C 语言实现：根据时间替大笨钟“敲钟”代码解析

```c #include <stdio.h> int main() { int hour, minute; scanf("%d:%d", &hour, &minute); if (hour < 12 || (hour == 12 && minute == 0)) { printf("only %02d:%02d. too early to dang.\n", hour, minute); return 0; } if (hour >= 12 && hour < 24) { hour -= 12; } for (int i = 0; i <= hour; i++) { printf("dang"); } printf("\n"); return 0; } ``` ######[AI写代码神器 | 162点数解答 | 2024-10-31 18:01:56]