如何计算未来3天的日期？

例如今天2011.06.28

输出：

2011.06.29

2011.06.30

2011.07.01

各位拜托啦 ！小弟不才，跪求指导！

是获取系统时间还是手动输入时间呢？

.....

#include <stdio.h>
#include <time.h>
#define MAXSIZE 21

int main(void) {
    char str[MAXSIZE];
    char * fmt = "%Y.%m.%d";
    time_t t;
    time(&t);
    strftime(str, MAXSIZE, fmt, localtime(&t));
    printf("%s\n", str);
    return 0;
}

[ 本帖最后由 lz1091914999 于 2011-6-26 11:12 编辑 ]

都可以的呀

以前做过的习题，加个循环就可以啦。

root@~ #cat lt94.c
#include <stdio.h>
#include <stdbool.h>

struct date {
        int month;
        int day;
        int year;
};

struct date dateUpdate (struct date today) {

        struct date tomorrow;
        int numberOfDays (struct date d);

        if(today.day != numberOfDays(today)) {
                tomorrow.day=today.day+1;
                tomorrow.month=today.month;
                tomorrow.year=today.year;
        }
        else if(today.month==12) {
                tomorrow.day=1;
                tomorrow.month=1;
                tomorrow.year=today.year+1;
        }else{
                tomorrow.day=1;
                tomorrow.month=today.month+1;
                tomorrow.year=today.year;
        }

        return tomorrow;

}

int numberOfDays (struct date d) {

        int days;
        bool isLeapYear (struct date d);

        const int daysPerMonth[12]={ 31,28,31,30,31,30,31,31,30,31,30,31 };

        if(isLeapYear(d)==true && d.month==2) {
                days=29;
        }else{
                days=daysPerMonth[d.month-1];
        }

        return days;

}

bool isLeapYear (struct date d) {

        bool leapYearFlag;

        if((d.year%4==0 && d.year%100!=0)||d.year%400==0) {
                leapYearFlag=true;
        }else{
                leapYearFlag=false;
        }

        return leapYearFlag;

}

int main (void) {
        int i;
        struct date dateUpdate (struct date today);
        struct date today,nextday;

        printf ("Enter today's date (mm dd yyyy) :");
        scanf ("%i%i%i",&today.month,&today.day,&today.year);

        for(i=0;i<3;i++) {
                nextday=dateUpdate(today);
                printf ("%.2i/%.2i/%i\n",nextday.month,nextday.day,nextday.year)
;
                today.day++;
        }

        return 0;

}

root@~ #./lt94
Enter today's date (mm dd yyyy) :12 29 2011
12/30/2011
12/31/2011
01/01/2012
root@~ #

这位大哥，stdbool.h是什么啊 ？我用的是c++求解。。。。