Update 0044.开发商购买土地.md

Author: Leehouc

problems/kamacoder/0044.开发商购买土地.md

@@ -388,3 +388,85 @@ if __name__ == "__main__":
     main()
 
 ```
+### C
+
+前缀和
+```c
+#include <stdlib.h>
+#include <stdio.h>
+
+int main()
+{
+    int n = 0, m = 0, ret_ver = 0, ret_hor = 0;
+
+    // 读取行和列的值
+    scanf("%d%d", &n, &m);
+    // 动态分配数组a(横)和b(纵)的空间
+    int *a = (int *)malloc(sizeof(int) * n);
+    int *b = (int *)malloc(sizeof(int) * m);
+
+    // 初始化数组a和b
+    for (int i = 0; i < n; i++)
+    {
+        a[i] = 0;
+    }
+    for (int i = 0; i < m; i++)
+    {
+        b[i] = 0;
+    }
+
+    // 读取区块权值并计算每行和每列的总权值
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < m; j++)
+        {
+            int tmp;
+            scanf("%d", &tmp);
+            a[i] += tmp;
+            b[j] += tmp;
+        }
+    }
+
+    // 计算每列以及每行的前缀和
+    for (int i = 1; i < n; i++)
+    {
+        a[i] += a[i - 1];
+    }
+    for (int i = 1; i < m; i++)
+    {
+        b[i] += b[i - 1];
+    }
+
+    // 初始化ret_ver和ret_hor为最大可能值
+    ret_hor = a[n - 1];
+    ret_ver = b[m - 1];
+
+    // 计算按行划分的最小差异
+    int ret2 = 0;
+    while (ret2 < n)
+    {
+        ret_hor = (ret_hor > abs(a[n - 1] - 2 * a[ret2])) ? abs(a[n - 1] - 2 * a[ret2]) : ret_hor;
+        // 原理同列，但更高级
+        ret2++;
+    }
+    // 计算按列划分的最小差异
+    int ret1 = 0;
+    while (ret1 < m)
+    {
+        if (ret_ver > abs(b[m - 1] - 2 * b[ret1]))
+        {
+            ret_ver = abs(b[m - 1] - 2 * b[ret1]);
+        }
+        ret1++;
+    }
+
+    // 输出最小差异
+    printf("%d\n", (ret_ver <= ret_hor) ? ret_ver : ret_hor);
+
+    // 释放分配的内存
+    free(a);
+    free(b);
+    return 0;
+}
+
+```