Codecheff may lunch time(TOM and Jerry)

   youtube channel  : https://youtube.com/c...

 

 youtube channel  : https://youtube.com/channel/UCYzdgz0Xhcb0tFp_b3gTesQ

Tom And Jerry 1 Problem Code: TANDJ1 

There is a grid of size 105×105

, covered completely in railway tracks. Tom is riding in a train, currently in cell (a,b), and Jerry is tied up in a different cell (c,d), unable to move. The train has no breaks. It shall move exactly K steps, and then its fuel will run out and it shall stop. In one step, the train must move to one of its neighboring cells, sharing a side. Tom can’t move without the train, as the grid is covered in tracks. Can Tom reach Jerry’s cell after exactly K

steps?

Note: Tom can go back to the same cell multiple times.

Input

• The first line contains an integer T

• .

Output

For each testcase, output in a single line "YES" if Tom can reach Jerry's cell in exactly K

moves and "NO" if not.

You may print each character of the string in uppercase or lowercase (for example, the strings "yEs", "yes", "Yes" and "YES" will all be treated as identical).

Constraints

• 1≤T≤105

• 
  

Subtasks

Subtask #1 (100 points): original constraints

Sample Input

3
1 1 2 2 2
1 1 2 3 4
1 1 1 0 3

Sample Output

YES
NO
YES

Explanation

Test Case 1

: A possible sequence of moves is (1,1)→(1,2)→(2,2)

.

Test Case 2

: There is a possible sequence in 3 moves, but not in exactly 4

moves.

Test Case 3

: A possible sequence of moves is (1,1)→(1,0)→(0,0)→(1,0)

===============================================

 

SOLUTION:

  #include "bits/stdc++.h"
  #include "ext/pb_ds/assoc_container.hpp"
 
  using namespace std;
  using namespace __gnu_pbds;
 
  #define int long long
  #define pi (3.141592653589)
  #define mod 1000000007
  #define float double
  #define t() int test;cin>>test;while(test--)
  #define pb push_back
  #define mp make_pair
  #define ii pair<int,int>  
  #define ff first
  #define ss second
  #define all(c) c.begin(), c.end()
  #define fr(i, n) for(int i=n-1;i>=0;i--)
  #define fo(i,n) for(int i=0;i<n;i++)
  #define f(i,a,n) for(int i=a;i<=n;i++)
  #define     endl            "\n"
  #define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
 
 
 
  bool isPrime(int n){
      if(n==1) return false;
      if(n==2) return true;
        for(int i=2;i*i<=n;i++){
          if(n%i==0)return false;
      }
      return true;
  }
  typedef tree< int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
 
  #define TRACE
  #ifdef TRACE
  #define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
      cout << name << " : " << arg1 << endl;
      //use cerr if u want to display at the bottom
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
      const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1 << " | "; __f(comma + 1, args...);
  }
  #else
  #define trace(...)
  #endif
 
  void init(){
  #ifndef ONLINE_JUDGE
      freopen("in.txt", "r", stdin);
      freopen("outp.txt", "w", stdout);
  #endif
      
  }
 
  int32_t main(){
    init();
    fast
    
    
    t(){
      int a,b,c,d,k;cin>>a>>b>>c>>d>>k;
      int diff = abs(b-d) + abs(a-c);
      int sum= (d+b)+(a+c);
      if(diff%2==0){
        if(k%2==0 && k>=diff){
          cout<<"YES"<<endl;
        }
        else {
          cout<<"NO"<endl;
        }
        
               
      }
      else {
        if(k%2!=0 && k>=diff){
          cout<<"YES"<<endl;
        }
        else {
          cout<<"NO"<<endl;
        }
      }
    }
      return 0;
 }