Merge pull request ashutosh97#63 from murtaza1112/master

When to take medicine-Codechef

Author: ashutosh97

When to take medicine/problem_statement.txt

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+                                                                                                             When to take medicine 
+
+All submissions for this problem are available.You visit a doctor on a date given in the format yyyy:mm:dd. Your doctor suggests you to take pills every alternate day starting from that day. You being a forgetful person are pretty sure won’t be able to remember the last day you took the medicine and would end up in taking the medicines on wrong days.
+
+So you come up with the idea of taking medicine on the dates whose day is odd or even depending on whether dd is odd or even. Calculate the number of pills you took on right time before messing up for the first time.
+
+

When to take medicine/when to take medicine_answer.txt

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+Language-c++
+
+
+Solution-
+
+#include <iostream>
+using namespace std;
+int get_year(long int year,int month,int day);
+int main() 
+{
+	// your code goes here
+	int test;
+	cin>>test;
+	while(test!=0)
+	{  int month,day;
+	   long int year;
+	  char a,b;
+	  cin>>year>>a>>month>>b>>day;
+	   int answer=get_year(year,month,day);
+	    cout<<answer<<endl;
+	    test--;
+	    
+	}
+	return 0;
+}
+
+int get_year(long int year,int month,int day)
+{   bool leap_year=false;
+     int answer{0};
+     
+    
+     if((year%4==0 && year%100!=0) ||year%400==0)
+        leap_year=true;
+    
+     for(;;)
+    {
+        bool odd=day%2;
+        
+        if(month ==1 ||month ==3 ||month ==5 ||month ==7 ||month ==8 ||
+    month ==10 ||month ==12)
+    { 
+         answer+=((31-day)/2) + 1;
+               break;
+        
+    }else if(month ==4 ||month ==6 ||month ==9 ||month ==11)
+    {    
+            answer+=((30-day)/2) + 1;
+               
+               ((odd)?day=1:day=2);
+               month++;
+               continue;
+    }else
+    {   
+        if(leap_year)
+        {    
+            answer+=((29-day)/2) + 1;
+               break;
+        }else
+        {
+            answer+=((28-day)/2) + 1;
+               
+              ((odd)?day=1:day=2);
+              month++;
+              continue;
+        }
+    
+    }
+    }
+    return answer;
+}