[백준]2290 - LCD Test (구현)

https://www.acmicpc.net/problem/2290

2290번: LCD Test

주어진 숫자를 그림과 같이 출력하는 문제

https://astrid-dm.tistory.com/372

 

도저히 감이 안와서 다른 블로그 글을 참조

위 그림과 같이 7구역으로 나누고 각 숫자가 지나가는 구역을 배열로 표시한다

 

#define _CRT_SECURE_NO_WARNINGS
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

int temp[10][7] = {
    {1,1,1,0,1,1,1}, //0
    {0,0,1,0,0,1,0}, //1
    {1,0,1,1,1,0,1}, //2
    {1,0,1,1,0,1,1}, //3
    {0,1,1,1,0,1,0}, //4
    {1,1,0,1,0,1,1}, //5
    {1,1,0,1,1,1,1}, //6
    {1,0,1,0,0,1,0}, //7
    {1,1,1,1,1,1,1}, //8
    {1,1,1,1,0,1,1}, //9
};

int main() {
    int N,k;
    cin >> N;
    cin >> k;
    vector<int> v;
    while (1) {
        int t;
        t = k % 10;
        v.push_back(t);
        k = k / 10;
        if (k == 0)
            break;

    }
    reverse(v.begin(), v.end());

    for (int i = 0; i < v.size(); i++) {
        if (temp[v[i]][0] == 0)
            for(int j=0 ; j<N+3 ; j++)
                cout << ' ';
        else {
            cout << ' ';
            for (int j = 0; j < N; j++)
                cout << '-';
            cout << ' ';
            cout << ' ';
        }
    }
    cout << '\n';

    for (int k = 0; k < N; k++) {
        for (int i = 0; i < v.size(); i++) {
            if (temp[v[i]][1] == 0 && temp[v[i]][2] == 0)
                for (int j = 0; j < N + 3; j++)
                    cout << ' ';
            else if (temp[v[i]][1] == 1 && temp[v[i]][2] == 0) {
                cout << '|';
                for (int j = 0; j < N + 2; j++)
                    cout << ' ';
            }
            else if (temp[v[i]][2] == 1 && temp[v[i]][1] == 0) {
                for (int j = 0; j < N + 1; j++)
                    cout << ' ';
                cout << '|';
                cout << ' ';
            }
            else {
                cout << '|';
                for (int j = 0; j < N; j++)
                    cout << ' ';
                cout << '|';
                cout << ' ';
            }
        }
        cout << '\n';
    }

    for (int i = 0; i < v.size(); i++) {
        if (temp[v[i]][3] == 0)
            for (int j = 0; j < N + 3; j++)
                cout << ' ';
        else {
            cout << ' ';
            for (int j = 0; j < N; j++)
                cout << '-';
            cout << ' ';
            cout << ' ';
        }
    }
    cout << '\n';

    for (int k = 0; k < N; k++) {
        for (int i = 0; i < v.size(); i++) {
            if (temp[v[i]][4] == 0 && temp[v[i]][5] == 0)
                for (int j = 0; j < N + 3; j++)
                    cout << ' ';
            else if (temp[v[i]][4] == 1 && temp[v[i]][5] == 0) {
                cout << '|';
                for (int j = 0; j < N + 2; j++)
                    cout << ' ';
            }
            else if (temp[v[i]][5] == 1 && temp[v[i]][4] == 0) {
                for (int j = 0; j < N + 1; j++)
                    cout << ' ';
                cout << '|';
                cout << ' ';
            }
            else {
                cout << '|';
                for (int j = 0; j < N; j++)
                    cout << ' ';
                cout << '|';
                cout << ' ';
            }
        }
        cout << '\n';
    }

    for (int i = 0; i < v.size(); i++) {
        if (temp[v[i]][6] == 0)
            for (int j = 0; j < N + 3; j++)
                cout << ' ';
        else {
            cout << ' ';
            for (int j = 0; j < N; j++)
                cout << '-';
            cout << ' ';
            cout << ' ';
        }
    }
}

이후엔 0번, 12번, 3번, 45번, 6번 구역을 값에 따라 출력해주면 끝