[Java] baekjoon 2293 (#93)

* boj 2293

* Update Main.java

---------

Co-authored-by: Minsang Kim <[email protected]>

Author: 336699go

solutions/baekjoon/2293/Main.java

@@ -0,0 +1,78 @@
+// Authored by: 336699go
+// Co-authored by: -
+// Link: http://boj.kr/06a33314bada4e6cbb75b91eb3b4f498
+
+import java.io.*;
+import java.util.*;
+
+public class Main {
+    public static void main(String[] args) throws IOException {
+        FastReader br = new FastReader();
+        int n = br.nextInt();
+        int k = br.nextInt();
+        int[] coins = new int[n];
+        int[] dp = new int[k + 1];
+
+        for (int i = 0; i < n; i++) {
+            coins[i] = br.nextInt();
+        }
+
+        dp[0] = 1;
+
+        for (int i = 0; i < n; i++) {
+            for (int j = coins[i]; j <= k; j++) {
+                dp[j] = dp[j] + dp[j - coins[i]];
+            }
+        }
+
+        System.out.println(dp[k]);
+    }
+
+    static class FastReader {
+        BufferedReader br;
+        StringTokenizer st;
+
+        public FastReader() {
+            br = new BufferedReader(new InputStreamReader(System.in));
+        }
+
+        String next() {
+            while (st == null || !st.hasMoreElements()) {
+                try {
+                    st = new StringTokenizer(br.readLine());
+                } catch (IOException e) {
+                    e.printStackTrace();
+                }
+            }
+            return st.nextToken();
+        }
+
+        int nextInt() {
+            return Integer.parseInt(next());
+        }
+
+        long nextLong() {
+            return Long.parseLong(next());
+        }
+
+        double nextDouble() {
+            return Double.parseDouble(next());
+        }
+
+        String nextLine() {
+            String str = "";
+            try {
+                str = br.readLine();
+            } catch (IOException e) {
+                e.printStackTrace();
+            }
+            return str;
+        }
+    }
+}
+
+/* Solution Description
+배낭 문제와 유사하나 1+2, 2+1 같은 중복을 신경 써야 한다.  
+점화식: dp[j] = dp[j] + dp[j - coins[i]] = i번째 동전까지만 사용해서 가치의 합이 j원이 되는 경우의 수  
+메모리 제한이 빡빡해서 dp 배열을 2차원이 아닌 1차원 배열로 해야 한다. 
+*/