大家看看这个问题出在哪了

Question

#include <stdio.h>

int main(void) {
    int n, m, z, k, sum, i;
    int number1[] = {}; /*输入存储数组*/
    int number2[] = {}; /*筛选后存储数组*/

    while (scanf("%d", &n) != EOF) {
        for (i = 0, m = 0; i < n; i++, m++) //赋值存储
            scanf("%d", &number1[m]);
        for (z = 0, k = 0; z < m; z++) { //赋值筛选
            if ((number1[z] % 2) != 0) {
                number2[k] = number1[z];
                k++;
            }
        }
        for (sum = 1; k >= 0; k--) //数值相乘
            sum *= number2[k];
        printf("%d\n", sum);
    }
    return 0;
}

这是写的整个程序，要求如下

看了其他写的答案有更简单的方法，但我没看出来我写的哪里做错了，结果差了很多，如下

希望大家帮我指出问题在哪，该如何修改？

Answer 1

修改如下，供参考：

#include <stdio.h>
#define  N  100    //修改
int main(void) {
    int n, m, z, k, sum, i;
    int number1[N] = {0}; /*输入存储数组*/  //修改
    int number2[N] = {0}; /*筛选后存储数组*///修改
    while (scanf("%d", &n) != EOF) {
        for (i = 0, m = 0; i < n; i++, m++) //赋值存储
            scanf("%d", &number1[m]);
        for (z = 0, k = 0; z < m; z++) { //赋值筛选
            if ((number1[z] % 2) != 0) {
                number2[k] = number1[z];
                k++;
            }
        }
        for (sum = 1, k--; k >= 0; k--) //数值相乘  //修改
            sum *= number2[k];
        printf("%d\n", sum);
    }
    return 0;
}