这里有一个问题：

z = 100 / x;
    b = n / z;

如果价格x大于100，则由于整数除法语义，z将具有值0，并且由于除法溢出，n / z将导致未定义的行为，这可能导致程序停止。
以下是我的建议：

• 避免被零除
• 使用浮点运算以避免过度舍入。
• 使用明确的名称，使您的程序更容易阅读和理解。
• 测试scanf()的返回值以检测无效或丢失的输入。

以下是修改后的版本：

#include <stdio.h>

int main(void) {
    int cost, sale;

    printf("Enter cost of item: ");
    if (scanf("%d", &cost) != 1) {
        printf("input error!\n");
        return 1;
    }
    printf("Enter selling price of item: ");
    if (scanf("%d", &sale) != 1) {
        printf("input error!\n");
        return 1;
    }
    if (cost == 0) {
        printf("no cost base, total profit of $%d\n", sale);
    } else
    if (cost == sale) {
        printf("it's a wash: no profit, no loss\n");
    } else
    if (cost < sale) {
        int profit = sale - cost;
        int margin = profit * 100 / cost;
        printf("You made a %d%% profit of $%d\n", margin, profit);
    } else {
        int loss = cost - sale;
        int margin = loss * 100 / cost;
        printf("You made a %d%% loss of $%d\n", margin, loss);
    }
    return 0;
}

这是给你的完整答案。你在计算中有错误

#include <stdio.h>

int main() {
    float x, y, z, b, n;
    char k = '%';
    char r = '$';
    printf("Enter cost of item: ");
    scanf("%f", &x);
    printf("Enter selling price of item: ");
    scanf("%f", &y);
    if (x <= y) {
        n = y - x;
        z = n/x*100;
      
        printf("You made %f %c profit \n", z, k);
        printf("or\n");
        printf("%f %c profits\n", n, r);
    }
    if (x > y) {
        n = x-y;
        z = n/x*100;
        printf("You made %f%c loss \n", z, k);
        printf("or\n");
        printf("%f %c loss\n", n, r);
    }
    return 0;
}