C语言实现输入两个大数加减乘除四个输出的代码

### 回答1：

可以使用以下代码：#include <stdio.h>int main()
{
int a, b;
int result;
char op; printf("输入第一个数字：");
scanf("%d", &a); printf("输入第二个数字：");
scanf("%d", &b); printf("输入运算符：");
scanf(" %c", &op); switch (op)
{
case '+':
result = a + b;
break;
case '-':
result = a - b;
break;
case '*':
result = a * b;
break;
case '/':
result = a / b;
break;
default:
printf("无效的运算符！\n");
break;
} printf("运算结果：%d\n", result); return 0;
}

### 回答2：
下面是用C语言实现输入两个大数进行加减乘除运算并输出结果的代码：

#include <stdio.h>
#include <string.h>

#define MAX_LENGTH 100    // 假设大数的最大长度为100

// 大数相加
char* add(char* num1, char* num2)
{
    static char result[MAX_LENGTH + 1];
    memset(result, '0', MAX_LENGTH + 1);
    
    int len1 = strlen(num1);
    int len2 = strlen(num2);
    int len = len1 > len2 ? len1 : len2;
    int carry = 0;
    
    for (int i = 0; i < len; i++) {
        int sum = (i < len1 ? num1[len1 - 1 - i] - '0' : 0) + (i < len2 ? num2[len2 - 1 - i] - '0' : 0) + carry;
        carry = sum / 10;
        result[MAX_LENGTH - i] = sum % 10 + '0';
    }
    
    if (carry > 0) {
        result[MAX_LENGTH - len - 1] = carry + '0';
    }
    
    return result + (result[0] == '0' ? 1 : 0);
}

// 大数相减
char* subtract(char* num1, char* num2)
{
    static char result[MAX_LENGTH + 1];
    memset(result, '0', MAX_LENGTH + 1);
    
    int len1 = strlen(num1);
    int len2 = strlen(num2);
    int len = len1 > len2 ? len1 : len2;
    int borrow = 0;
    
    for (int i = 0; i < len; i++) {
        int diff = (i < len1 ? num1[len1 - 1 - i] - '0' : 0) - (i < len2 ? num2[len2 - 1 - i] - '0' : 0) - borrow;
        borrow = diff < 0 ? 1 : 0;
        result[MAX_LENGTH - i] = diff < 0 ? diff + 10 + '0' : diff + '0';
    }
    
    return result + (result[0] == '0' ? 1 : 0);
}

// 大数相乘
char* multiply(char* num1, char* num2)
{
    static char result[MAX_LENGTH + MAX_LENGTH];
    memset(result, '0', MAX_LENGTH + MAX_LENGTH);
    
    int len1 = strlen(num1);
    int len2 = strlen(num2);
    
    for (int i = 0; i < len1; i++) {
        int carry = 0;
        
        for (int j = 0; j < len2; j++) {
            int product = (num1[len1 - 1 - i] - '0') * (num2[len2 - 1 - j] - '0') + carry + (result[MAX_LENGTH + MAX_LENGTH - 1 - i - j] - '0');
            carry = product / 10;
            result[MAX_LENGTH + MAX_LENGTH - 1 - i - j] = product % 10 + '0';
        }
        
        if (carry > 0) {
            result[MAX_LENGTH + MAX_LENGTH - 1 - i - len2] = carry + '0';
        }
    }
    
    return result + (result[0] == '0' ? 1 : 0);
}

// 大数相除
char* divide(char* num1, char* num2)
{
    static char result[MAX_LENGTH + 1];
    memset(result, '0', MAX_LENGTH + 1);
    
    int len1 = strlen(num1);
    int len2 = strlen(num2);
    
    if (len2 == 1 && num2[0] == '0') {
        return "NaN";    // 除数为0，结果为无穷大
    }
    
    if (len1 < len2 || (len1 == len2 && strcmp(num1, num2) < 0)) {
        return "0";    // 被除数小于除数，结果为0
    }
    
    char temp[MAX_LENGTH + 1];
    strcpy(temp, num1);
    
    for (int i = len1 - len2; i >= 0; i--) {
        while (strcmp(temp, num2) >= 0) {
            strcpy(temp, subtract(temp, num2));
            result[i]++;
        }
    }
    
    return result + (result[0] == '0' ? 1 : 0);
}

int main()
{
    char num1[MAX_LENGTH];
    char num2[MAX_LENGTH];
    
    printf("请输入第一个大数：");
    scanf("%s", num1);
    
    printf("请输入第二个大数：");
    scanf("%s", num2);
    
    printf("加法结果：%s\n", add(num1, num2));
    printf("减法结果：%s\n", subtract(num1, num2));
    printf("乘法结果：%s\n", multiply(num1, num2));
    printf("除法结果：%s\n", divide(num1, num2));
    
    return 0;
}

这段代码实现了大数的加法（add）、减法（subtract）、乘法（multiply）和除法（divide）运算。用户可以输入两个大数，然后计算并输出它们的加、减、乘、除运算结果。代码使用静态字符数组来保存结果，并返回一个指向结果数组的指针，同时处理结果数组前多余的0。