i can’t build logic , don’t know how to approach this ques
#include <bits/stdc++.h>
using namespace std;
int main() {
// your code goes here
int t;
cin>>t;
while(t--){
int A,B;
cin>>A>>B;
if(A<=B){
cout<<"";
}
}
}
Problem Link: Extreme Basketball Practice Coding Problem - CodeChef
Alice has A amount of points
Bob has B amount of points
What is the smallest amount of shots Alice has to do to have a dominant winning (winning for at least 10 points)?
Alice has to have:
A >= B +10
Imagine that Alice has 10 points and Bob 0. She has a dominant winning, but if Alice has 10 points and Bob 1, that’s not a dominant winning.
Let’s say this:
Alice needs at least B + 10 points. Did she already have them? How many does she need if not?
If she already have the points, then you doesn’t need more shots.
If she need more points, she needs more shots.
// Does she already have them or not?
if (A >= B + 10){
cout << "0\n";
}
else{
left = B + 10 - A;
...
}
If she needs more shots (else case), what the minimal amount of shots she needs to do? If she can make 2 and 3 point shots, well… let’s make her try 3 shots. It’s not prohibited by the problem anyway.
shots = left / 3
But what if she needs 10 points?
10 / 3 is 3.3333
She needs 4 shots, not “3.333” shots. That’s senseless. You you needed to ceil your division.
Heeeence:
#include <bits/stdc++.h>
using namespace std;
long long Ceil(long long A, long long B){
if (B == 0){
return NULL;
}
long long res = A / B;
if (A % B > 0){
res++;
}
return res;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int T;
cin >> T;
while(T--){
long long A,B;
cin >> A >> B;
if (A >= B + 10){
cout << "0\n";
}
else{
long long left, shots;
left = B + 10 - A;
shots = Ceil(left,3);
cout << shots << "\n";
}
}
}