0. [c++] sw 7812


https://swexpertacademy.com/main/code/problem/problemDetail.do?contestProbId=AWslJ57qFS0DFASy&categoryId=AWslJ57qFS0DFASy&categoryType=CODE


 1. 풀이





 2. 소스코드


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#include<iostream>
#include<cstring>
#include<algorithm>
 
using namespace std;
 
int arr[10001];
 
int makePortfolio(const int& N, const int& M) {
    int count = 0;
    pair<int, int> idx;
    idx = { 0,0 };
    int sum = 0;
    for (int i = 0; i < N; i++) {
        if (sum < M) {
            sum += arr[i];
            idx.second = i;
        }
        if (sum == M) {
            count++;
            sum -= arr[idx.first];
            idx.first++;
        }
        if (sum > M)
        {
            while (1)
            {
                if (sum <= M)
                    break;
                sum -= arr[idx.first];
                idx.first++;
            }
            
        }
        if (sum == M) {
            count++;
            sum -= arr[idx.first];
            idx.first++;
        }
    }
    return count;
}
 
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int T, N, M;
    cin >> T;
    for (int i = 1; i <= T; i++) {
        cin >> N >> M;
        for (int j = 0; j < N; j++) {
            cin >> arr[j];
        }
        cout << "#" << i << " " << makePortfolio(N, M) << "\n";
    }
    return 0;
}
cs


 3. 참고




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sw 7731 복불복 다트판 만들기(이후에 계속)  (0) 2019.05.23

 0. [c++] 백준  - 


https://swexpertacademy.com/main/code/problem/problemDetail.do?contestProbId=AWq44zy6L_kDFARG&categoryId=AWq44zy6L_kDFARG&categoryType=CODE


 1. 풀이


1) 

2)

3)



 2. 소스코드


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#include<algorithm>
#include<vector>
#include<iostream>
#include<string>
 
using namespace std;
 
string Board;
int Size, Trans;
 
string conversion() {
    string ret;
    int count = 0;
    ret = Board;
    for (count; count < Trans; count++) {
        string thisTime = ret;
        for (int i = 0; i < Size; i++) {
            if (i == 0) {
                //만일 3개가 전부 같다면 B로 변환한다.
                if ((ret[0] == ret[Size - 1] == ret[1]) || ((ret[0] != ret[Size - 1])&&(ret[0] != ret[1])&&(ret[1] !=ret[Size-1])))
                    thisTime[i] = 'B';
                else {
                    int countB, countR, countG;
                    countB = countR = countG = 0;
                    for (int j = i; j <= i + 1; j++) {
                        if (ret[j] == 'R')
                            countR++;
                        else if (ret[j] == 'G')
                            countG++;
                        else
                            countB++;
                    }
                    if (ret[Size - 1] == 'R')
                        countR++;
                    else if (ret[Size - 1] == 'G')
                        countG++;
                    else
                        countB++;
                    //변환해준다.
                    if ((countR == 2 && countG == 1) || (countG == 2 && countB == 1) || (countB == 2 && countR == 1))
                        thisTime[i] = 'R';
                    else
                        thisTime[i] = 'G';
                }
            }
            else if (i == Size - 1) {
                //만일 3개가 전부 같다면 B로 변환한다.
                if ((ret[0] == ret[Size - 1] == ret[Size - 2])||((ret[0] != ret[Size - 1])&&(ret[0] != ret[Size - 2])&&(ret[Size-1] != ret[Size-2])))
                    thisTime[i] = 'B';
                else {
                    int countB, countR, countG;
                    countB = countR = countG = 0;
                    for (int j = i - 1; j <= i; j++) {
                        if (ret[j] == 'R')
                            countR++;
                        else if (ret[j] == 'G')
                            countG++;
                        else
                            countB++;
                    }
                    if (ret[0] == 'R')
                        countR++;
                    else if (ret[0] == 'G')
                        countG++;
                    else
                        countB++;
                    //변환해준다.
                    if ((countR == 2 && countG == 1) || (countG == 2 && countB == 1) || (countB == 2 && countR == 1))
                        thisTime[i] = 'R';
                    else
                        thisTime[i] = 'G';
                }
            }
            else {
                //만일 3개가 전부 같다면 B로 변환한다.
                if ((ret[i] == ret[i + 1] == ret[i - 1])|| ((ret[i] != ret[i + 1])&&(ret[i] != ret[i - 1])&&(ret[i-1] != ret[i+1])))
                    thisTime[i] = 'B';
                else {
                    int countB, countR, countG;
                    countB = countR = countG = 0;
                    for (int j = i - 1; j <= i + 1; j++) {
                        if (ret[j] == 'R')
                            countR++;
                        else if (ret[j] == 'G')
                            countG++;
                        else
                            countB++;
                    }
                    //변환해준다.
                    if ((countR == 2 && countG == 1) || (countG == 2 && countB == 1) || (countB == 2 && countR == 1))
                        thisTime[i] = 'R';
                    else
                        thisTime[i] = 'G';
                }
            }
 
        }
        ret = thisTime;
    }
 
    return ret;
}
 
int main() {
    int T;
    cin >> T;
    for (int i = 1; i <= T; i++) {
        cin >> Size >> Trans;
        cin >> Board;
        cout << "#" << i << " " << conversion() << endl;
    }
}
cs


 3. 참고


구종만, 「프로그래밍 대회에서 배우는 알고리즘 문제 해결 전략」, 인사이트, 2012, p.216~236.



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