From 59c95ff24d60ec7809aa80c632b3728291892077 Mon Sep 17 00:00:00 2001
From: Guanzhong Pan <gpan20020205@gmail.com>
Date: Thu, 20 Jan 2022 08:29:02 +0000
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200343.=E6=95=B4=E6=95=B0?=
 =?UTF-8?q?=E6=8B=86=E5=88=86.md=20C=E8=AF=AD=E8=A8=80=E8=A7=A3=E6=B3=95?=
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---
 problems/0343.整数拆分.md | 35 +++++++++++++++++++++++++++++++++++
 1 file changed, 35 insertions(+)

diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md
index 5d11f670..9882c6ea 100644
--- a/problems/0343.整数拆分.md
+++ b/problems/0343.整数拆分.md
@@ -271,5 +271,40 @@ var integerBreak = function(n) {
 };
 ```
 
+C:
+```c
+//初始化DP数组
+int *initDP(int num) {
+    int* dp = (int*)malloc(sizeof(int) * (num + 1));
+    int i;
+    for(i = 0; i < num + 1; ++i) {
+        dp[i] = 0;
+    }
+    return dp;
+}
+
+//取三数最大值
+int max(int num1, int num2, int num3) {
+    int tempMax = num1 > num2 ? num1 : num2;
+    return tempMax > num3 ? tempMax : num3;
+}
+
+int integerBreak(int n){
+    int *dp = initDP(n);
+    //初始化dp[2]为1
+    dp[2] = 1;
+
+    int i;
+    for(i = 3; i <= n; ++i) {
+        int j;
+        for(j = 1; j < i - 1; ++j) {
+            //取得上次循环:dp[i]，原数相乘，或j*dp[]i-j] 三数中的最大值
+            dp[i] = max(dp[i], j * (i - j), j * dp[i - j]);
+        }
+    }
+    return dp[n];
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>