大数（高精度）加减乘除

Posted 2020-10-09 萌新上路

1、加减法间是可以相互转化的

2、大数加减乘除要靠字符串实现

#include <iostream>   
#include <string>   
using namespace std;   
  
inline int compare(string str1, string str2)   
{   
      if(str1.size() > str2.size()) //长度长的整数大于长度小的整数   
            return 1;   
      else if(str1.size() < str2.size())   
            return -1;   
      else  
            return str1.compare(str2); //若长度相等，从头到尾按位比较，compare函数：相等返回0，大于返回1，小于返回－1   
}   
//高精度加法   
string ADD_INT(string str1, string str2)   
{   
      string MINUS_INT(string str1, string str2);   //定义减法函数 
      int sign = 1; //sign 为符号位   
      string str;   
      
      //加减法转换阶段 
      /***********************************************/
      if(str1[0] == ‘-‘) 
      {   
           if(str2[0] == ‘-‘) 
           {   
                 sign = -1;   
                 str = ADD_INT(str1.erase(0, 1), str2.erase(0, 1));   //erase(iterator first, iterator last);//删除[first，last）之间的所有字符，返回删除后迭代器的位置 
           }
           else 
           {   
                 str = MINUS_INT(str2, str1.erase(0, 1));   
           }   
      }
      else 
      {   
           if(str2[0] == ‘-‘)   
                 str = MINUS_INT(str1, str2.erase(0, 1));   
    /***********************************************/
    
           else 
           {   
                 //把两个整数对齐，短整数前面加0补齐  
                 /***********************************************/ 
                 string::size_type l1, l2;   
                 int i;   
                 l1 = str1.size(); l2 = str2.size();   
                 if(l1 < l2) 
                 {   
                       for(i = 1; i <= l2 - l1; i++)   
                       str1 = "0" + str1;   
                 }
                 else 
                 {   
                       for(i = 1; i <= l1 - l2; i++)   
                       str2 = "0" + str2;   
                 }   
                 /************************************************/ 
                 
                 int int1 = 0, int2 = 0; //int2 记录进位   
                 for(i = str1.size() - 1; i >= 0; i--) 
                 {   
                       int1 = (int(str1[i]) - 48 + int(str2[i]) - 48 + int2) % 10;  //48 为 ‘0‘ 的ASCII 码   
                       int2 = (int(str1[i]) - 48 + int(str2[i]) - 48 +int2) / 10;   
                       str = char(int1 + 48) + str;   
                 }   
                 if(int2 != 0) str = char(int2 + 48) + str;   
          }   
     }   
     //运算后处理符号位   
     if((sign == -1) && (str[0] != ‘0‘))   
          str = "-" + str;   
     return str;   
}   
  
  
//高精度减法   
string MINUS_INT(string str1, string str2)   
{   
     string MULTIPLY_INT(string str1, string str2);   
     int sign = 1; //sign 为符号位   
     string str;   
     if(str2[0] == ‘-‘)   
            str = ADD_INT(str1, str2.erase(0, 1));   
     else {   
            int res = compare(str1, str2);   
            if(res == 0) return "0";   
            if(res < 0) {   
                  sign = -1;   
                  string temp = str1;   
                  str1 = str2;   
                  str2 = temp;   
            }   
            string::size_type tempint;   
            tempint = str1.size() - str2.size();   
            for(int i = str2.size() - 1; i >= 0; i--) {   
                 if(str1[i + tempint] < str2[i]) {   
                       str1[i + tempint - 1] = char(int(str1[i + tempint - 1]) - 1);   
                       str = char(str1[i + tempint] - str2[i] + 58) + str;   
                 }   
                 else  
                       str = char(str1[i + tempint] - str2[i] + 48) + str;   
            }   
           for(int i = tempint - 1; i >= 0; i--)   
                str = str1[i] + str;   
     }   
     //去除结果中多余的前导0   
     str.erase(0, str.find_first_not_of(‘0‘));   
     if(str.empty()) str = "0";   
     if((sign == -1) && (str[0] != ‘0‘))   
          str = "-" + str;   
     return str;   
}   
  
