From 702dfcf536e15ea064bc884a7872fac74db5d933 Mon Sep 17 00:00:00 2001
From: leebyeongman <ssey0921@gmail.com>
Date: Sun, 19 May 2019 18:39:23 +0900
Subject: [PATCH] =?UTF-8?q?[=EC=9D=B4=EB=B3=91=EB=A7=8C]=20=ED=83=80?=
 =?UTF-8?q?=EA=B2=9F=20=EB=84=98=EB=B2=84(Programmers)?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 .../README.md"                                | 11 ++++++++++
 .../solution.py"                              | 22 +++++++++++++++++++
 2 files changed, 33 insertions(+)
 create mode 100644 "\354\235\264\353\263\221\353\247\214/Programmers/\355\203\200\352\262\237 \353\204\230\353\262\204/README.md"
 create mode 100644 "\354\235\264\353\263\221\353\247\214/Programmers/\355\203\200\352\262\237 \353\204\230\353\262\204/solution.py"

diff --git "a/\354\235\264\353\263\221\353\247\214/Programmers/\355\203\200\352\262\237 \353\204\230\353\262\204/README.md" "b/\354\235\264\353\263\221\353\247\214/Programmers/\355\203\200\352\262\237 \353\204\230\353\262\204/README.md"
new file mode 100644
index 0000000..2950a07
--- /dev/null
+++ "b/\354\235\264\353\263\221\353\247\214/Programmers/\355\203\200\352\262\237 \353\204\230\353\262\204/README.md"	
@@ -0,0 +1,11 @@
+[Programmers](https://programmers.co.kr/learn/courses/30/lessons/43165). 
+
+Problem : 타겟 넘버 **DFS/BFS**
+
+Flow :
+
+1. n개의 음이 아닌 정수가 있습니다. 이 수를 적절히 더하거나 빼서 타겟 넘버를 만든다.
+
+Solution :
+1. 배열에 있는 값으로 만들수 있는 모든 경우를 구해야할 것 같다.
+2. 음... 경우의 수로 해야하나.. 
diff --git "a/\354\235\264\353\263\221\353\247\214/Programmers/\355\203\200\352\262\237 \353\204\230\353\262\204/solution.py" "b/\354\235\264\353\263\221\353\247\214/Programmers/\355\203\200\352\262\237 \353\204\230\353\262\204/solution.py"
new file mode 100644
index 0000000..3248c07
--- /dev/null
+++ "b/\354\235\264\353\263\221\353\247\214/Programmers/\355\203\200\352\262\237 \353\204\230\353\262\204/solution.py"	
@@ -0,0 +1,22 @@
+cnt = 0
+
+
+def func(numbers, target, curr, idx):
+    global cnt
+    if idx == len(numbers):
+        if curr == target:
+            cnt += 1
+        return
+    else:
+        func(numbers, target, curr + numbers[idx], idx+1)
+        func(numbers, target, curr - numbers[idx], idx+1)
+
+
+def solution(numbers, target):
+    global cnt
+    func(numbers, target, 0, 0)
+    answer = cnt
+    return answer
+
+if __name__ == '__main__':
+    solution([1, 1, 1, 1, 1], 3)