diff --git a/cpp/Algorithm.md b/cpp/Algorithm.md
index 23b52f6..3d03e53 100644
--- a/cpp/Algorithm.md
+++ b/cpp/Algorithm.md
@@ -95,6 +95,8 @@
 		- [12-7.电话号码的字母组合](#12-7电话号码的字母组合)
 		- [12-8.组合问题](#12-8组合问题)
 		- [12-9.分割回文串](#12-9分割回文串)
+		- [12-10.N皇后问题](#12-10n皇后问题)
+		- [12-10.解数独问题](#12-10解数独问题)
 
 # 1.时间复杂度和空间复杂度
 ## 1-1.概念
diff --git a/cpp/cpp_learn.md b/cpp/cpp_learn.md
index faaa190..fbe9c33 100644
--- a/cpp/cpp_learn.md
+++ b/cpp/cpp_learn.md
@@ -15254,6 +15254,7 @@ void print(const T& firstArg, const Types&... args) {
  */
 
 
+<<<<<<< Updated upstream
 template <typename... Types>
 void print(const Types&... args) {
   std::cout << "print(...)" << std::endl;
@@ -15307,3 +15308,122 @@ int main(int argc, char *argv[]) {
 2. queue = [0, 1, 2, 3] (head = 0, tail = 3, 插入在3， 删除在0)
 3. stack = [0, 1, 2, 3] (head = 0, tail = 3, 插入和删除在3)
 
+=======
+```
+
+## 21.字符串的加减法
+```cpp
+#include <iostream>
+#include <string>
+#include <algorithm>
+#include <vector>
+using namespace std;
+
+string addString(string s1, string s2) {
+    int i = s1.length() - 1;
+    int j = s2.length() - 1;
+    int add = 0;
+    string ss = "";
+    while (i >= 0 || j >= 0|| add != 0) {
+        if (i >= 0)
+            add = add + s1[i--] - '0';
+        if (j >= 0)
+            add = add + s2[j--] - '0';
+        ss = ss + to_string(add % 10);
+        add = add / 10;
+    }
+    reverse(ss.begin(), ss.end());
+    return ss;
+}
+//大数减法 
+string subString(string s1,string s2){
+	bool minus = false;
+	if( s1.size() < s2.size() || ( s1.size() == s2.size() && s1 < s2 ) ){
+		swap(s1,s2);
+		minus = true; 
+	}
+	int i = s1.size() - 1;
+	int j = s2.size() - 1;
+	int flag = 0;  
+	string ans = "";
+	while( i >= 0 && j >= 0 ){
+		int tmp = 0;
+		if( s1[i] >= s2[j] ){
+			tmp = s1[i] - s2[j];
+			tmp = tmp - flag;
+			ans += to_string(tmp);
+			flag = 0;
+		}
+		else{
+			tmp = s1[i] - s2[j] + 10;
+			tmp = tmp - flag;
+			ans += to_string(tmp);
+			flag = 1;	
+		}
+		i--;
+		j--; 
+	}
+	// 处理较大数的剩余部分 
+	while( i >= 0 ){ 
+		if(flag == 0){
+			ans += s1[i];
+		}
+		else{
+			int tmp = s1[i] - '0' - flag;
+			ans += to_string(tmp);
+			flag = 0;
+		}
+		i--; 
+	}
+	 // 翻转 
+	reverse(ans.begin(),ans.end());
+	//去除前导 0 
+	int k = 0;
+	while( k < ans.size() && ans[k] == '0') k++;
+	if( k == ans.size() ){
+		ans = "0";	
+	}
+	else ans = ans.substr(k);
+	// 结果是否为负 
+	return minus?"-" + ans:ans;
+} 
+//使用大小为 n + m 的数组，num1的第i位乘以num2的第j位，结果
+// 对应存放在数组的i+j+1的位置。
+string multiplyString(string num1, string num2) {
+    int n = num1.size();
+    int m = num2.size();
+    vector<int> result(n + m,0);
+    for(int i = n - 1; i >= 0; i--)
+    {
+        for(int j = m - 1;j >= 0; j--)
+        {
+            int tmp = (num1[i] - '0') * (num2[j] - '0');
+            tmp += result[i + j + 1];
+            result[i + j] += tmp / 10;  //注意 这里是 +=
+            result[i + j + 1] = tmp % 10;
+        }
+    }
+    int i = 0;
+    while( i < n + m && result[i] == 0 )
+    {
+        i++;
+    }
+    string ans;
+    for(; i < n + m; i++){
+        ans.push_back(result[i] + '0');
+    }
+    return ans.size() == 0 ? "0" : ans;
+}
+int main()
+{
+    string s1 = "abc";
+    string s2 =  "a";
+    string s_add = addString(s1, s2);
+    string s_sub = subString(s1, s2);
+    string s_mul = multiplyString(s1, s2);
+    cout << s_add << endl;
+    cout << s_sub << endl;
+    cout << s_mul << endl;
+}
+```
+>>>>>>> Stashed changes