From a5176bebffdb7ec3e6a60c853ebb5e7f046dfd51 Mon Sep 17 00:00:00 2001
From: yanzhuang-h <126091390+yanzhuang-h@users.noreply.github.com>
Date: Sun, 9 Feb 2025 10:10:59 +0800
Subject: [PATCH] =?UTF-8?q?Update=200044.=E5=BC=80=E5=8F=91=E5=95=86?=
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提供一个新的思路求解二维前缀和
---
 ...55\344\271\260\345\234\237\345\234\260.md" | 39 ++++++++++++++++++-
 1 file changed, 38 insertions(+), 1 deletion(-)

diff --git "a/problems/kamacoder/0044.\345\274\200\345\217\221\345\225\206\350\264\255\344\271\260\345\234\237\345\234\260.md" "b/problems/kamacoder/0044.\345\274\200\345\217\221\345\225\206\350\264\255\344\271\260\345\234\237\345\234\260.md"
index 64804842f2..3b1d208b95 100644
--- "a/problems/kamacoder/0044.\345\274\200\345\217\221\345\225\206\350\264\255\344\271\260\345\234\237\345\234\260.md"
+++ "b/problems/kamacoder/0044.\345\274\200\345\217\221\345\225\206\350\264\255\344\271\260\345\234\237\345\234\260.md"
@@ -166,8 +166,45 @@ int main () {
 }
  
 ```
+除了上述做法，还可采用的思路是直接求二维数组的前缀和，时间复杂度为 O(n^2)，
+```c++
+#include <iostream>
+#include <vector>
 
-
+using namespace std;
+int main() {
+    int n, m;
+    scanf("%d%d", &n, &m);
+    // 这里采用n+1,m+1，是为了避免处理边界时潜在的溢出问题
+    vector<vector<int>> val(n+1, vector<int>(m+1, 0));
+    vector<vector<int>> sum(n+1, vector<int>(m+1, 0));
+    for(int i = 1; i <= n; i++) {
+        for(int j = 1; j <= m; j++) {
+            scanf("%d", &val[i][j]);
+            
+            // 代码的核心，初始化前缀和数组
+            sum[i][j] = val[i][j] + sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];
+        }
+    }
+    
+    int min = INT32_MAX;
+    //按行进行划分求最小值
+    for(int i = 1; i <= n; i++) {
+        int sub = sum[n][m] - 2 * sum[i][m];
+        sub = sub > 0 ? sub : (-sub);
+        min = min < sub ? min : sub;
+    }
+    //按列进行划分求最小值
+    for(int j = 1; j <= m; j++) {
+        int sub = sum[n][m] - 2 * sum[n][j];
+        sub = sub > 0 ? sub : (-sub);
+        min = min < sub ? min : sub;
+    }
+    
+    printf("%d", min);
+    return 0;
+}
+```
 
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