runtime terror

An Interesting Sequence Problem Solution Code: CodeChef #include   <bits/stdc++.h> using   namespace   std ; #define   endl   " \n " int   main () {     cin. tie ( 0 );      int  N  =   4 e 6   +   5 ;      int  phi[N], ans[N];      for  ( int  i  =   0 ; i  <  N; i ++ )     {         phi[i]  =  i;         ans[i]  =   0 ;     }      for  ( int  p  =   2 ; p  <  N; p ++ )     {          if  (phi[p]  ==  p)         {             phi[p]  =  p  -   1 ;              for  ( int  i  =   2   *  p; i  <  N; i  +=  p)             {                 phi[i]  =  (phi[i]  /  p)  *  (p  -   1 );             }         }     }      for  ( int  i  =   1 ; i  <  N; i ++ )     {         ans[i]  +=  i  -   1 ;          for  ( int  j  =   2   *  i; j  <  N; j  +=  i)         {             ans[j]  +=  i  *  (( 1   +  phi[j  /  i])  /   2 );         }     }      int  t;     cin  >>  t;      while  (t -- )     {          int  k;         cin  >>  k;         cout  <&l

 Valid Paths Problem Solution Code: CodeChef Solution Code #include <bits/stdc++.h> using   namespace   std ; #define   ll   long   long #define   f ( i ,  a ,  b )  for ( int  i = a; i < b; i ++ ) #define   mod   1000000007 #define   mk   make_pair #define   uniq ( v ) (v). erase ( unique ( all (v)), (v). end ()) #define   ff  first #define   ss  second #define   rf ( i ,  a ,  b )  for ( int  i = a;i >= b;i -- ) #define   sc ( a )  scanf ( " %lld " ,  & a) #define   pf   printf #define   sz ( a ) ( int )(a. size ()) #define   psf  push_front #define   ppf  pop_front #define   ppb  pop_back #define   pb  push_back #define   pq  priority_queue #define   all ( s ) s. begin (),s. end () #define   sp ( a )  setprecision (a) #define   rz  resize #define   ld   long   double #define   inf  ( ll ) 1 e 18 #define   ub  upper_bound #define   lb  lower_bound #define   bs  binary_search #define   eb  emplace_back const   double  pi  =   acos ( - 1 ); ll   binpow ( ll  

 Tree House Problem Solution Code: CodeChef Solution Code #pragma   GCC   optimize (" Ofast ", " unroll-loops ") #include   <bits/stdc++.h> using   namespace   std ; #define   int   long   long   int #define   double   long   double using   pii   =   pair < int ,  int >; template  < typename   T > using   Prior   =   std :: priority_queue < T ,  vector < T >,  greater < T >>; #define   X  first #define   Y  second #define   eb  emplace_back #define   ALL ( x )  begin (x),  end (x) #define   RALL ( x )  rbegin (x),  rend (x) #define   fastIO ()  ios_base ::sync_with_stdio, cin. tie ( 0 ) mt19937_64  rng( chrono :: steady_clock :: now (). time_since_epoch (). count ()); inline   int   getRand ( int   L ,  int   R ) {      if  ( L   >   R )          swap ( L ,  R );      return  ( int )(rng()  %  ( uint64_t )( R   -   L   +   1 )  +   L ); } template  < typename   T1 ,  typename   T2 > ostream   &operator <<( ostre

 Modular Equation Problem Code: CodeChef Solution Code #include <bits/stdc++.h> using   namespace   std ; #define   int   long   long   int #define   endl   " \n " int32_t   main () {      ios_base :: sync_with_stdio ( false );     cin. tie ( 0 );      int  t;cin >> t;      while (t -- )     {          int  n, m;cin >> n >> m;          int  count_of_pair  =   0 ;          vector < int >  modular_equation(n + 1 ,  1 );          for ( int  a  =   2 ;a <= n;a ++ )         {              int  x  =  m % a;             count_of_pair  +=  modular_equation [ x ] ;              for ( int  b  =  x;b <= n;b += a)             {                 modular_equation [ b ] ++ ;             }         }         cout << count_of_pair << endl ;     }      return   0 ; }

 Tic Tac Toe Problem Solution Code:  CodeChef Solution Code #include <bits/stdc++.h> using   namespace   std ; #define   ll   long   long   int int   main () {      ll  t;     cin >> t;      while (t -- )     {          ll  cx  =   0 , co  = 0 , c_ =   0 ;          char  a[ 3 ][ 3 ];          for ( ll  i = 0 ;i < 3 ;i ++ )         {              for ( ll  j = 0 ;j < 3 ;j ++ )             {                 cin >> a[i][j];                  if (a[i][j] == 'X' )cx ++ ;                  if (a[i][j] == 'O' )co ++ ;                  if (a[i][j] == '_' )c_ ++ ;             }         }          ll  wx  =   0 , wo  =   0 ;          if (a[ 0 ][ 0 ]  ==   'X'   &&  a[ 1 ][ 0 ]  ==   'X'   &&  a[ 2 ][ 0 ]  ==   'X' )wx = 1 ;          if (a[ 0 ][ 1 ]  ==   'X'   &&  a[ 1 ][ 1 ]  ==   'X'   &&  a[ 2 ][ 1 ]  ==   'X' )wx = 1 ;          if (a[ 0 ][ 2 ]  ==   'X'   &

 Xor Equality Problem Solution Code: CodeChef Solution Code #include <bits/stdc++.h> using   namespace   std ; #define   int   long   long   int #define   endl   " \n " #define   m   1000000007 int   power ( int   x ,  unsigned   int   y ,  int   p ) {      int  res  =   1 ;       x   =   x   %   p ;      if  ( x   ==   0 )  return   0 ;      while  ( y   >   0 )     {          if  ( y   &   1 )             res  =  (res * x )  %   p ;          y   =   y >> 1 ;  // y = y/2          x   =  ( x * x )  %   p ;     }      return  res; } int32_t   main () {      int  t;cin >> t;      while (t -- )     {          int  n;cin >> n;         cout << power ( 2 , n - 1 ,  m ) << endl ;     }      return   0 ; }