[BOJ][2225] 합분해

1. 문제 링크

https://www.acmicpc.net/problem/2225

2. 사용 알고리즘

DP

3. 풀이

d[k][n] : 0 ~ N 정수 k개를 더했을 때 합이 n이 되는 경우의 수

• k번째 수가 0인 경우의 수 : d[k - 1][n - 0]

• k번째 수가 1인 경우의 수 : d[k - 1][n - 1]

• …

• k번째 수가 n인 경우의 수 : d[k - 1][n - n]

∴ d[k][n] = d[k - 1][n - 0] + d[k - 1][n - 1] + … + d[k - 1][n - n];

4. 소스 코드

4-1. C++

4-1-1. Top-Down

https://github.com/dev-aiden/problem-solving/blob/main/boj/2225.cpp

#include <iostream>

using namespace std;

long long d[203][203];

long long solve(int k, int n) {
    if (k == 1) return 1LL;
    if (d[k][n] > 0) return d[k][n];
    for (int i = 0; i <= n; ++i) {
        d[k][n] += solve(k - 1, i);
        d[k][n] %= 1000000000;
    }
    return d[k][n];
}

int main(void) {
    ios_base::sync_with_stdio(false);
    int n, k; cin >> n >> k;
    cout << solve(k, n) << "\n";
    return 0;
}

4-1-2. Bottom-Up

https://github.com/dev-aiden/problem-solving/blob/main/boj/2225_2.cpp

#include <iostream>

using namespace std;

long long d[203][203];

int main(void) {
    ios_base::sync_with_stdio(false);
    int n, k; cin >> n >> k;
    d[0][0] = 1LL;
    for (int i = 1; i <= k; ++i) {
        for (int j = 0; j <= n; ++j) {
            for (int l = 0; l <= j; ++l) {
                d[i][j] += d[i - 1][j - l];
                d[i][j] %= 1000000000;
            }
        }
    }
    cout << d[k][n] << "\n";
    return 0;
}

4-2. JAVA

4-2-1. Top-Down

https://github.com/dev-aiden/problem-solving/blob/main/boj/2225.java

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

public class Main {

    static long[][] d = new long[203][203];

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());
        System.out.println(solve(k, n));
    }

    public static Long solve(int k, int n) {
        if (k == 1) return 1L;
        if (d[k][n] > 0) return d[k][n];
        for (int i = 0; i <= n; ++i) {
            d[k][n] += solve(k - 1, i);
            d[k][n] %= 1000000000;
        }
        return d[k][n];
    }
}

4-2-2. Bottom-Up

https://github.com/dev-aiden/problem-solving/blob/main/boj/2225_2.java

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

public class Main {

    static long[][] d = new long[203][203];

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());
        d[0][0] = 1L;
        for (int i = 1; i <= k; ++i) {
            for (int j = 0; j <= n; ++j) {
                for (int l = 0; l <= j; ++l) {
                    d[i][j] += d[i - 1][j - l];
                    d[i][j] %= 1000000000;
                }
            }
        }
        System.out.println(d[k][n]);
    }
}