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+# step1
+思考ログ
+- subarrayは空ではないと問題文にある
+- subarrayだから、2 pointer, スライスを使って解けそう
+- 負の数も考えないといけないからO(n^2)は避けて通れない?
+- 愚直にやる方法だとtime limit exceededとでた(以下その実装)
+```python
+class Solution:
+    def subarraySum(self, nums: List[int], k: int) -> int:
+        num_subarrays = 0
+
+        for i in range(len(nums)):
+            complement = k - nums[i]
+            if complement == 0:
+                num_subarrays += 1
+            for j in range(i+1, len(nums)):
+                complement -= nums[j]
+                if complement == 0:
+                    num_subarrays += 1
+
+        return num_subarrays
+```
+- 1回計算した箇所は保存しておく方法でできるのではないか→実装がうまくいかなかったので断念答えを見る
+```python
+class Solution:
+    def subarraySum(self, nums: List[int], k: int) -> int:
+        num_subarrays = 0
+        cumsum = 0
+        cumsum_to_count = defaultdict(int)
+        cumsum_to_count[0] = 1
+        for num in nums:
+            cumsum += num
+            if cumsum - k in cumsum_to_count:
+                num_subarrays += cumsum_to_count[cumsum - k]
+            cumsum_to_count[cumsum] += 1
+
+        return num_subarrays
+```
+# step2
+参考にした方のPR
+- https://github.com/Satorien/LeetCode/pull/16/files?short_path=567cd09#diff-567cd09815edd159a86e52be227aee58958dcad0ffe5c15ec6141af5b929797b
+- "累積和はCountをHashMapに入れておく"というのがかなりしっくりきた
+
+- https://github.com/potrue/leetcode/pull/16/files?short_path=9af6c29#diff-9af6c29cdcdd4f79ab8b011317151f6cd8b13f570a036eb4d7473388bc3f0d56
+- chainを使って解いているがchainについて初めて見たので[ドキュメント]()https://docs.python.org/ja/3.13/library/itertools.html#itertools.chainを見る
+
+- https://github.com/shintaroyoshida20/leetcode/pull/22/files#diff-5db2f0c29e6198a74e1b6caeaddee45eda91677924fcd859ee7060d0bfc08c85
+c++の実装
+
+- https://github.com/tokuhirat/LeetCode/pull/16/files#diff-d4900f989c6f9680b8e8144658ef8f10d6025523b2c0c63bed653dcdcc4fc290
+- 変数について, resultを許容してもよい, countに対してfrequencyも候補である.
+
+- https://github.com/katataku/leetcode/pull/15/files
+- https://github.com/t0hsumi/leetcode/pull/13#discussion_r1902635530
+ここら辺を見ていて, この[サイト](https://source.chromium.org/search?q=%22return%20res;%22%20filepath:.*%5C.cc$)を知った. 慣習的に使われている変数名とか検索できそう
+
+
+# step 3
+```python
+class Solution:
+    def subarraySum(self, nums: List[int], k: int) -> int:
+        num_subarrays = 0
+        cumsum = 0
+        cumsum_to_count = defaultdict(int)
+        cumsum_to_count[0] = 1
+
+        for num in nums:
+            cumsum += num
+            if cumsum - k in cumsum_to_count:
+                num_subarrays += cumsum_to_count[cumsum - k]
+            cumsum_to_count[cumsum] += 1
+
+        return num_subarrays
+```