blocks|key|2743760|text|您可以使用嵌套的switch+case来避免使用if-else。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2743761|2743762|找出今年是否是闰年。|ordered-list-item|2743763|创建四个月桶(1-99)、(100-199)..存储桶编号将用作箱号。|2743764|现在检查一天中最左边的位，并写入一个开关箱体，将其分配到右侧存储桶中。|2743765|每个存储桶也可以被分成4个以上的存储桶(将使用闰年信息)。|2743766|对中间位重复步骤3，并根据结果(情况)切换到适当的存储桶。|2743767|2743768|简而言之，逻辑可能如下：|2743769|isLeapYear+=+year+%25+4

Switch(isLeapYear)
Case+0:+{
++first_bucket+=+days/4
++Switch(first_bucket)
++{
+++Case+0:+{
++++++++++++days_left+=+days+%25+100
++++++++++++second_bucket+=+days+/+50;
//+...
//+...
}
Case+1,+2,+3:+{
//+Similar+logic+for+non-leap+year
//+...
}|code-block|syntax|javascript|2743770|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|Z|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|10|8|@]|9|@]|A|$]]|$1|H|3|I|5|E|7|11|8|@]|9|@]|A|$]]|$1|J|3|K|5|E|7|12|8|@]|9|@]|A|$]]|$1|L|3|M|5|E|7|13|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|14|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|15|8|@]|9|@]|A|$]]|$1|Q|3|R|5|S|7|16|8|@]|9|@]|A|$T|U]]|$1|V|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|W|$]]

You can use nested switch case to avoid using if-else. 

<ol>
<li>Find out whether it is a leap year or not. </li>
<li>Create four buckets of months (1-99), (100-199).. etc. Bucket numbers will be used as case numbers.</li>
<li>Now check the left most bit of the day, and write a switch case for assigning it into the right bucket. </li>
<li>Each bucket could be divided into 4 more buckets too (the leap year information will be used). </li>
<li>Repeat step 3 for the middle bit and based on the result (case) switch to appropriate bucket.</li>
</ol>

Briefly, logic could be as follows:

<pre><code>isLeapYear = year % 4

Switch(isLeapYear)
Case 0: {
 first_bucket = days/4
 Switch(first_bucket)
 {
 Case 0: {
 days_left = days % 100
 second_bucket = days / 50;
// ...
// ...
}
Case 1, 2, 3: {
// Similar logic for non-leap year
// ...
}
</code></pre>

blocks|key|814609|text|对于1901到2099年，可以使用时间函数轻松地完成。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|814610|void+Makedate(int+year,+int+day,+struct+tm+*dest)+{
++struct+tm+tm1+=+{+0+};
++//+Leap+year+every+4+years,+lets+do+calc+referencing+2000-2003
++tm1.tm_year+=+2000+-+1900+%2B+year+%25+4;
++tm1.tm_mday+=+day;+//+Make+the+Jan+1st+to+Jan+366th.++OK+to+be+out+of+range.
++//+Avoid+day+changes+due+to+DST+by+using+Noon.+BTW+I+doubt+this+is+needed+(CYA)
++tm1.tm_hour+=+12;++
++//+mktime+adjusts+our+fields+for+us,+setting+the+month,+mday,+dow,+etc.
++mktime(&tm1);
++tm1.tm_year+=+year+-+1900;++//+set+to+selected+year
++*dest+=+tm1;
}

int+main()+{
++struct+tm+tm0;
++Makedate(2013,+++1,+&+tm0);+printf("y:%254d+m:%252d+d:%252d\n",+tm0.tm_year+%2B+1900,+tm0.tm_mon+%2B+1,+tm0.tm_mday);
++Makedate(2013,+365,+&+tm0);+printf("y:%254d+m:%252d+d:%252d\n",+tm0.tm_year+%2B+1900,+tm0.tm_mon+%2B+1,+tm0.tm_mday);
++Makedate(2020,+++1,+&+tm0);+printf("y:%254d+m:%252d+d:%252d\n",+tm0.tm_year+%2B+1900,+tm0.tm_mon+%2B+1,+tm0.tm_mday);
++Makedate(2020,+366,+&+tm0);+printf("y:%254d+m:%252d+d:%252d\n",+tm0.tm_year+%2B+1900,+tm0.tm_mon+%2B+1,+tm0.tm_mday);
++return+0;
}

