1.普通的写法

#include

int leapyear(int year)

{

if((year%4==0 && year%100!=0) || year%400==0)

return 1;

else

return 0;

}

int days(int *day1, int *day2)

{

int i=0;

int *tmp;

int diff = 0;

const int month[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

if(day1[0] == day2[0])

{

if(day1[1] == day2[1])

{

diff = day1[2] - day2[2];

diff = (diff < 0)?(-diff):diff;

}

else

{

if(day1[1] < day2[1])    //day1=1991-5-8   day2=1991-6-2

{

tmp = day1;      //day1=1991-6-2   day2=1991-5-8

day1 = day2;

day2 = tmp;

}

for(i=day2[1]+1; i

{

diff +=  month[i];

}

diff += month[day2[1]] - day2[2] + day1[2];

if(day2[1] <= 2 && day1[1] >2)

if(leapyear(day2[0]))

diff++;

}

}

else

{

if(day1[0] < day2[0])

{

tmp = day1;

day1 = day2;

day2 = tmp;

}

for(i=day2[0]+1; i

{

if(leapyear(i))

diff += 366;

else

diff += 365;

}

for(i=day2[1]+1; i<=12; i++)     //day1=1992-1-1   day2=1991-1-1

{

diff += month[i];

}

diff += (month[day2[1]] - day2[2]);

if(day2[1] <= 2)

if(leapyear(day2[0]))

diff++;

for(i=1; i

{

diff += month[i];

}

diff += day1[2];

if(day1[1] > 2)

if(leapyear(day1[0]))

diff++;

}

return diff;

}

int main()

{

int day1[3], day2[3];

int day = 0;

printf("输入日期：");

scanf("%d-%d-%d",&day1[0], &day1[1], &day1[2]);

printf("输入另一个日期：");

scanf("%d-%d-%d",&day2[0], &day2[1], &day2[2]);

day = days(day1, day2);

printf("两个日期之间共有%d天。\n",day);

return 0;

}

2.利用结构体，代码更整洁一些

#include

typedef struct date

{

int year;

int month;

int day;

}DATE;

int leapyear(int year)

{

if((year%4==0 && year%100!=0) || year%400==0)

return 1;

else

return 0;

}

int compare(DATE *d1, DATE *d2)    //如果第一个日期比第二个日期大，交换日期

{

DATE *tmp;

if(d1->year == d2->year)    //年数相等

{

if(d1->month > d2->month)   //月数相等

{

tmp = d1;

d1 = d2;

d2 = d1;

}

else if(d1->month == d2->month)  //日期相等

{

if(d1->day > d2->day)

{

tmp = d1;

d1 = d2;

d2 = d1;

}

}

}

else if(d1->year > d2->year)

{

tmp = d1;

d1 = d2;

d2 = tmp;

}

return 0;

}

int diff(DATE *date1, DATE *date2)

{

int i;

int diff = 0;

const int month[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

if(date1->year == date2->year)

{

if(date1->month == date2->month)

{

diff = date2->day - date1->day;

}

else

{

for(i=date1->month+1; imonth; i++)

{

diff += month[i];

}

diff += month[date1->month] - date1->day + date2->day;

if(leapyear(date1->year))

if(date1->month <=2 && date2->month >2)

diff++;

}

}

else

{

for(i=date1->year+1; iyear; i++)

{

if(leapyear(i))

diff += 366;

else

diff += 365;

}

for(i=date1->month+1; i<=12; i++)    //date1距离年末多少天

{

diff += month[i];

}

diff += month[date1->month] - date1->day;

if(date1->month <= 2)

if(leapyear(date1->year))

diff++;

for(i=1; imonth; i++)     //date2距离年初多少天

{

diff += month[i];

}

diff += date2->day;

if(date1->month > 2)

if(leapyear(date2->year))

diff++;

}

return diff;

}

int main()

{

int days = 0;

DATE day1, day2;

DATE *date1, *date2;

date1 = &day1;

date2 = &day2;

printf("输入日期：");

scanf("%d-%d-%d",&(date1->year), &(date1->month), &(date1->day));

printf("输入另一个日期：");

scanf("%d-%d-%d",&date2->year, &date2->month, &date2->day);

compare(date1, date2);

days = diff(date1, date2);

printf("两个日期之间共有%d天。\n",days);

return 0;

}