Calculate two days separated by the number of ideas: Suppose 1998-10-10 2010-5-5 first get 1889-10-10 This date how many days left in this year again get 2010-5-5 this day has been over in this year

Class FunDemo6

{

public static void Main (string[] args)

{

test function getDays

System.out.println (GetDays (1992,4,20));

System.out.println (Subdays (1999,1,5,2001,3,10));

}

determine if a leap year

public static Boolean isleap (int y)

{

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

return true;

Else

return false;

}

get how many days this year has passed

public static int getDays (int y,int m,int D)

{

int sum=0;

Switch (m-1) {

Case 11:sum=sum+30;

Case 10:sum=sum+31;

Case 9:sum=sum+30;

Case 8:sum=sum+31;

Case 7:sum=sum+31;

Case 6:sum=sum+30;

Case 5:sum=sum+31;

Case 4:sum=sum+30;

Case 3:sum=sum+31;

Case 2:if (Isleap (y))// a function called another function

sum+=29;// functions cannot nest declarations

Else

sum+=28;

Case 1:sum=sum+31;

}

return sum+d;

}

Write a function that calculates how many days are left in the year for this date

public static int otherdays (int y,int m,int D)

{

if (Isleap (y))

Return 366-getdays (Y,M,D);

else// can be omitted

Return 365-getdays (Y,M,D);

}

write a function directly to calculate the number of days separated by two dates

public static int subdays (int y1,int m1,int d1,int y2,int m2,int D2)

{

if (y1==y2)

{

int Days1=getdays (Y1,M1,D1)-getdays (Y2,M2,D2);

Return ABS (DAYS1);

}

else if (y1<y2)

{

Small year remaining + big year already over + Middle year

int sum1=otherdays (Y1,M1,D1);

int sum2=getdays (Y2,M2,D2);

int sum3=0;

for (int i=y1+1;i<y2;i++)

{

if (Isleap (i))

sum3+=366;

Else

sum3+=365;

}

return sum1+sum2+sum3;

}

when else//y2>y1

{

int sum1=otherdays (Y1,M1,D1);

int sum2=getdays (Y2,M2,D2);

int sum3=0;

for (int i=y2+1;i<y1;i++)

{

if (Isleap (i))

sum3+=366;

Else

sum+=365;

}

return sum1+sum2+sum3;

}

}

public static int abs (int n) {

Return n>=0?n:-n;

}

}

Calculate two days separated by the number of ideas: Suppose 1998-10-10 2010-5-5 first get 1889-10-10 This date how many days left in this year again get 2010-5-5 this day has been over in this year