(SWEA) 5658. [모의 SW 역량테스트] 보물상자 비밀번호

https://swexpertacademy.com/main/code/problem/problemDetail.do?contestProbId=AWXRUN9KfZ8DFAUo

SW Expert Academy

 

문제는 링크를 통해 확인해 주시기 바랍니다. 

 

#include <iostream>
#include <cstring>
#include <algorithm>
 
using namespace std;
 
char ch[35];
char cv1[8]; // 0~n/4 
char cv2[8]; // n/4 ~ n/4*2;
char cv3[8]; // n/4*2 ~n/4*3
char cv4[8]; // n/4*2 ~ n
 
int result[40] = { 0, };
int tc, n, k;
 
int main() {
  ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cin >> tc;
 
    for (int t = 1; t <= tc; t++) {
 
        cin >> n >> k;
        cin >> ch;
 
        // for(int i = 0; i<n; i++){
        //   cout<<ch[i];
        // }
 
        // cout<<endl;
        int index = 0;
        for (int r = 0; r < n / 4; r++) {
            int temp = ch[n - 1];
 
            for (int i = n - 1; i > 0; i--) {
                ch[i] = ch[i - 1];
            }
            ch[0] = temp;
            //cout<<ch<<endl;
 
            memset(cv1, NULL, sizeof(cv1));
            for (int i = 0; i < n / 4; i++) {
                cv1[i] = ch[i];
                //cout<<cv1[i];
            }//cout<<endl;
            //cout<<cv1<<endl;
 
            memset(cv2, NULL, sizeof(cv2));
            for (int i = n / 4; i < (n / 4) * 2; i++) {
                cv2[i - n / 4] = ch[i];
                //cout<<cv2[i-n/4];
            }//cout<<endl;
           // cout<<cv2<<endl;
 
            memset(cv3, NULL, sizeof(cv3));
            for (int i = (n / 4) * 2; i < (n / 4) * 3; i++) {
                cv3[i - (n / 4) * 2] = ch[i];
                // cout<<cv3[i-(n/4)*2];
            }//cout<<endl;
           // cout<<cv3<<endl;
 
            memset(cv4, NULL, sizeof(cv4));
            for (int i = (n / 4) * 3; i < n; i++) {
                cv4[i - (n / 4) * 3] = ch[i];
                //cout<<cv4[i-(n/4)*3];
            }//cout<<endl;
            //cout<<cv4<<endl;
            result[index] = (int)strtol(cv1, NULL, 16); index++;
            result[index] = (int)strtol(cv2, NULL, 16); index++;
            result[index] = (int)strtol(cv3, NULL, 16); index++;
            result[index] = (int)strtol(cv4, NULL, 16); index++;
 
 
            // cout<<cv1<<cv2<<cv3<<cv4<<endl;
            // cout<<cv1<<endl;
            // cout<<cv2<<endl;
            // cout<<cv3<<endl;
            // cout<<cv4<<endl;
 
          // //10진수로 변환
        //   int nDec = (int)strtol(cv1, NULL, 16);
 
        //   //10진수 출력
        //   cout << nDec << endl;
 
        }
        sort(result, result + 30, greater<int>());
        int descResult[100] = { 0, };
        int inx = 0;
        for (int i = 0; i < n; i++) {
            if (result[i] != result[i + 1]) {
                descResult[inx] = result[i];
                inx++;
            }
        }
 
        // for(int i = 0; i<n; i++){
        //   cout<<descResult[i]<<endl;
        // }
        cout <<"#"<<t<<" "<< descResult[k - 1] << endl;
        memset(descResult, 0, sizeof(descResult));
        memset(result, 0, sizeof(result));
        memset(ch, NULL, sizeof(ch));
 
    }
 
    return 0;
}
 
/*
1
12 10
1B3B3B81F75E
 
2
12 10
1B3B3B81F75E
16 2
F53586D76286B2D8
*/
 
/*
int main(){
char ch[100] = "FF";
 
    //10진수로 변환
    int nDec = (int)strtol(ch, NULL, 16);
 
    //10진수 출력
    cout << nDec << endl;
 
    //대문자 출력
    //cout.setf(ios::uppercase);
 
    //16진수 출력
    cout << hex << nDec <<endl;
}
*/

사각형의 각 변을 구간별로 나누어 한칸씩 회전시킨 다음 값을 10진수로 변환하여 출력 하도록 했습니다. 

구간을 네가지로 나누고 값을 이동할줄만 안다면 어려운 문제는 아닌거 같습니다.