是的,我有一個想法。我的想法是,你回去重新思考你選擇的算法來確定一天是星期四。這是大錯特錯了:-)
現在,像斷線的時鐘是正確的,一天兩次,你可能會發現輸入參數給你正確的答案,但他們會是例外,而不是規則。
如果你想弄清楚的時候,下週四是從給定日期,C提供日期和時間功能,正是出於這個目的:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
static char *textday[] = {"Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat"};
int main (int argc, char *argv[]) {
int year, month, day, today, thursday;
struct tm *mytm;
time_t mytime;
// Get all arguments (minimal error checks).
if (argc != 3) {
printf ("Usage: next_thursday <year> <month>\n");
return -1;
}
year = atoi (argv[1]);
month = atoi (argv[2]);
// Do first fourteen days of the month.
for (day = 1; day <= 14; day++) {
直到有,它只是越來越參數和啓動循環。計算的結果如下,建立一個有用的struct tm
,然後強制我們的年,月和日進入它。然後mktime
將填寫我們的tm_wday
(星期幾)字段,我們可以使用它來計算直到星期四的天數。
// Make the tm structure based on date (and midday).
mytime = time (0);
mytm = localtime (&mytime);
mytm->tm_year = year - 1900;
mytm->tm_mon = month - 1;
mytm->tm_mday = day;
mytm->tm_hour = 12;
mytime = mktime (mytm);
// Output filled in fields and days till next Thursday.
today = mytm->tm_wday;
thursday = (11 - today) % 7;
if (thursday == 0)
thursday = 7;
printf ("%04d-%02d-%02d, weekday = %d (%s), days till Thu = %d\n",
mytm->tm_year + 1900, mytm->tm_mon + 1, mytm->tm_mday,
today, textday[today], thursday);
}
return 0;
}
注意,thursday
計算是位模數弄虛作假的 - 它只是用來給我們幾天的基於下表上的數字:
today thursday
------- --------
0 (sun) 4
1 (mon) 3
2 (tue) 2
3 (wed) 1
4 (thu) 7
5 (fri) 6
6 (sat) 5
如果你想要一個更可讀的解決方案,你可以使用:
if (today < 4) thursday = 4 - today;
else thursday = 11 - today;
該程序輸出爲2011-03
如下:
2011-03-01, weekday = 2 (Tue), days till Thu = 2
2011-03-02, weekday = 3 (Wed), days till Thu = 1
2011-03-03, weekday = 4 (Thu), days till Thu = 7
2011-03-04, weekday = 5 (Fri), days till Thu = 6
2011-03-05, weekday = 6 (Sat), days till Thu = 5
2011-03-06, weekday = 0 (Sun), days till Thu = 4
2011-03-07, weekday = 1 (Mon), days till Thu = 3
2011-03-08, weekday = 2 (Tue), days till Thu = 2
2011-03-09, weekday = 3 (Wed), days till Thu = 1
2011-03-10, weekday = 4 (Thu), days till Thu = 7
2011-03-11, weekday = 5 (Fri), days till Thu = 6
2011-03-12, weekday = 6 (Sat), days till Thu = 5
2011-03-13, weekday = 0 (Sun), days till Thu = 4
2011-03-14, weekday = 1 (Mon), days till Thu = 3
我不明白它是如何工作的,甚至你的例子。 daysToNextThursday(2011,3,17)週四(7)減去日(17)。 7 - 17 = -10,而不是7 – Nikolay 2012-03-24 06:57:30
很難告訴你與你提供的東西少的代碼做什麼,但當然'週四 - day'將是錯的,如果'day'是天而不是一週中的某一天... – jamesdlin 2012-03-24 06:57:33
我猜這是作業嗎?如果是這樣,請將作業標籤添加到此問題。除此之外,算法是有缺陷的。你可能想重新思考每個變量扮演的角色。 – Bart 2012-03-24 07:09:54