Codeforces 71B Progress Bar Solution & Explanation

Difficulty : 1300

Problem Description

A progress bar is an element of graphical interface that displays the progress of a process for this very moment before it is completed. Let’s take a look at the following form of such a bar.

A bar is represented as n squares, located in line. To add clarity, let’s number them with positive integers from 1 to n from the left to the right. Each square has saturation (a_i for the i-th square), which is measured by an integer from 0 to k. When the bar for some i (1 ≤ i ≤ n) is displayed, squares 1, 2, … , i - 1 has the saturation k, squares i + 1, i + 2, … , n has the saturation 0, and the saturation of the square i can have any value from 0 to k.

So some first squares of the progress bar always have the saturation k. Some last squares always have the saturation 0. And there is no more than one square that has the saturation different from 0 and k.

The degree of the process’s completion is measured in percents. Let the process be t% completed. Then the following inequation is fulfilled:

An example of such a bar can be seen on the picture.

For the given n, k, t determine the measures of saturation for all the squares a_i of the progress bar.

Input

We are given 3 space-separated integers n, k, t (1 ≤ n, k ≤ 100, 0 ≤ t ≤ 100).

Output

Print n numbers. The i-th of them should be equal to a_i.

Examples

Solution

C# Solution

Solution1

string[] first = Console.ReadLine().Split(' '); //split by space
int n = int.Parse(first[0]);
int k = int.Parse(first[1]);
int t = int.Parse(first[2]);

int res = n*k*t/100;
int full = res/k;
int left = res%k;

int[]resArr = new int[n];

for(int i=0; i<n; i++){
    resArr[i] = i<full ? k : (i==full ? left : 0);
}

Console.WriteLine(String.Join(" ", resArr));

This C# program generates a progress bar with n squares, each having saturation values according to specific rules.

The first full squares have saturation k, the next square has a partial saturation (left), and the remaining squares have saturation 0. The resulting saturation values are then printed to the console.

Java Solution

Solution1

import java.util.Scanner;
import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String[] first  = scanner.nextLine().split(" ");
        
        int n = Integer.parseInt(first[0]);
        int k = Integer.parseInt(first[1]);
        int t = Integer.parseInt(first[2]); 
        
        int res = n*k*t/100;
        int full = res/k;
        int left = res%k;
        
        int[]resArr = new int[n];

        for(int i=0; i<n; i++){
            resArr[i] = i<full ? k : (i==full ? left : 0);
        }
        System.out.println(String.join(" ", Arrays.stream(resArr).mapToObj(String::valueOf).toArray(String[]::new)));
    }
}

Conclusion

This problem is the second problem from codeforces contest – Codeforces Beta Round 65 (Div. 2)

You can find the solution for the first problem here – Codeforces 71A Way Too Long Words Solution & Explanation

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The problem link : Problem – 71B – Codeforces

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