就不copy人家原题了。

【题目大意】

公司按照每小时 wage 元给员工发工资，员工上班打卡记录一个时间，下班记录一个时间，时间格式为 "hh:mm:ss"。其中，18:00:00 到 23:59:59 和 00:00:00 到 05:59:59 之间算加班时间，工资为1.5倍。计算工资时，分钟和秒都换算成小时来计算，最终的工资取整数，舍弃后面的小数。

【函数格式】

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int howMuch(vector <string> arrival, vector <string> departure, int wage)

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【注意】

arrival和departure元素个数一样，且arrival[0]

【解题思路】

思路不难，只是时间区间判断比较麻烦，有两个时间关键点06:00:00和18:00:00。

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【我的代码】

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class Salary {public:	int compareTime(int hh1, int mm1, int ss1, int hh2, int mm2, int ss2) {		if (hh1 > hh2 || (hh1 == hh2 && mm1 > mm2) || (hh1 == hh2 && mm1 == mm2 && ss1 > ss2)) {			return 1;		}		if (hh1 < hh2 || (hh1 == hh2 && mm1 < mm2) || (hh1 == hh2 && mm1 == mm2 && ss1 < ss2)) {			return -1;		}		return 0;	}	double duringTime(int hh1, int mm1, int ss1, int hh2, int mm2, int ss2) {		double time = 0;		if (ss2 >= ss1) {			time += (ss2 - ss1) / 3600.0;		} else {			time += (ss2 + 60 - ss1) / 3600.0;			mm2 -= 1;		}		if (mm2 >= mm1) {			time += (mm2 - mm1) / 60.0;		} else {			time += (mm2 + 60 - mm1) / 60.0;			hh2 -= 1;		}		time += hh2 - hh1;		return time;	}	int howMuch(vector <string> arrival, vector <string> departure, int wage) {		int beginHour, endHour, beginMinute, endMinute, beginSecond, endSecond;		stringstream ss;		double total = 0;		for (int i = 0; i < arrival.size(); i++) {			string time1 = arrival[i];			ss.clear();			ss << time1.substr(0, 2);			ss >> beginHour;			ss.clear();			ss << time1.substr(3, 2);			ss >> beginMinute;			ss.clear();			ss << time1.substr(6, 2);			ss >> beginSecond;			string time2 = departure[i];			ss.clear();			ss << time2.substr(0, 2);			ss >> endHour;			ss.clear();			ss << time2.substr(3, 2);			ss >> endMinute;			ss.clear();			ss << time2.substr(6, 2);			ss >> endSecond;			if (compareTime(beginHour, beginMinute, beginSecond, 6, 0, 0) < 0) {				if (compareTime(endHour, endMinute, endSecond, 6, 0, 0) < 0) {					total += duringTime(beginHour, beginMinute, beginSecond, endHour, endMinute, endSecond) * wage * 1.5;				} else if (compareTime(endHour, endMinute, endSecond, 6, 0, 0) >= 0						&& compareTime(endHour, endMinute, endSecond, 18, 0, 0) < 0) {					total += duringTime(beginHour, beginMinute, beginSecond, 6, 0, 0) * wage * 1.5;					total += duringTime(6, 0, 0, endHour, endMinute, endSecond) * wage;				} else if (compareTime(endHour, endMinute, endSecond, 18, 0, 0) >= 0) {					total += duringTime(beginHour, beginMinute, beginSecond, 6, 0, 0) * wage * 1.5;					total += duringTime(6, 0, 0, 18, 0, 0) * wage;					total += duringTime(18, 0, 0, endHour, endMinute, endSecond) * wage * 1.5;				}			} else if (compareTime(beginHour, beginMinute, beginSecond, 18, 0, 0) < 0) {				if (compareTime(endHour, endMinute, endSecond, 18, 0, 0) < 0) {					total += duringTime(beginHour, beginMinute, beginSecond, endHour, endMinute, endSecond) * wage;				} else if (compareTime(endHour, endMinute, endSecond, 18, 0, 0) >= 0) {					total += duringTime(beginHour, beginMinute, beginSecond, 18, 0, 0) * wage;					total += duringTime(18, 0, 0, endHour, endMinute, endSecond) * wage * 1.5;				}			} else {				total += duringTime(beginHour, beginMinute, beginSecond, endHour, endMinute, endSecond) * wage * 1.5;			}		}		return (int)total;	}};

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首先是字符串转换成时分秒整数值。两个自定义函数功能分别为：比较两个时间点先后，主要是和两个关键时间点06:00:00和18:00:00比较；计算两个时间点之间的小时数。

然后一大段逻辑判断，判断员工工作时间是不是在00:00:00和06:00:00之间、06:00:00和18:00:00之间、18:00:00和00:00:00之间，还是说跨时间段。

由于基础不好，代码写得比较?嗦，欢迎大家改进。