C语音20以内加减乘除自动输出，并计算正确率

在C语言中实现一个简单的加减乘除自动计算器，并统计正确率，你需要编写一个控制循环的程序，接收用户输入的数字和操作符，然后进行相应的计算。这里是一个基础的例子：

#include <stdio.h>

// 函数声明
int add(int a, int b);
int subtract(int a, int b);
int multiply(int a, int b);
int divide(int a, int b);

int main() {
    int correct_count = 0;
    int total_attempts = 0;

    while (1) {
        printf("\nEnter an operation (1 for addition, 2 for subtraction, 3 for multiplication, or 4 for division): ");
        int choice;
        scanf("%d", &choice);

        if (choice < 1 || choice > 4) {
            printf("Invalid choice. Please try again.\n");
            continue;
        }

        printf("Enter the first number: ");
        int num1;
        scanf("%d", &num1);

        printf("Enter the second number: ");
        int num2;
        scanf("%d", &num2);

        switch (choice) {
            case 1:
                int result = add(num1, num2);
                break;
            case 2:
                result = subtract(num1, num2);
                break;
            case 3:
                result = multiply(num1, num2);
                break;
            case 4:
                if (num2 == 0) {
                    printf("Error: Division by zero is not allowed.\n");
                    continue;
                }
                result = divide(num1, num2);
                break;
        }

        if (choice == 1 && result == num1 + num2) {
            correct_count++;
        } else if (choice == 2 && result == num1 - num2) {
            correct_count++;
        } else if (choice == 3 && result == num1 * num2) {
            correct_count++;
        } else if (choice == 4 && !((num2 != 0) && (result == num1 / num2))) {
            correct_count++; // 注意这里的判断，因为除法结果可能有小数部分，此处仅作为演示
        }

        total_attempts++;

        printf("Result: %d\n", result);

        printf("Correct rate: %.2f%%\n", (correct_count * 100.0) / total_attempts);

        printf("Do you want to continue? (y/n): ");
        char cont;
        scanf(" %c", &cont);
        if (cont != 'y' && cont != 'Y') {
            break;
        }
    }

    return 0;
}

// 定义实际的加、减、乘、除函数
int add(int a, int b) { return a + b; }
int subtract(int a, int b) { return a - b; }
int multiply(int a, int b) { return a * b; }
int divide(int a, int b) { return a / b; }

这个示例中，我们首先询问用户的操作选择和数字，然后根据选择进行计算并比较结果。如果计算结果正确，就增加正确计数。最后，我们会显示当前的正确率，并询问用户是否继续。注意，此示例仅用于演示目的，实际应用中可能需要考虑更多的边界条件和浮点数处理。