大数运算

学校大作业要求实现任意大数（可带小数局部）的加减乘除，如果有小数则须要保留10位小数，我采纳了vector容器把每一位的数字的贮存下来，再通过如同小学竖式计算来实现。那么实际上实现难度不高，惟一难度就是大数除法的实现。而通过对齐除数与被除数，逐位屡次相减，则可得出对应位的商。保留10位小数，那么求商就要求到第11位，则被除数后要接上11减去被除数小数长度加上除数小数长度个0，求出后果后再四舍五入即可。

除法运算实现代码

vector<short> divide(vector<short>number1, vector<short>number2, short decimal_length1, short decimal_length2, short decimal_length) { vector<short> number; number.reserve(60); while (!number2[0]) number2.erase(number2.begin()); short extra_zero = decimal_length - decimal_length1 + decimal_length2; short former_number2_size = number2.size(); for (short i = 0;i < extra_zero;i++) number1.insert(number1.end(), 0); while (number2.size() < number1.size()) number2.insert(number2.end(), 0); while (number2.size() >= former_number2_size) { short digit_result = 0; short former_number1_size = number1.size(); while (1) { if (number1.size() == number2.size()) { short i = 0; for (;i < number1.size() - 1 && number1[i] == number2[i];i++); if (number1[i] - number2[i] < 0) break; } if (number1.size() < number2.size()) break; number1 = subtract_for_divide(number1, number2); carry(number1); digit_result++; } number.push_back(digit_result); number2.pop_back(); } return number;}void carry_for_mutiple_and_divide(vector<short>& number, short& decimal_length) { for (short j = number.size() - 1;j >= 0;j--) { if (number[j] >= 10) { if (!j) { number.insert(number.begin(), number[j] / 10); number[1] %= 10; j++; } else { number[j - 1] += number[j] / 10; number[j] %= 10; } } else if (number[j] < 0) { number[j] += 10; number[j - 1]--; } } while (!number[0] && number.size() > 0 && number.size() - decimal_length > 1) number.erase(number.begin()); while (decimal_length > 0 && *(number.end() - 1) == 0) { number.pop_back(); decimal_length--; }}

大数运算实现代码

Source
办法挺笨的，写了差不多500行。（用c++实现）

发明的vector数组都应用了reserve，尽管会占用稍多的内存，但能够缩小裁减数组（须要将整个vector数组块复制到另一块更大的内存）的次数，从而进步运行速度。