VALENTINE - Editorial

editorial
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PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

Setter: Utkarsh Gupta
Tester: Abhinav Sharma
Editorialist: Kanhaiya Mohan

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

Chef has a total of X Rs and one chocolate costs Y Rs. What is the maximum number of chocolates Chef can buy.

EXPLANATION:

We are given that one chocolate costs Y rupees. Thus, in X rupees, we can buy ⌊\frac{X}{Y}⌋ chocolates.
Here, ⌊N⌋ is floor(N) which represents the greatest integer not greater than N.
Note that in languages like C++ and Java, a floating point number is automatically rounded down to the nearest integer.

Examples
  • X = 7, Y = 2: We know that 1 chocolate costs 2 rupees. Then, from 7 rupees, we can buy ⌊\frac{7}{2}⌋ = ⌊3.5⌋ = 3 chocolates.
  • X = 15, Y = 3: Single chocolate costs 3 rupees. Then, from 15 rupees, we can buy ⌊\frac{15}{3}⌋ = ⌊5⌋ = 5 chocolates.

TIME COMPLEXITY:

The time complexity is O(1) per test case.

SOLUTION:

Tester's Solution 
#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define db double
#define el "\n"
#define ld long double
#define rep(i,n) for(int i=0;i<n;i++)
#define rev(i,n) for(int i=n;i>=0;i--)
#define rep_a(i,a,n) for(int i=a;i<n;i++)


int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r" , stdin);
    freopen("output.txt", "w" , stdout);
    #endif
    int T=1;
    cin >> T;
    while(T--){
    
        int x,y;
        cin>>x>>y;

        cout<<x/y<<'\n';
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
    return 0;
}
Editorialist's Solution 
#include <iostream>
using namespace std;

int main() {
	int t;
	cin>>t;
	while(t--){
	    int x, y;
	    cin>>x>>y;
	    cout<<(x/y)<<endl;
	}
	return 0;
}