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; +} + +``` diff --git a/problems/kamacoder/0058.区间和.md b/problems/kamacoder/0058.区间和.md index f5ce08dc..c5a84a29 100644 --- a/problems/kamacoder/0058.区间和.md +++ b/problems/kamacoder/0058.区间和.md @@ -263,3 +263,45 @@ if __name__ == "__main__": main() ``` +### C + +```C +#include +#include + +int main(int argc, char *argv[]) +{ + int num; + // 读取数组长度 + scanf("%d", &num); + + // 使用动态内存分配而不是静态数组,以适应不同的输入大小 + int *a = (int *)malloc((num + 1) * sizeof(int)); + + // 初始化前缀和数组的第一个元素为0 + a[0] = 0; + + // 读取数组元素并计算前缀和 + for (int i = 1; i <= num; i++) + { + int mm; + scanf("%d", &mm); + // 累加前缀和 + a[i] = a[i - 1] + mm; + } + + int m, n; + // 循环读取区间并计算区间和,直到输入结束 + // scanf()返回成功匹配和赋值的个数,到达文件末尾则返回 EOF + while (scanf("%d%d", &m, &n) == 2) + { + // 输出区间和,注意区间是左闭右开,因此a[n+1]是包含n的元素的前缀和 + printf("%d\n", a[n+1] - a[m]); + } + + // 释放之前分配的内存 + free(a); + return 0; +} + +```