Answer:
C++:
C++ Code:
#include <iostream>
#include <string>
using namespace std;
struct date
{
int d,m,y;
};
int isLeap(int y)
{
if(y%100==0)
{
if(y%400==0)
return 1;
return 0;
}
if(y%4==0)
return 1;
return 0;
}
int day_no(date D)
{
int m = D.m;
int y = D.y;
int d = D.d;
int i;
int mn[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
for(i=0;i<m;i++)
{
d += mn[i];
}
if(isLeap(y))
{
if(m>2)
d++;
}
return d;
}
date get_info(string s)
{
date D;
int i,p1,p2,l = s.length();
for(i=0;i<l;i++)
{
if(s[i] == '-')
{
p1 = i;
break ;
}
}
for(i=p1+1;i<l;i++)
{
if(s[i] == '-')
{
p2 = i;
break ;
}
}
D.m = 0;
for(i=0;i<p1;i++)
D.m = (D.m)*10 + (s[i]-'0');
D.d = 0;
for(i=p1+1;i<p2;i++)
D.d = (D.d)*10 + (s[i]-'0');
D.y = 0;
for(i=p2+1;i<l;i++)
D.y = (D.y)*10 + (s[i]-'0');
return D;
}
int main()
{
string s1 = "4-5-2008";
string s2 = "12-30-1995";
string s3 = "6-21-2000";
string s4 = "1-31-1500";
string s5 = "7-19-1983";
string s6 = "2-29-1976";
cout<<"Date\t\tDay no\n\n";
cout<<s1<<"\t"<<day_no(get_info(s1))<<endl;
cout<<s2<<"\t"<<day_no(get_info(s2))<<endl;
cout<<s3<<"\t"<<day_no(get_info(s3))<<endl;
cout<<s4<<"\t"<<day_no(get_info(s4))<<endl;
cout<<s5<<"\t"<<day_no(get_info(s5))<<endl;
cout<<s6<<"\t"<<day_no(get_info(s6))<<endl;
return 0;
}
Explanation:
