diff --git a/problems/0017.电话号码的字母组合.md b/problems/0017.电话号码的字母组合.md
index dfd0e875..1221115c 100644
--- a/problems/0017.电话号码的字母组合.md
+++ b/problems/0017.电话号码的字母组合.md
@@ -410,6 +410,66 @@ var letterCombinations = function(digits) {
 };
 ```
 
+C:
+```c
+char* path;
+int pathTop;
+char** result;
+int resultTop;
+char* letterMap[10] = {"", //0
+        "", //1
+        "abc", //2
+        "def", //3
+        "ghi", //4
+        "jkl", //5
+        "mno", //6
+        "pqrs", //7
+        "tuv", //8
+        "wxyz", //9
+};
+void backTracking(char* digits, int index) {
+    //若当前下标等于digits数组长度
+    if(index == strlen(digits)) {
+        //复制digits数组，因为最后要多存储一个0，所以数组长度要+1
+        char* tempString = (char*)malloc(sizeof(char) * strlen(digits) + 1);
+        int j;
+        for(j = 0; j < strlen(digits); j++) {
+            tempString[j] = path[j];
+        }
+        //char数组最后要以0结尾
+        tempString[strlen(digits)] = 0;
+        result[resultTop++] = tempString;
+        return ;
+    }
+    //将字符数字转换为真的数字
+    int digit = digits[index] - '0';
+    //找到letterMap中对应的字符串
+    char* letters = letterMap[digit];
+    int i;
+    for(i = 0; i < strlen(letters); i++) {
+        path[pathTop++] = letters[i];
+        //递归，处理下一层数字
+        backTracking(digits, index+1);
+        pathTop--;
+    }
+}
+
+char ** letterCombinations(char * digits, int* returnSize){
+    //初始化path和result
+    path = (char*)malloc(sizeof(char) * strlen(digits));
+    result = (char**)malloc(sizeof(char*) * 300);
+
+    *returnSize = 0;
+    //若digits数组中元素个数为0，返回空集
+    if(strlen(digits) == 0) 
+        return result;
+    pathTop = resultTop = 0;
+    backTracking(digits, 0);
+    *returnSize = resultTop;
+
+    return result;
+}
+```
 
 
 -----------------------
diff --git a/problems/0039.组合总和.md b/problems/0039.组合总和.md
index 8cea3b9e..97987e6d 100644
--- a/problems/0039.组合总和.md
+++ b/problems/0039.组合总和.md
@@ -346,6 +346,60 @@ var combinationSum = function(candidates, target) {
 };
 ```
 
+C:
+```c
+int* path;
+int pathTop;
+int** ans;
+int ansTop;
+//记录每一个和等于target的path数组长度
+int* length;
+
+void backTracking(int target, int index, int* candidates, int candidatesSize, int sum) {
+    //若sum>=target就应该终止遍历
+    if(sum >= target) {
+        //若sum等于target，将当前的组合放入ans数组中
+        if(sum == target) {
+            int* tempPath = (int*)malloc(sizeof(int) * pathTop);
+            int j;
+            for(j = 0; j < pathTop; j++) {
+                tempPath[j] = path[j];
+            }
+            ans[ansTop] = tempPath;
+            length[ansTop++] = pathTop;
+        }
+        return ;
+    }
+
+    int i;
+    for(i = index; i < candidatesSize; i++) {
+        //将当前数字大小加入sum
+        sum+=candidates[i];
+        path[pathTop++] = candidates[i];
+        backTracking(target, i, candidates, candidatesSize, sum);
+        sum-=candidates[i];
+        pathTop--;
+    }
+}
+
+int** combinationSum(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes){
+    //初始化变量
+    path = (int*)malloc(sizeof(int) * 50);
+    ans = (int**)malloc(sizeof(int*) * 200);
+    length = (int*)malloc(sizeof(int) * 200);
+    ansTop = pathTop = 0;
+    backTracking(target, 0, candidates, candidatesSize, 0);
+
+    //设置返回的数组大小
+    *returnSize = ansTop;
+    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
+    int i;
+    for(i = 0; i < ansTop; i++) {
+        (*returnColumnSizes)[i] = length[i];
+    }
+    return ans;
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)