y:2013+m:+1+d:+1
y:2013+m:12+d:31
y:2020+m:+1+d:+1
y:2020+m:12+d:31|code-block|syntax|javascript|814611|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

For years 1901 to 2099, can easily be done using time functions.

<pre><code>void Makedate(int year, int day, struct tm *dest) {
 struct tm tm1 = { 0 };
 // Leap year every 4 years, lets do calc referencing 2000-2003
 tm1.tm_year = 2000 - 1900 + year % 4;
 tm1.tm_mday = day; // Make the Jan 1st to Jan 366th. OK to be out of range.
 // Avoid day changes due to DST by using Noon. BTW I doubt this is needed (CYA)
 tm1.tm_hour = 12; 
 // mktime adjusts our fields for us, setting the month, mday, dow, etc.
 mktime(&amp;tm1);
 tm1.tm_year = year - 1900; // set to selected year
 *dest = tm1;
}

int main() {
 struct tm tm0;
 Makedate(2013, 1, &amp; tm0); printf("y:%4d m:%2d d:%2d\n", tm0.tm_year + 1900, tm0.tm_mon + 1, tm0.tm_mday);
 Makedate(2013, 365, &amp; tm0); printf("y:%4d m:%2d d:%2d\n", tm0.tm_year + 1900, tm0.tm_mon + 1, tm0.tm_mday);
 Makedate(2020, 1, &amp; tm0); printf("y:%4d m:%2d d:%2d\n", tm0.tm_year + 1900, tm0.tm_mon + 1, tm0.tm_mday);
 Makedate(2020, 366, &amp; tm0); printf("y:%4d m:%2d d:%2d\n", tm0.tm_year + 1900, tm0.tm_mon + 1, tm0.tm_mday);
 return 0;
}

y:2013 m: 1 d: 1
y:2013 m:12 d:31
y:2020 m: 1 d: 1
y:2020 m:12 d:31
</code></pre>

blocks|key|808431|text|您可以在不使用任何类型的流控制语句(if、switch)，也不需要自己计算闰年的情况下完成此操作。获取与所需年份的1月1日对应的时间戳。然后，您可以在时间戳上添加一次，以获得正确的日期，并将时间戳转换为您想要的任何格式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|808432|我会提供代码，具体的函数名，并解释你需要在时间戳中添加什么，但因为这显然是一个家庭作业问题，所以我不会提供。如果有人在两周后提醒我(即作业的最后期限最有可能过了)，我会很高兴地发布示例代码。|808433|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

You can do it without any kind of flow control statements (if, switch) and without doing leap year calculation yourself. Get the timestamp corresponding to January 1st of the desired year. Then, you can use a single addition on the timestamp to get to the correct day, and convert the timestamp to whatever format you want.

I would provide code, specific function names and explain what you need to add to the timestamp, but since this is obviously a homework question, I won't. If anyone reminds me in two weeks (i.e. when the deadline for the assignment is most likely over) I'll happily post example code.

blocks|key|428023|text|如果允许使用库函数(我对此表示怀疑)，我会这样做(懒惰的乐高积木)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|428024|#define+_XOPEN_SOURCE+/*+glibc2+needs+this+for+strptime+*/
#include+<stdio.h>
#include+<stdlib.h>
#include+<time.h>++++
#include+<errno.h>

int+to_date(
++char+*+date,+
++const+size_t+size,+
++const+char+*+fmt,+
++const+short+unsigned+int+day_of_year,+
++const+short+unsigned+int+year)
{
++char+buffer[16]+=+"";

