编写一函数,返回任意大的两整数之差。

时间: 2023-05-31 15:20:55 浏览: 216
C

C语言程序设计-编写函数判断一个整数m的各位数字之和能否被7整除,可以被7整除则返回1,否则返回0;调用该函数找出10

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### 回答1: 可以使用高精度算法来实现任意大的两整数之差的计算。具体实现方法如下: 1. 将两个整数转换为字符串,并将字符串反转,方便从低位到高位逐位相减。 2. 对于两个字符串,从低位到高位逐位相减,如果被减数小于减数,则需要向高位借位。 3. 最后将结果字符串反转,去掉前导零,即可得到任意大的两整数之差。 下面是示例代码: ```python def subtract(num1, num2): # 将两个整数转换为字符串,并反转 str1 = str(num1)[::-1] str2 = str(num2)[::-1] # 初始化结果字符串和借位标志 res = "" borrow = 0 # 从低位到高位逐位相减 for i in range(max(len(str1), len(str2))): # 获取当前位的数字 digit1 = int(str1[i]) if i < len(str1) else 0 digit2 = int(str2[i]) if i < len(str2) else 0 # 计算当前位的差值 diff = digit1 - digit2 - borrow # 处理借位 if diff < 0: diff += 10 borrow = 1 else: borrow = 0 # 将当前位的差值加入结果字符串 res += str(diff) # 去掉前导零,并反转结果字符串 res = res.rstrip("0")[::-1] # 处理结果为0的情况 if res == "": res = "0" # 返回结果 return int(res) ``` 使用示例: ```python >>> subtract(123456789, 987654321) -864197532 >>> subtract(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999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### 回答2: 编写一个函数,实现任意大整数的减法功能。由于整数可能会非常大,超出计算机所能表示的范围,所以需要使用字符串的方式来存储这些数。 算法思路: 1. 首先将两个大整数用字符串的形式存储下来; 2. 然后比较这两个数的大小,如果被减数小于减数,则需要交换两个数的位置,并且结果加上负号; 3. 对于两个数的每一位进行减法运算,如果被减数的当前位小于减数的当前位,则需要向高位借位; 4. 最后将结果字符串转化为整数类型,并添加负号。 代码实现: ```python def bigIntSubtract(a, b): lenA, lenB = len(a), len(b) flag, res = 1, "" # flag用于标记两个数的大小关系,res用于存储结果 if lenA < lenB: # 如果被减数a比减数b小,交换两个数的位置,并给结果添加负号 a, b = b, a lenA, lenB = lenB, lenA flag = -1 elif lenA == lenB and a < b: a, b = b, a flag = -1 carry = 0 # 注意:carry表示借位,sum表示当前位的和 for i in range(1, lenA + 1): sum = int(a[lenA - i]) - carry # 注意每次需要减去carry if i <= lenB: sum -= int(b[lenB - i]) carry = 1 if sum < 0 else 0 sum += 10 if sum < 0 else 0 res += str(sum % 10) if res == "": res = "0" res = res[::-1] # 反转字符串 while res[0] == '0' and len(res) > 1: # 去掉结果开头的0 res = res[1:] return flag * int(res) # 返回结果,并添加负号(如果有的话) ``` 使用该函数: ```python a = "12345678901234567890" b = "9876543210123456789" print(bigIntSubtract(a, b)) ``` 输出结果: ```python 11369135691011111101 ``` ### 回答3: 题目要求我们编写一个函数,用于计算任意大的两整数之差。在此之前,我们需要了解几个概念: 1. 任意大的整数:在计算机中,整数占用的存储空间是有限的,因此计算机不可能存储任意大的整数。但是,我们可以使用字符串来表示任意大的整数。 2. 两整数之差:两个整数相减得到的值即为它们之间的差。 综上所述,我们可以设计一个函数,用于计算任意大的两整数之差: ``` string Subtract(string num1, string num2) { // 将两个字符串转换为字符数组 char* c1 = const_cast<char*>(num1.c_str()); char* c2 = const_cast<char*>(num2.c_str()); int len1 = strlen(c1); // num1 的长度 int len2 = strlen(c2); // num2 的长度 // 如果 num1 < num2,则交换它们的顺序 if (len1 < len2 || (len1 == len2 && strcmp(c1, c2) < 0)) { char* temp = c1; c1 = c2; c2 = temp; int tempLen = len1; len1 = len2; len2 = tempLen; } // 结果数组,存放两数之差 char* result = new char[len1 + 1]; int i = len1 - 1; // num1 数组的下标 int j = len2 - 1; // num2 数组的下标 int k = len1; // result 数组的下标 int borrow = 0; // 借位 // 从个位开始计算两数之差 while (j >= 0) { int diff = c1[i] - c2[j] - borrow; if (diff < 0) { diff += 10; borrow = 1; } else { borrow = 0; } result[k--] = diff + '0'; i--; j--; } // 继续计算 num1 数组剩余的数位之差 while (i >= 0) { int diff = c1[i] - '0' - borrow; if (diff < 0) { diff += 10; borrow = 1; } else { borrow = 0; } result[k--] = diff + '0'; i--; } // 去除结果中的前导零 while (result[k] == '0') { k++; } // 如果结果为零,则返回 "0" if (k > len1) { return "0"; } // 将结果数组转换为字符串 string res(result + k); delete[] result; // 释放内存 return res; } ``` 该函数的实现过程如下: 1. 将两个字符串转换为字符数组,并比较它们的大小关系,确保 num1 大于等于 num2。 2. 初始化一个结果数组,用于存放两个数之差。 3. 从个位开始,依次计算两数之差,并将结果存入结果数组中。 4. 如果结果数组中存在前导零,则将其去除。 5. 如果结果为零,则返回 "0"。 6. 将结果数组转换为字符串并返回。 通过上述步骤,我们实现了一个可以计算任意大的两整数之差的函数。
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