sorry i forgot to upload the explanation :
in this problem the in every 400 years no. of days on which cookoff and long challenge strikes are 101. so, we divide the diff b/w the years with 400 and multiply it with 101 . and then shift the year y1 by y1 + ((y2-y1)/400)*400 which will shift year y1 near to y2 and diff b/w these years is now less than 400 now. it will not give you “TLE” and after that you just have to find the days on which cookoff and long intersect b/w y1 and y2.
for that you see that video https://www.youtube.com/watch?v=45vy5OBw4k4&t=168s
and apply some trick to just find it in feb= month and date = 1 and if you cant understand why feb as month and date =1 only then ask it to me.

#include <bits/stdc++.h>
using namespace std;
int dayofweek(int d, int m, long int y)
{
static int t[] = { 0, 3, 2, 5, 0, 3,
5, 1, 4, 6, 2, 4 };
y -= m < 3;
return ( y + y / 4 - y / 100 +
y / 400 + t[m - 1] + d) % 7;
}
int main()
{
int t;
cin>>t;
for(int i=0;i<t;i++)
{
long int m1,m2,y1,y2;
cin>>m1>>y1>>m2>>y2;
if(m1>2)
{y1 = y1+1;}
if(m2<2)
{y2 = y2-1;}
long int dif=0;
dif = ((y2 - y1)/400)*101;
y1 = y1+(((y2-y1)/400)*400);
for(long int j=y1;j<=y2;j++)
{
int c=0;
int day = dayofweek(1, 2, j);
if(((j% 4 == 0) && (j% 100 != 0)) || (j% 400 == 0))
{c=1;}
if(day == 6)
{dif++;}
else if(day==0 && c!=1)
{dif++;}
}
cout<<dif<<endl;
}
return 0;
}

sorry bro, i can tell you the concept behind that.
in this problem the in every 400 years no. of days on which cookoff and long challenge strikes are 101. so, we divide the diff b/w the years with 400 and multiply it with 101 . and then shift the year y1 by y1 + ((y2-y1)/400)*400 which will shift year y1 near to y2 and diff b/w these years is now less than 400 now. it will not give you “TLE” and after that you just have to find the days on which cookoff and long intersect b/w y1 and y2.
for that you see that video https://www.youtube.com/watch?v=45vy5OBw4k4&t=168s
and apply some trick to just find it in feb= month and date = 1 and if you cant understand why feb as month and date =1 only then ask it to me.
and another sorry for inconvinence.

in this problem the in every 400 years no. of days on which cookoff and long challenge strikes are 101. so, we divide the diff b/w the years with 400 and multiply it with 101 . and then shift the year y1 by y1 + ((y2-y1)/400)*400 which will shift year y1 near to y2 and diff b/w these years is now less than 400 now. it will not give you “TLE” and after that you just have to find the days on which cookoff and long intersect b/w y1 and y2.
for that you see that video https://www.youtube.com/watch?v=45vy5OBw4k4&t=168s
and apply some trick to just find it in feb= month and date = 1 and if you cant understand why feb as month and date =1 only then ask it to me.