//高精度乘法   
string MULTIPLY_INT(string str1, string str2)   
{   
     int sign = 1; //sign 为符号位   
     string str;   
     if(str1[0] == ‘-‘) {   
           sign *= -1;   
           str1 = str1.erase(0, 1);   
     }   
     if(str2[0] == ‘-‘) {   
           sign *= -1;   
           str2 = str2.erase(0, 1);   
     }   
     int i, j;   
     string::size_type l1, l2;   
     l1 = str1.size(); l2 = str2.size();   
     for(i = l2 - 1; i >= 0; i --) {  //实现手工乘法   
           string tempstr;   
           int int1 = 0, int2 = 0, int3 = int(str2[i]) - 48;   
           if(int3 != 0) {   
                  for(j = 1; j <= (int)(l2 - 1 - i); j++)   
                        tempstr = "0" + tempstr;   
                  for(j = l1 - 1; j >= 0; j--) {   
                        int1 = (int3 * (int(str1[j]) - 48) + int2) % 10;   
                        int2 = (int3 * (int(str1[j]) - 48) + int2) / 10;   
                        tempstr = char(int1 + 48) + tempstr;   
                  }   
                  if(int2 != 0) tempstr = char(int2 + 48) + tempstr;   
           }   
           str = ADD_INT(str, tempstr);   
     }   
     //去除结果中的前导0   
     str.erase(0, str.find_first_not_of(‘0‘));   
     if(str.empty()) str = "0";   
     if((sign == -1) && (str[0] != ‘0‘))   
           str = "-" + str;   
     return str;   
}   
//高精度除法   
string DIVIDE_INT(string str1, string str2, int flag)   
{   
     //flag = 1时,返回商; flag = 0时,返回余数   
     string quotient, residue; //定义商和余数   
     int sign1 = 1, sign2 = 1;   
     if(str2 == "0") {  //判断除数是否为0   
           quotient = "ERROR!";   
           residue = "ERROR!";   
           if(flag == 1) return quotient;   
           else return residue;   
     }   
     if(str1 == "0") { //判断被除数是否为0   
           quotient = "0";   
           residue = "0";   
     }   
     if(str1[0] == ‘-‘) {   
           str1 = str1.erase(0, 1);   
           sign1 *= -1;   
           sign2 = -1;   
     }   
     if(str2[0] == ‘-‘) {   
           str2 = str2.erase(0, 1);   
           sign1 *= -1;   
     }   
     int res = compare(str1, str2);   
     if(res < 0) {   
           quotient = "0";   
           residue = str1;   
     }else if(res == 0) {   
           quotient = "1";   
           residue = "0";   
     }else {   
           string::size_type l1, l2;   
           l1 = str1.size(); l2 = str2.size();   
           string tempstr;   
           tempstr.append(str1, 0, l2 - 1);   
           //模拟手工除法   
           for(int i = l2 - 1; i < l1; i++) {   
                 tempstr = tempstr + str1[i];   
                 for(char ch = ‘9‘; ch >= ‘0‘; ch --) { //试商   
                       string str;   
                       str = str + ch;   
                       if(compare(MULTIPLY_INT(str2, str), tempstr) <= 0) {   
                              quotient = quotient + ch;   
                              tempstr = MINUS_INT(tempstr, MULTIPLY_INT(str2, str));   
                              break;   
                       }   
                 }   
           }   
           residue = tempstr;   
     }   
     //去除结果中的前导0   
     quotient.erase(0, quotient.find_first_not_of(‘0‘));   
     if(quotient.empty()) quotient = "0";   
     if((sign1 == -1) && (quotient[0] != ‘0‘))   
     quotient = "-" + quotient;   
     if((sign2 == -1) && (residue[0] != ‘0‘))   
     residue = "-" + residue;   
     if(flag == 1) return quotient;   
     else return residue;   
}   
  
//高精度除法,返回商   
string DIV_INT(string str1, string str2)   
{   
      return DIVIDE_INT(str1, str2, 1);   
}   
//高精度除法,返回余数   
string MOD_INT(string str1, string str2)   
{   
      return DIVIDE_INT(str1, str2, 0);   
}   
                     
int main()   
{   
      char ch;   
      string s1, s2, res;   
      while(cin >> ch) {   
           cin >> s1 >> s2;   
           switch(ch) {   
                case ‘+‘:  res = ADD_INT(s1, s2); break;   //高精度加法   
                case ‘-‘:  res = MINUS_INT(s1, s2); break; //高精度减法   
                case ‘*‘:  res = MULTIPLY_INT(s1, s2); break; //高精度乘法   
                case ‘/‘:  res = DIV_INT(s1, s2); break; //高精度除法, 返回商   
                case ‘m‘:  res = MOD_INT(s1, s2); break; //高精度除法, 返回余数   
                default :  break;   
           }   
           cout << res << endl;   
      }   
      return(0);   
}