巧用函数：求解 2 - 10 行矩阵不靠边元素之和

```c #include<stdio.h> #define M 10 #define N 4 void Input(float a[][N], int m); float Sum(float a[][N], int m); void Input(float a[][N], int m) { for (int i = 0; i < m; i++) { for (int j = 0; j < N; j++) { scanf("%f", &a[i][j]); } } } float Sum(float a[][N], int m) { float sum = 0; for (int i = 1; i < m - 1; i++) { for (int j = 1; j < N - 1; j++) { sum = sum + a[i][j]; } } return sum; } int main(void) { float num[M][N], sum; int m; scanf("%d", &m); Input(num, m); sum = Sum(num, m); printf("sum = %.2f\n", sum); return 0; } ``` ######[AI问答 | 266点数解答 | 2025-12-10 19:43:19]