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请问这一题怎么解决?(用c语言)(并解决闰年的问题)
9.2. Exercise
Calling time (NULL) gives you a large integer. Use this value to calculate today's year, date, hours, and minutes.
Simplifying assumption:we pretend that each year has exactly365 days. The 15th of March is therefore day 83 in year 2011. (use that date to figure out today's "day number")(hint: how many seconds are there in a day?)
You cannot use struct for this assignment.
You cannot use any time-based functions from the standard library such asasctime or strftime.
You must use one function to calculate the year and day(pass inthe total seconds, and a reference to the year and day).
You must use a separate function to calculate the hours andminutes.(again, pass the total seconds, and a reference to the hours and minutes)
You must use a separate function to print the time; this functioncan only take as arguments the year, day, hour, and minutes.
handle leap-years correctly, and print the month as well.
首先,对于计算年份和日期,我们可以使用以下方法:
1. 获取当前时间的总秒数:
- time_t currentTime = time(NULL);
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2. 计算今天是一年中的第几天:
- int dayOfYear = ((currentTime / 60) / 60) / 24;
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3. 计算今年的年份:
- int currentYear = 2011 + dayOfYear / 365;
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4. 计算月份:
- int month;
- if (dayOfYear >= 0 && dayOfYear < 31)
- month = 1;
- else if (dayOfYear >= 31 && dayOfYear < 59)
- month = 2;
- else if (dayOfYear >= 59 && dayOfYear < 90)
- month = 3;
- // 其他月份省略...
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5. 计算日期:
- int dayOfMonth;
- switch (month) {
- case 1:
- dayOfMonth = dayOfYear + 1;
- break;
- case 2:
- dayOfMonth = dayOfYear - 30;
- break;
- case 3:
- dayOfMonth = dayOfYear - 58;
- break;
- // 其他月份省略...
- }
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接下来,我们来处理小时和分钟的计算。我们需要一个函数来计算这个时间段中的小时和分钟数。代码示例如下:
- void calcHoursMinutes(int totalSeconds, int* hours, int* minutes) {
- *hours = (totalSeconds/ 3600) % 24;
- *minutes = (totalSeconds / 60) % 60;
- }
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最后,我们需要编写一个函数来打印时间。代码示例如下:
- void printTime(int year, int month, int day, int hour, int minute) {
- printf("当前时间:%d年%d月%d日 %d时%d分\n", year, month, day, hour, minute);
- }
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至于闰年的问题,可以使用以下方法进行判断:
- int isLeapYear(int year) {
- if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
- return 1;
- else
- return 0;
- }
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在计算年份时,可以考虑闰年对天数的影响,例如2月份有28或29天。
希望这些代码对你有帮助!
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