挑戰面試編程：大整數的加、減、乘、除，面試整數

挑戰面試編程：大整數的加、減、乘、除，面試整數

                     挑戰面試編程：大整數的加、減、乘、除

    一切都是有限的，哪怕是看起來無限的時間或空間都很可能是有限的。在電腦中內建類型的加、減、乘、除都是有限的。我們來實現一個“無限”的大整數加、減、乘、除。

以下使用C++代碼實現

#include <iostream>#include <string>using namespace std;//大整數的加減乘除string add_int(string, string);string sub_int(string, string);string mul_int(string, string);string div_int(string, string);string mod_int(string, string);string divide_int(string, string, int);inline int compare(string s1, string s2){if (s1.size() < s2.size())return -1;else if (s1.size() > s2.size())return 1;elsereturn s1.compare(s2);}/*大整數加法本質上只處理：兩個正數相加，如 "123" + "234"其它情況需轉化1. 正加負 => "123" + "-234"  = "123" - "234" 轉化為減法2. 負加正 => "-234" + "123"  = "123" - "234"3. 負加負 => "-123" + "-234" = -("123" + "234")*/string add_int(string s1, string s2){if (s1 == "0")return s2;if (s2 == "0")return s1;if (s1[0] == '-'){if (s2[0] == '-'){return "-" + add_int(s1.erase(0, 1), s2.erase(0, 1));  //情況三}else{return sub_int(s2, s1.erase(0, 1));  //情況二}}if (s2[0] == '-'){return sub_int(s1, s2.erase(0, 1));  //情況一}//處理本質情況string::size_type i, size1, size2;size1 = s1.size();size2 = s2.size();if (size1 < size2){for (i = 0; i < size2 - size1; i++)   //在s1左邊補零s1 = "0" + s1;}else{for (i = 0; i < size1 - size2; i++)   //在s2左邊補零s2 = "0" + s2;}int n1, n2;n2 = 0;size1 = s1.size();size2 = s2.size();string res;for (i = size1 - 1; i != 0; i--)   //從最低位加起{n1 = (s1[i] - '0' + s2[i] - '0' + n2) % 10;  //n1代表當前位的值n2 = (s1[i] - '0' + s2[i] - '0' + n2) / 10;  //n2代表進位res = char(n1 + '0') + res;}/*上述迴圈不能處理第0位的原因在於i的類型是string::size_type，它是非負類型*///對於第0位單獨處理n1 = (s1[0] - '0' + s2[0] - '0' + n2) % 10;n2 = (s1[0] - '0' + s2[0] - '0' + n2) / 10;res = char(n1 + '0') + res;if (n2 == 1)res = "1" + res;return res;}/*大整數減法本質上只處理：兩整數相減，並且是一大減一小："1234" - "234"其它情況需轉化1. 小正減大正 => "234" - "1234" = -("1234" - "234")2. 正減負 => "1234" - "-234" = "1234" + "234"3. 負減正 => "-1234" - "234" = -("1234" + "234")4. 負減負 => "-1234" - "-234" = "234" - "1234" = -("1234" - "234")*/string sub_int(string s1, string s2){if (s2 == "0")return s1;if (s1 == "0"){if (s2[0] == '-')return s2.erase(0, 1);return "-" + s2;}if (s1[0] == '-'){if (s2[0] == '-'){return sub_int(s2.erase(0, 1), s1.erase(0, 1));   //情況四}return "-" + add_int(s1.erase(0, 1), s2);   //情況三}if (s2[0] == '-')return add_int(s1, s2.erase(0, 1));  //情況二//調整s1與s2的長度string::size_type i, size1, size2;size1 = s1.size();size2 = s2.size();if (size1 < size2){for (i = 0; i < size2 - size1; i++)   //在s1左邊補零s1 = "0" + s1;}else{for (i = 0; i < size1 - size2; i++)   //在s2左邊補零s2 = "0" + s2;}int t = s1.compare(s2);if (t < 0)   //s1與s2的size相同，但 s1 < s2return "-" + sub_int(s2, s1);if (t == 0)return "0";//處理本質情況：s1 > s2string res;string::size_type j;for (i = s1.size() - 1; i != 0; i--){if (s1[i] < s2[i])   //不足，需向前借一位{j = 1;while (s1[i - j] == '0'){s1[i - j] = '9';j++;}s1[i - j] -= 1;res = char(s1[i] + ':' - s2[i]) + res;}else{res = char(s1[i] - s2[i] + '0') + res;}}res = char(s1[0] - s2[0] + '0') + res;//去掉前置字元為零res.erase(0, res.find_first_not_of('0'));return res;}string mul_int(string s1, string s2){if (s1 == "0" || s2 == "0")return "0";//sign是符號位int sign = 1;if (s1[0] == '-'){sign *= -1;s1.erase(0, 1);}if (s2[0] == '-'){sign *= -1;s2.erase(0, 1);}string::size_type size1, size2;string res, temp;size1 = s1.size();size2 = s2.size();//讓s1的長度最長if (size1 < size2){temp = s1;s1 = s2;s2 = temp;size1 = s1.size();size2 = s2.size();}int i, j, n1, n2, n3, t;for (i = size2 - 1; i >= 0; i--){temp = "";n1 = n2 = n3 = 0;for (j = 0; j < size2 - 1 - i; j++) temp = "0" + temp;n3 = s2[i] - '0';for (j = size1 - 1; j >= 0; j--){t = (n3*(s1[j] - '0') + n2);n1 = t % 10;  //n1記錄當前位置的值n2 = t / 10;  //n2記錄進位的值temp = char(n1 + '0') + temp;}if (n2)temp = char(n2 + '0') + temp;res = add_int(res, temp);}if (sign == -1)return "-" + res;return res;}string divide_int(string s1, string s2, int flag)  //flag=1,返回商;flag=0,返回餘數{string quotient, residue;if (s2 == "0"){quotient = residue = "error";if (flag == 1)return quotient;elsereturn residue;}if (s1 == "0"){quotient = residue = "0";if (flag == 1)return quotient;elsereturn residue;}//sign1是商的符號，sign2是餘數的符號int sign1, sign2;sign1 = sign2 = 1;if (s1[0] == '-'){sign1 *= -1;sign2 = -1;s1.erase(0, 1);}if (s2[0] == '-'){sign1 *= -1;s2.erase(0, 1);}if (compare(s1, s2) < 0){quotient = "0";residue = s1;}else if (compare(s1, s2) == 0){quotient = "1";residue = "0";}else{string temp;string::size_type size1, size2;size1 = s1.size();size2 = s2.size();int i;if (size2 > 1) temp.append(s1, 0, size2 - 1);for (i = size2 - 1; i < size1; i++){temp = temp + s1[i];//試商for (char c = '9'; c >= '0' ; c--){string t = mul_int(s2, string(1, c));string s = sub_int(temp, t);if (s == "0" || s[0] != '-'){temp = s;quotient = quotient + c;break;}}}residue = temp;}//去除前置字元為零quotient.erase(0, quotient.find_first_not_of('0'));residue.erase(0, residue.find_first_not_of('0'));if (sign1 == -1){quotient = "-" + quotient;}if (sign2 == -1){if (residue.empty())residue = "0";elseresidue = "-" + residue;}if (flag == 1) return quotient;else return residue;}string div_int(string s1, string s2){return divide_int(s1, s2, 1);}string mod_int(string s1, string s2){return divide_int(s1, s2, 0);}int main(void){string s1, s2;char op;while (cin >> s1 >> op >> s2){switch (op){case '+':cout << add_int(s1, s2) << endl; break;case '-':cout << sub_int(s1, s2) << endl; break;case '*':cout << mul_int(s1, s2) << endl; break;case '/':cout << div_int(s1, s2) << endl; break;case '%':cout << mod_int(s1, s2) << endl; break;default:cout << "The operator is error!" << endl; break;}}return 0;}

隨後來一個C代碼的……

代碼下載

大整數的加、減、乘、除 C++

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