被这个题搞自闭了
编写函数,输入年、月、日,输出当年剩余天数。在main中输入今天的日期,调用函数并输出今年已过的天数和剩余天数。
void foo( unsigned y, unsigned m, unsigned d ) { static const unsigned s[2][12] = { {0,31,59,90,120,151,181,212,243,273,304,334} , {0,31,60,91,121,152,182,213,244,274,305,335} }; _Bool isleap = (y%400==0 || (y%4==0 && y%100!=0)); unsigned a = s[isleap][m-1] + d; unsigned b = (isleap?366:365) - a; printf( "%u, %u\n", a, b ); }
int sumday(int y, int m, int d) { m--; return m * 30 + d + (m > 1 ? -2 : 0) + (m > 7 ? m / 2 + 1 : (m + 1) / 2); }
void theDiff(unsigned year, unsigned month, unsigned day) { year -= 1900; struct tm start = { 0 }; struct tm end = { 0 }; start.tm_year = year; start.tm_mon = 1; start.tm_mday = 1; end.tm_year = year; end.tm_mon = 12; end.tm_mday = 31; end.tm_hour = 23; end.tm_min = 59; end.tm_sec = 59; time_t startT, currentT, endT; startT = mktime(&start); currentT = time(NULL); endT = mktime(&end); double diffPast = difftime(currentT ,startT); double diffHas = difftime(endT , currentT); printf("%f \t %f\n", diffPast / (24 * 3600), diffHas / (24 * 3600)); }
[此贴子已经被作者于2020-4-15 09:01编辑过]