SPCP2 - Editorial

editorial
,
1

PROBLEM LINK:

Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4

Author: kg_26, alpha_ashwin
Tester: raysh07
Editorialist: iceknight1093

DIFFICULTY:

TBD

PREREQUISITES:

None

PROBLEM:

An airline has X flights, each of which can carry 100 people.
How many more do they need to carry N people in total?

EXPLANATION:

Each flight can hold 100 people, so initially the airline can carry 100\cdot X people.
So, if 100\cdot X \geq N, no more aircraft are needed and the answer is 0.

Otherwise, N - 100\cdot X people are remaining.
Since 100 of them can be placed on a single flight, the minimum number of flights needed is

\left\lceil \frac{N-100\cdot X}{100} \right\rceil

where \left\lceil \ \ \right\rceil denotes the ceiling function.

TIME COMPLEXITY:

\mathcal{O}(1) per testcase.

CODE:

Editorialist's code (Python) 
from math import ceil
for _ in range(int(input())):
    x, n = map(int, input().split())
    print(max(0, ceil((n - 100*x)/100)))
2 
#include <bits/stdc++.h>
using namespace std;

int main() {
	// your code goes here
	int t, x, n;
	
	cin >> t;
	
	while(t--){
	    cin >> x >> n;
	    int totalPlanes = (n+99)/100;
	    if(totalPlanes > x){
	        cout << totalPlanes - x << endl;
	    }else{
	        cout << 0 << endl;
	    }
	}
	
	return 0;
	
}