[BOJ/백준][Silver3] 19941 : 햄버거분배 (C++)

https://www.acmicpc.net/problem/19941

19941번: 햄버거 분배

 

 

[난이도] Silver3
[유형] Greedy

[풀이]
P인 점을 기준으로 i-K~i+K 만큼을 확인하여 햄버거가 있으면 그 햄버거를 invalid 처리하고
정답에 1을 추가해주면 됩니다.

 

#include <iostream>
#include <string>
using namespace std;
int N,K,ans;
string s;
int main(){
    cin >> N >> K >> s;
    for(int i=0;i<N;i++){
        if(s[i]=='P'){
            for(int j=i-K;j<=i+K && j<N;j++){
                if(j<0) continue;
                if(s[j]=='H'){
                    ans++;
                    s[j]='O';
                    break;
                }
            }
        }
    }
    printf("%d",ans);
}

https://github.com/has2/Problem-Solving/blob/master/boj-solved.ac/Silver3/19941.cpp