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DallasAutumn · Mar 1, 2020 · 14f3df4 · 14f3df4
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+# A1092 To Buy or Not to Buy
+
+## 1.题意理解
+给出两个串，第一个是提供的，第二个是需要的，要求判断供给是否能满足需求，若能满足多多少，不能满足差多少。
+
+## 2.题目分析
+直接把串中每个字符的个数统计起来即可，可以视作hash的应用，使用map简单实现
+
+## 3.参考代码
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+
+const int MAXL = 1024;
+
+map<char, int> mp;
+
+int main()
+{
+    string buy, want;
+    int extra = 0, miss = 0;
+    bool flag = true;
+    cin >> buy;
+    cin >> want;
+    for (char c : buy)
+        mp[c]++;
+    for (char c : want)
+    {
+        if (!mp[c])
+        {
+            flag = false;
+            miss++;
+        }
+        else
+            mp[c]--;
+    }
+    if (flag)
+    {
+        for (auto it : mp)
+            extra += it.second;
+        printf("Yes %d\n", extra);
+    }
+    else
+        printf("No %d\n", miss);
+    system("pause");
+    return 0;
+}
+```
diff --git a/rebuild_btree.pdf b/rebuild_btree.pdf