++sprintf(buffer,+"%25hu+%25hu",+day_of_year,+year);

++{
++++struct+tm+t+=+{0};
++++char+*+presult+=+strptime(buffer,+"%25j+%25Y",+&t);

++++if+((NULL+==+presult)+%7C%7C+('\0'+!=+*presult))
++++{
++++++errno+=+EINVAL;
++++++return+-1;
++++}

++++strftime(date,+size,+fmt,+&t);
++}

++return+0;
}

int+main(int+argc,+char+**+argv)
{
++if+(2+>+argc)
++{
++++fprintf(stderr,+"Missing+arguments.+Usage:+%25s+day-of-year+year\n",+argv[0]);
++++return+EXIT_FAILURE;
++}

++short+unsigned+int+day_of_year+=+atoi(argv[1]);
++short+unsigned+int+year+=+atoi(argv[2]);
++char+date[16]+=+"";

++if+(-1+==+to_date(date,+sizeof(date),+"%25m/%25d/%25Y",+day_of_year,+year))
++{
++++perror("to_date()+failed");
++++return+EXIT_FAILURE;
++}

++printf("Result:+day+%25d+of+year+%25d+is+'%25s'.\n",+day_of_year,+year,+date);

++return+EXIT_SUCCESS;
}|code-block|syntax|javascript|428025|这样叫它|428026|$+./main+2+2000|428027|要获得|428028|Result:+day+2+of+year+2000+is+'01/02/2000'.++++|428029|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

In case library functions were allowed to be used (which I doubt) I'd do it this (lazy lego) way:

<pre><code>#define _XOPEN_SOURCE /* glibc2 needs this for strptime */
#include &lt;stdio.h&gt;
#include &lt;stdlib.h&gt;
#include &lt;time.h&gt; 
#include &lt;errno.h&gt;

int to_date(
 char * date, 
 const size_t size, 
 const char * fmt, 
 const short unsigned int day_of_year, 
 const short unsigned int year)
{
 char buffer[16] = "";

 sprintf(buffer, "%hu %hu", day_of_year, year);

 {
 struct tm t = {0};
 char * presult = strptime(buffer, "%j %Y", &amp;t);

 if ((NULL == presult) || ('\0' != *presult))
 {
 errno = EINVAL;
 return -1;
 }

 strftime(date, size, fmt, &amp;t);
 }

 return 0;
}

int main(int argc, char ** argv)
{
 if (2 &gt; argc)
 {
 fprintf(stderr, "Missing arguments. Usage: %s day-of-year year\n", argv[0]);
 return EXIT_FAILURE;
 }

 short unsigned int day_of_year = atoi(argv[1]);
 short unsigned int year = atoi(argv[2]);
 char date[16] = "";

 if (-1 == to_date(date, sizeof(date), "%m/%d/%Y", day_of_year, year))
 {
 perror("to_date() failed");
 return EXIT_FAILURE;
 }

 printf("Result: day %d of year %d is '%s'.\n", day_of_year, year, date);

 return EXIT_SUCCESS;
}
</code></pre>

Call it like this

<pre><code>$ ./main 2 2000
</code></pre>

to get

<pre><code>Result: day 2 of year 2000 is '01/02/2000'. 
</code></pre>

I am trying to write a program in C given these two data sets from a user. Year:[Range 1901-2099] and Day in year: [range 1-366] I need a formula to calculate the date in MM/DD/YYYY format. One more thing. No IF/ELSE statements. No AND/OR or GREATER THAN or LESS THAN are allowed.

calculate a date given day of year and year

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试用C语言编写一个程序，给定来自用户的这两个数据集。年份:范围1901-2099和年中的日期:范围1-366我需要一个公式来计算MM/DD/YYYY格式的日期。还有一件事。没有IF/ELSE语句。不允许AND/OR或大于或小于。

问计算一年中的某一天
EN

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