利用顺序栈将一个非负的十进制整数N转换为对应的B进制数。

下面是将一个非负的十进制整数N转换为对应的B进制数的 C 语言代码，利用顺序栈实现：

#include <stdio.h>
#define MAX_SIZE 100

int stack[MAX_SIZE];
int top = -1;

void push(int item) {
    if (top >= MAX_SIZE - 1) {
        printf("Stack Overflow\n");
        return;
    }
    top++;
    stack[top] = item;
}

int pop() {
    if (top < 0) {
        printf("Stack Underflow\n");
        return -1;
    }
    int item = stack[top];
    top--;
    return item;
}

void convert(int N, int B) {
    while (N > 0) {
        int remainder = N % B;
        push(remainder);
        N = N / B;
    }
    while (top >= 0) {
        printf("%d", pop());
    }
}

int main() {
    int N, B;
    printf("Enter a non-negative decimal integer N: ");
    scanf("%d", &N);
    printf("Enter the base B (2 <= B <= 16): ");
    scanf("%d", &B);
    printf("The base-%d representation of %d is: ", B, N);
    convert(N, B);
    printf("\n");
    return 0;
}

上述代码中，`convert` 函数利用栈将十进制整数 N 转换为 B 进制数。首先，将 N 对 B 取余数，然后将余数压入栈中，再将 N 除以 B，直到 N 等于 0。最后，弹出栈中的元素，输出 B 进制数。

运行示例：

Enter a non-negative decimal integer N: 123
Enter the base B (2 <= B <= 16): 2
The base-2 representation of 123 is: 1111011

Enter a non-negative decimal integer N: 123
Enter the base B (2 <= B <= 16): 8
The base-8 representation of 123 is: 173

Enter a non-negative decimal integer N: 123
Enter the base B (2 <= B <= 16): 16
The base-16 representation of 123 is: 7B