백준 31822 재수강 python / c++

문제링크 : https://www.acmicpc.net/problem/31822

 

recourse = input()

N = int(input())
lst = []
result =0

for _ in range(N):
    lst.append(input())



for l in lst:
    if recourse[:5] == l[:5]:
        result = result + 1

print(result)

 

#include <iostream>
#include <vector>
#include <string>

using namespace std;

int main() {
    string recourse;
    int N, result = 0;
    
    cin >> recourse;
    cin >> N;
    
    vector<string> v(N);

    for (int i = 0; i < N; ++i) {
        cin >> v[i];
    }

    for (int i = 0; i < N; ++i) {
        if (recourse.substr(0, 5) == v[i].substr(0, 5)) {
            result++;
        }
    }

    cout << result << endl;

    return 0;
}

 

첫번째 입력라인의 5글자만 잘라서, N줄에 거쳐 입력받는줄을 순회하며 비교하여 카운팅합니다