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教你如何用C语言实现的一个万年历(二)
2017-12-29 06:06:41 】 浏览:561
Tags:何用 语言 实现 一个万年历
break; case 6: return "六月"; break; case 7: return "七月"; break; case 8: return "八月"; break; case 9: return "九月"; break; case 10: return "十月"; break; case 11: return "十一月"; break; case 12: return "十二月"; break; default: break; } } /*按格式打印某年某月名称*/ void first_line_print(int month, int year) { //printf("%20s %d \n",month_name(month),year); printf("%8d年 %s \n",year, month_name(month)); } /*按格式打印星期名称*/ void week_print() { //printf("%-6s%-6s%-6s%-6s%-6s%-6s%-6s\n","Sun.", "Mon.", "Tues.", "Wed.", "Thur.", "Fri.", "Sat."); printf("%-3s%-3s%-3s%-3s%-3s%-3s%-3s\n","日", "一", "二", "三", "四", "五", "六"); } /*计算该年该月份与已知日子之间的距离天数*/ int date_distance_count(int month, int year) { int leap_year_count = 0; int i; int distance; if (year > known_year) { for (i=known_year; i { if(((i%4 == 0) && (i%100 != 0) ) || (i%400 == 0)) { leap_year_count++; } } if (month > 2) { if(((year%4 == 0) && (year%100 != 0) ) || (year%400 == 0)) { leap_year_count++; } } } else if (year == known_year) { if (month > 2) { leap_year_count = 1; } } distance = (year - known_year)*365 + leap_year_count + day_count(month); return distance; } /*确定该月份第一天是星期几*/ int makesure_firstday_weekday(int month, int year) { int date_distance = 0; int weekday; date_distance = date_distance_count(month, year); weekday = (known_weekday + date_distance)%7; return weekday; } /*依次打印出该月份的日子*/ void print_in_turn(int month, int firstday, int year) { int i = 1; int weekday; switch(firstday) { case Sun: break; case Mon: printf("%-3s",""); break; case Tues: printf("%-6s",""); break; case Wed: printf("%-9s",""); break; case Thur: printf("%-12s",""); break; case Fri: printf("%-15s",""); break; case Sat: printf("%-18s",""); break; } switch(month) { case 1: case 3: case 5: case 7: case 8: case 10: case 12: { for(i=0; i<31; i++) { weekday = (firstday + i)%7; printf("%-3d",i+1); if(weekday == Sat) { printf("\n"); } } break; } case 2: { if(((year%4 == 0) && (year%100 != 0) ) || (year%400 == 0)) /*闰年*/ { for(i=0; i<29; i++) { weekday = (firstday + i)%7; printf("%-3d",i+1); if(weekday == Sat) { printf("\n"); } } break; } else /*平年*/ { for(i=0; i<28; i++) { weekday = (firstday + i)%7; printf("%-3d",i+1); if(weekday == Sat) { printf("\n"); } } break; } } case 4: case 6:
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