diff --git a/problems/kamacoder/0044.开发商购买土地.md b/problems/kamacoder/0044.开发商购买土地.md index 37bb98ed..ea2c696e 100644 --- a/problems/kamacoder/0044.开发商购买土地.md +++ b/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 +#include + +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; +} + +```