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  1. #include <stdio.h>
  2.  
  3.  
  4. int Leap(int year) {
  5.     if (year % 400 == 0) return 1;
  6.     else if (year % 100 == 0) return 0;
  7.     else if (year % 4 == 0) return 1;
  8.     else return 0;
  9. }
  10.  
  11.  
  12. int DayMonth(int year, int month) {
  13.     if (month == 4 || month == 6 || month == 9 || month == 11)
  14.         return 30;
  15.     else if (month == 2)
  16.         return Leap(year) ? 29 : 28;
  17.     else
  18.         return 31;
  19. }
  20.  
  21. int Zeller(int year, int month, int day) {
  22.     if (month == 1 || month == 2) {
  23.         month += 12;
  24.         year--;
  25.     }
  26.     int C = year / 100;
  27.     int Y = year % 100;
  28.  
  29.     int weekday = ((day + (26 * (month + 1) / 10) + Y + Y / 4 + 5 * C + C / 4) + 5) % 7;
  30.     return weekday; // 月曜=0, ..., 日曜=6
  31. }
  32.  
  33. int main() {
  34.     int year, month;
  35.     printf("西暦年と月を入力してください（例: 2024 6）: ");
  36.     scanf("%d %d", &year, &month);
  37.  
  38.     if (month < 1 || month > 12) {
  39.         printf("エラー: 月は1〜12の範囲で入力してください。\n");
  40.         return 1;
  41.     }
  42.  
  43.     int days = DayMonth(year, month);
  44.     int weekday = Zeller(year, month, 1);  // その月の1日の曜日
  45.  
  46.     printf("\n  月  火  水  木  金  土  日\n");
  47.  
  48.  
  49.     int i;
  50.     for (i = 0; i < weekday; i++) {
  51.         printf("   ");
  52.     }
  53.  
  54.  
  55.     for (i = 1; i <= days; i++) {
  56.         printf("%3d", i);
  57.         if (weekday == 6)
  58.             printf("\n");
  59.         weekday++;
  60.         weekday %= 7; // 0〜6 に戻す
  61.     }
  62.  
  63.     if (weekday != 0) printf("\n");
  64.  
  65.     return 0;
  66. }
Success #stdin #stdout 0.01s 5308KB
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2025 6
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西暦年と月を入力してください（例: 2024 6）: 
  月  火  水  木  金  土  日
                    1
  2  3  4  5  6  7  8
  9 10 11 12 13 14 15
 16 17 18 19 20 21 22
 23 24 25 26 27 28 29
 30
https://ideone.com/fnJjpb
language:
C (gcc 8.3)
created:
2 days ago
visibility:
public

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