写代码:用线性分布编写只有1-7的计算器
时间: 2024-05-08 18:20:54 浏览: 5
以下是使用线性分布的只能计算1-7的计算器代码:
```
#include <stdio.h>
int main() {
int num1, num2, result;
char operator;
printf("Enter first number (1-7): ");
scanf("%d", &num1);
printf("Enter second number (1-7): ");
scanf("%d", &num2);
printf("Enter operator (+, -, *, /): ");
scanf(" %c", &operator);
switch(operator) {
case '+':
result = num1 + num2;
break;
case '-':
result = num1 - num2;
break;
case '*':
result = num1 * num2;
break;
case '/':
result = num1 / num2;
break;
default:
printf("Invalid operator\n");
return 1;
}
if(result < 1 || result > 7) {
printf("Result is out of range (1-7)\n");
return 1;
}
printf("Result: %d\n", result);
return 0;
}
```
这个计算器只能接受1-7之间的数字作为输入,如果输入的数字超过了范围,将会输出错误信息。