[백준/BOJ] [C++] [6069번] Switching Lights

문제 페이지

https://www.acmicpc.net/problem/6069

 

 

 

문제

Farmer John tries to keep the cows sharp by letting them play with intellectual toys. One of the larger toys is the lights in the barn. Each of the N (2 <= N <= 500) cow stalls conveniently numbered 1..N has a colorful light above it.

At the beginning of the evening, all the lights are off. The cows control the lights with a set of N pushbutton switches that toggle the lights; pushing switch i changes the state of light i from off to on or from on to off.

The cows read and execute a list of M (1 <= M <= 2,000) operations expressed as one of two integers (0 <= operation <= 1).

The first operation (denoted by a 0 command) includes two subsequent integers S_i and E_i (1 <= S_i <= E_i <= N) that indicate a starting switch and ending switch. They execute the operation by pushing each pushbutton from S_i through E_i inclusive exactly once.

The second operation (denoted by a 1 command) asks the cows to count how many lights are on in the range given by two integers S_i and E_i (1 <= S_i <= E_i <= N) which specify the inclusive range in which the cows should count the number of lights that are on.

Help FJ ensure the cows are getting the correct answer by processing the list and producing the proper counts.

입력

• Line 1: Two space-separated integers: N and M
• Lines 2..M+1: Each line represents an operation with three space-separated integers: operation, S_i, and E_i

 

출력

• Lines 1..number of queries: For each output query, print the count as an integer by itself on a single line.

 

코드

#include <bits/stdc++.h> 
using namespace std;

int main()
{
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int n, m; 
    cin >> n >> m;
    
    vector<bool> switches(n+1, false);
    
    for (int i = 0; i < m; ++i)
    {
        int op, b, c;
        cin >> op >> b >> c;
        
        if (op == 0)
        {
            for (int j = b; j <= c; ++j)
            {
                switches[j] = !switches[j];
            }
        }
        else
        {
            int cnt = 0;
            for (int j = b; j <= c; ++j)
            {
                if (switches[j])
                {
                    ++cnt;
                }
            }
            cout << cnt << '\n';
        }
    }

    return 0;
}

 

코드 설명

1. n과 m이 입력된다. n은 기본적으로 꺼져있는 스위치의 길이이며, m은 명령의 개수이다.

2. m회 명령을 입력받는데, 명령은 operation, start, end 값의 구조로 구성된다.

3. operation은 0과 1이 있고, 0이 입력되면 start부터 end까지 위치 스위치의 상태를 변경한다. 꺼져있다면 켜고, 켜져있다면 끈다. 1이 입력되면 start부터 end 위치 스위치에 대해 켜져있는 스위치의 개수를 출력한다.

4. 문제에 맞춰 코드를 작성한다.