以及距他下一次生日的天数。是C语言，不是C++

Question

希望能给完整的代码，谢谢
从键盘输入一个人的出生年月日和当前日期，计算他的实际年龄，以及距他下一次生日的天数。C语言程序

Answer 1

代码如下，最下面有用例示范：

#include <stdio.h>

#include <math.h>

int bigMonth[7]={1,3,5,7,8,10,12};

int smallMonth[4]={4,6,9,11};

int sumOfDays(int year1,int year2,int month1,int month2,int date1,int date2 )

{

int totalDays=0;

int dif_year,dif_month,dif_date;//difference between strating ana ending year/month/day

int startingDayOfYear,endingDayOfYear;//起始日期在当年是第多少天

dif_year=year2-year1;

dif_month=month2-month1;

dif_date=date2-date1;

switch(month1)

{

case(1):

startingDayOfYear=date1;break;

case(2):

startingDayOfYear=31+date1;break;

case(3):

startingDayOfYear=60+date1;break;

case(4):

startingDayOfYear=91+date1;break;

case(5):

startingDayOfYear=121+date1;break;

case(6):

startingDayOfYear=152+date1;break;

case(7):

startingDayOfYear=182+date1;break;

case(8):

startingDayOfYear=213+date1;break;

case(9):

startingDayOfYear=244+date1;break;

case(10):

startingDayOfYear=274+date1;break;

case(11):

startingDayOfYear=305+date1;break;

case(12):

startingDayOfYear=335+date1;break;

default:break;

}

if(month1>2 && year1%4!=0)

startingDayOfYear--;

switch(month2)

{

case(1):

endingDayOfYear=date2;break;

case(2):

endingDayOfYear=31+date2;break;

case(3):

endingDayOfYear=60+date2;break;

case(4):

endingDayOfYear=91+date2;break;

case(5):

endingDayOfYear=121+date2;break;

case(6):

endingDayOfYear=152+date2;break;

case(7):

endingDayOfYear=182+date2;break;

case(8):

endingDayOfYear=213+date2;break;

case(9):

endingDayOfYear=244+date2;break;

case(10):

endingDayOfYear=274+date2;break;

case(11):

endingDayOfYear=305+date1;break;

case(12):

endingDayOfYear=335+date2;break;

default:break;

}

if(month2>2 && year2%4!=0)

endingDayOfYear--;

totalDays=endingDayOfYear-startingDayOfYear+365*(year2-year1);

for(int i=year1;i<year2;i++)

{

if(i%4==0)

totalDays++;

}

return totalDays;

}

void main()

{

int year1,year2,month1,month2,date1,date2;

printf("请输入出生日期:\n");

scanf("%d %d %d",&year1,&month1,&date1);

printf("请输入当前日期:\n");

scanf("%d %d %d",&year2,&month2,&date2);

if(month2>month1 || (month2==month1 && date2>=date1))

{

printf("实际年龄为: %d\n",year2-year1);

printf("距离下一次生日还有 %d 天\n",sumOfDays(year2,year2+1,month2,month1,date2,date1));

}

else

{

printf("实际年龄为: %d\n",year2-year1-1);

printf("距离下一次生日还有 %d 天\n",sumOfDays(year2,year2,month2,month1,date2,date1));

}

}

追问

非常感谢

Answer 2

#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#include <time.h>

int verify(const struct tm *ptm)
{
int leap;
int year;
int mday;
year = ptm->tm_year;
mday = ptm->tm_mday;
leap = (year%4 == 0 && year%100 != 0) || (year%400 == 0);
switch(ptm->tm_mon)
{
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:
return mday <= 31;
case 4:
case 6:
case 9:
case 11:
return mday <= 30;
case 2:
if(leap)
return mday <= 29;
else
return mday <= 28;
default:
return 0;
}
}

int main()
{
int age;
double secs;
time_t birth;
time_t today;
struct tm tm1;
struct tm tm2;
struct tm *ptm;
memset(&tm1, 0, sizeof(struct tm));
memset(&tm2, 0, sizeof(struct tm));
today = time(NULL);
ptm = localtime(&today);
tm2.tm_year = ptm->tm_year;
tm2.tm_mon = ptm->tm_mon;
tm2.tm_mday = ptm->tm_mday;
today = mktime(&tm2);
while(1)
{
printf("请输入生日(如：1990-3-15)：\n");
scanf("%d-%d-%d", &tm1.tm_year, &tm1.tm_mon, &tm1.tm_mday);
if(!verify(&tm1))
{
printf("输入的日期不合法，请重新输入。\n");
continue;
}
tm1.tm_year -= 1900;
tm1.tm_mon -= 1;
birth = mktime(&tm1);
if(today > birth)
break;
printf("您输入的出生日期大于或等于当前日期。\n");
}
printf("出生日期：%d-%d-%d\n", tm1.tm_year+1900, tm1.tm_mon+1, tm1.tm_mday);
printf("当前日期：%d-%d-%d\n", tm2.tm_year+1900, tm2.tm_mon+1, tm2.tm_mday);
age = tm2.tm_year - tm1.tm_year;
tm1.tm_year = tm2.tm_year;
birth = mktime(&tm1);
if(today >= birth)
{
age++;
tm1.tm_year++;
birth = mktime(&tm1);
}
secs = difftime(birth, today);
printf("年龄(周岁加一)：%d\n", age);
printf("距离下次生日%d-%d-%d还有%d天。\n",
tm1.tm_year+1900, tm1.tm_mon+1, tm1.tm_mday, (int)(secs/24/3600));
return 0;
}本回答被提问者采纳

Answer 3

这个问题很简单么，其实就一个判断平年和瑞年就好了，别的都是纯数字计算的。你是刚学C还是学了很久了？追问

学了不久，懂的不多

Answer 4

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