Create qqchangE_problem.md

src/qqchangE_problem.md

+```python
+def canPartition(nums):
+    def dfs(nums, target, num):
+        for i in range(n):
+            B = nums[:i] + nums[i + 1:]
+            if num + nums[i] == target:
+                return True
+            elif num + nums[i] < target:
+                if dfs(B, target, num + nums[i]):
+                    return True
+            elif num == 0:  # 有一个数比sum/2还大，直接返回False
+                return False
+        return False
+
+    n, total = len(nums), sum(nums)
+    if total % 2 != 0: return False
+    target = total // 2
+    nums.sort(reverse=True)  # 逆序排序，先从大的开始判断，速度会更快
+    res = dfs(nums, target, 0)
+    return res
+
+
+print(canPartition([3, 3, 3, 4, 5]))
+```