[BOJ][1912] 연속합

1. 문제 링크

https://www.acmicpc.net/problem/1912

2. 사용 알고리즘

DP

3. 풀이

d[n] : a[n]을 마지막으로 하는 최대 연속합

• n - 1 번째 수를 포함하는 경우 : d[n - 1] + a[n]

• n - 1 번째 수를 포함하지 않는 경우 : a[n]

∴ d[n] = max(d[n - 1] + a[n], a[n])

∴ max(d[i]) (1 <= i <= n)

4. 소스 코드

4-1. C++

4-1-1. Top-Down

https://github.com/dev-aiden/problem-solving/blob/main/boj/1912.cpp

#include <iostream>
#include <algorithm>

using namespace std;

int a[100003], d[100003];

int solve(int n) {
    if (d[n] != -2147483648) return d[n];
    return d[n] = max(solve(n - 1) + a[n], a[n]);
}

int main(void) {
    ios_base::sync_with_stdio(false);
    int n; cin >> n;
    int ans;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        if (i == 1) {
            ans = d[1] = a[1];
            continue;
        }
        d[i] = -2147483648;
        solve(i);
        if (ans < d[i]) ans = d[i];
    }
    cout << ans << "\n";
    return 0;
}

4-1-2. Bottom-Up

https://github.com/dev-aiden/problem-solving/blob/main/boj/1912_2.cpp

#include <iostream>
#include <algorithm>

using namespace std;

int a[100003], d[100003];

int main(void) {
    ios_base::sync_with_stdio(false);
    int n; cin >> n;
    int ans;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        if (i == 1) {
            ans = d[1] = a[1];
            continue;
        }
        d[i] = max(d[i - 1] + a[i], a[i]);
        if (ans < d[i]) ans = d[i];
    }
    cout << ans << "\n";
    return 0;
}

4-2. JAVA

4-2-1. Top-Down

https://github.com/dev-aiden/problem-solving/blob/main/boj/1912.java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

import static java.lang.Math.max;

public class Main {

    static int a[] = new int[100003];
    static int d[] = new int[100003];

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int ans = 0;
        StringTokenizer st = new StringTokenizer(br.readLine());
        for(int i = 1; i <= n; ++i) {
            a[i] = Integer.parseInt(st.nextToken());
            if (i == 1) {
                ans = d[1] = a[1];
                continue;
            }
            d[i] = -2147483648;
            solve(i);
            if (ans < d[i]) ans = d[i];
        }
        System.out.println(ans);
    }

    public static int solve(int n) {
        if (d[n] != -2147483648) return d[n];
        return d[n] = max(solve(n - 1) + a[n], a[n]);
    }
}

4-2-2. Bottom-Up

https://github.com/dev-aiden/problem-solving/blob/main/boj/1912_2.java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

import static java.lang.Math.max;

public class Main {

    static int a[] = new int[100003];
    static int d[] = new int[100003];

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int ans = 0;
        StringTokenizer st = new StringTokenizer(br.readLine());
        for(int i = 1; i <= n; ++i) {
            a[i] = Integer.parseInt(st.nextToken());
            if (i == 1) {
                ans = d[1] = a[1];
                continue;
            }
            d[i] = max(d[i - 1] + a[i], a[i]);
            if (ans < d[i]) ans = d[i];
        }
        System.out.println(ans);
    }
}