本文共 2026 字，大约阅读时间需要 6 分钟。

写的比较粗陋，编程能力还需要多多磨练啊

struct SDate{	int Leapday[12]; //闰年的每月天数//		int NoLeapDay[12]; //非闰年的每月天数	int year;	int month;	int day;	int hour; 	int minute;	int second;} sDate = { 31,29,31,30,31,30,31,31,30,31,30,31 };bool isLeap(int year) //判断是否闰年{	if (year%4==0 && year%100 || year%400==0)		return true;	else return false;}void getNextSecond(SDate *in, SDate &out) //获取当前秒的下一秒{	if (!in) return;	memcpy(&out, in, sizeof(SDate));	if (in->second==59) 	{		out.second=0;		if (in->minute==59) 		{			out.minute=0;			if (in->hour==23) 			{				out.hour=0;				if (in->day==in->Leapday[in->month-1]) 				{					out.day=1;					if (in->month==12) 					{						out.month=1;						out.year+=1;					}					else out.month+=1;				}				else out.day+=1;			}			else out.hour += 1;		}		else out.minute +=1;	}	else out.second += 1;}bool fillData(SDate *in){	if (in==NULL) return false;	printf("请依次输入：年 月 日 时 分 秒");	while (1)	{		puts("年： ");		scanf("%d", &in->year);		if (in->year<=0) {			puts("年必须大于0,请重新输入年份：");			continue;		}		break;	}	while (1)	{		printf("月： ");		scanf("%d", &in->month);		if (in->month<1 || in->month >12) {			puts("月范围是<1-12>,请重新输入月份：");			continue;		}		break;	}	bool leap;	leap = isLeap(in->year);	if (leap)		;	else in->Leapday[1]=28; 	while (1)	{		printf("日： ");		scanf("%d", &in->day);		if (in->day < 0 || in->day > in->Leapday[in->month-1]) {			puts("日子大小不对,请重新输入日：");			continue;		}		break;	}	while (1)	{		puts("时： ");		scanf("%d", &in->hour);		if (in->hour <0 || in->hour > 24) {			puts("小时数不对,请重新输入：");			continue;		}		break;	}	while (1)	{		printf("分： ");		scanf("%d", &in->minute);		if (in->hour <0 || in->minute > 60) {			puts("分钟数不对,请重新输入：");			continue;		}		break;	}	while (1)	{		printf("秒： ");		scanf("%d", &in->second);		if (in->second <0 || in->second > 60) {			puts("秒数不对,请重新输入：");			continue;		}		break;	}}void main(){	SDate outDate;	fillData(&sDate);	getNextSecond(&sDate, outDate);	printf("下一秒为：%d 年 %d 月 %d 日 %d 时 %d 分 %d 秒\n", \		outDate.year, outDate.month, outDate.day, outDate.hour, outDate.minute, outDate.second);}