大数运算

Posted 2021-10-05 BLIIIIIIND

学校大作业要求实现任意大数（可带小数部分）的加减乘除，如果有小数则需要保留10位小数，我采用了vector容器把每一位的数字的储存下来，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，唯一难度就是大数除法的实现。而通过对齐除数与被除数，逐位多次相减，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，求出结果后再四舍五入即可。

更新：老师不允许用stl，还只能用链表，只好手写一个跟list差不多的类出来，也改进了下代码。

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--;
}
}
//双向链表版：
My_List divide(My_List number1, My_List number2, short decimal_length1, short decimal_length2) {
My_List number;
while (!number2.head->val) number2.pop_head();
short extra_zero = 11 - decimal_length1 + decimal_length2;
short former_number2_size = number2.size;
for (short i = 0;i < extra_zero;i++)
number1.push_back(0);
while (number2.size < number1.size)
number2.push_back(0);
while (number2.size >= former_number2_size) {
short digit_result = 0;
short former_number1_size = number1.size;
while (1) {
if (number1.size < number2.size)
break;
number1 = subtract(number1, number2);
carry(number1);
bool positive = 1;
for (node* p = number1.head;p;p = p->next)
if (p->val < 0) {
positive = 0;
break;
}
if (positive) digit_result++;
else {
number1 = sum(number1, number2);
break;
}
}
number.push_back(digit_result);
number2.pop_back();
}
return number;
}
​
void carry_for_md(My_List& number, short& decimal_length) {
for (node* p = number.end;p;p = p->pre) {
if (p->val >= 10) {
if (p == number.head)
number.push_head(p->val / 10);
else p->pre->val += p->val / 10;
p->val %= 10;
}
}
while (!number.head->val && number.size > 0 && number.size - decimal_length > 1)
number.pop_head();
while (number.end && decimal_length > 0 && !number.end->val) {
number.pop_back();
decimal_length--;
}
if (decimal_length > 10) {
node* p = number.head;
for (short i = 0;i < number.size - decimal_length + 10;i++)
p = p->next;
if (p->val >= 5)
p->pre->val++;
if (p->pre->val >= 10) {
p->pre->val -= 10;
p->pre->val++;
}
}
}

大数运算实现代码

Source

创造的vector数组都使用了reserve，虽然会占用稍多的内存，但可以减少扩充数组（需要将整个vector数组块复制到另一块更大的内存）的次数，从而提高运行速度。

更新：把用双向链表的版本也放进去了，改进了逻辑，感觉代码也没这么乱了，虽然还是接近500行，但是如果直接用stl里的list能删掉差不多一百行，在oj上能跑0s。