Codechef Rating Improvement problem solution

In this Codechef Rating Improvement problem solution Chef's current rating is XX, and he wants to improve it. It is generally recommended that a person with rating XX should solve problems whose difficulty lies in the range [X, X+200][X,X+200], i.e, problems whose difficulty is at least XX and at most X+200X+200.

You find out that Chef is currently solving problems with a difficulty of YY.

Is Chef following the recommended practice or not?

Problem solution in Python.

# cook your dish here
T = int(input())
for i in range(T):
    X,Y = map(int,input().split())
    if (X<=Y<=X+200):
        print("Yes")
    else:
        print("No")

Problem solution in Java.

import java.io.*;
import java.util.*;

class codechef {
    public static void main (String[] args) throws IOException {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new PrintWriter(System.out)));
        StringBuilder sb = new StringBuilder();
        
        StringTokenizer st = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(st.nextToken());

        while (t --> 0) {

            st = new StringTokenizer(br.readLine());
            int x = Integer.parseInt(st.nextToken());
            int y = Integer.parseInt(st.nextToken());

            sb.append((y >= x && y <= x + 200 ? "Yes" : "No") + "\n");
        }

        pw.println(sb.toString().trim());

        pw.close();
    }
}

Problem solution in C++.

#include <iostream>
using namespace std;

int main() {
	// your code goes here
	int T;
	cin >> T;
	for(int i =0; i < T; i++){
	    int X,Y,Z;
	    (cin >> X ) >> Y;
	    Z = X+200;
	    if(X <= Y && Z >= Y){
	        cout << "YES";
	    }else{
	        cout << "NO";
	    }
	    cout << "\n";
	}
	return 0;
}

Problem solution in C.

#include <stdio.h>

int main(void) {
	// your code goes here
	int P;
	scanf("%d",&P);
	int S,F;
	while(P--)
	{
	    scanf("%d %d",&S,&F);
	    if(F>=S && F<=(S+200))
	        printf("YES\n");
	    else 
	        printf("NO\n");
	}
	
	return 0;
}