大数（高精度）加减乘除取模运算

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千辛万苦找到了大数（高精度）加减乘除取模运算的算法，有的地方还需要再消化消化，代码先贴出来~
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);      int sign = 1; //sign 为符号位      string str;      if(str1[0] == '-') {           if(str2[0] == '-') {                 sign = -1;                 str = ADD_INT(str1.erase(0, 1), str2.erase(0, 1));           }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);}
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