时间处理：计算下一天日期,如输入"2004/12/31"(注释2014年12月31日)，则输出"2005/1/1".

/*
 ============================================================================
 Name         : Exercise.c
 Author       : haier
 Version      : 0.01
 Copyright   : Your copyright notice
 Description : Ansi-style, Compile by Code:Blocks, Platform Linux
 ============================================================================
 */

/*
思路：首先求出输入时间到1970-1-1总秒数，然后加上一天时间后转换成日历时间并输出。
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <ctype.h>

#define SECONDS_PER_DAY (60*60*24)

struct date
{
    unsigned year;
    unsigned month;
    unsigned day;
} myDate;

void nextDay(struct date myDate)
{
    struct tm time,*tm_ptr;
    time_t seconds;

    /*填充struct tm*/
    time.tm_year = myDate.year - 1900;
    time.tm_mon  = myDate.month - 1;
    time.tm_mday = myDate.day;
    time.tm_hour = 12; //默认为该日12:00:00
    time.tm_min  = 0;
    time.tm_sec  = 0;
    seconds=mktime(&time)+SECONDS_PER_DAY;  //转换tm结构为time_t类型值，并加一天

    /*转换并打印日期*/
    tm_ptr = gmtime(&seconds);
    printf("%d/%d/%d\n",tm_ptr->tm_year+1900, tm_ptr->tm_mon+1, tm_ptr->tm_mday);
}

int main()
{
    char strTime[20];
    char *p;
    int inputTimes=3;

    /*输入处理，允许输入3次*/
    do
    {
        inputTimes--;
        printf("Please input the date(YYYY/MM/DD) : ");
        scanf("%s",strTime);

        /*转换时间字符串*/
        p = strtok(strTime, "/");
        myDate.year=atoi(p);
        p = strtok(NULL, "/");
        myDate.month=atoi(p);
        p = strtok(NULL, "/");
        myDate.day=atoi(p);

        if(myDate.year<1970 || myDate.month<=0 || myDate.month>12 || myDate.day<=0 || myDate.day >31)
        {
            printf("error in input, please try again !\n");
        }
        else
        {
             break;
        }

    }while(inputTimes>0);


    nextDay(myDate);

    return 0;
}