高精度（大数）的四则运算与逻辑运算---c++ struct版

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因为刚好做一个高精度的加法题，就顺便把高精度的四则运算都写了。

以下代码是用struct实现的，下次再用类实现吧。

ps：保证大数都是正整数（有负数或带小数的写不来23333OTZ）。

直接上代码吧。。

#include <iostream>#include <string>#include <string.h>#include <stdio.h>using namespace std;const int maxn = 1000;                   // 最大运算位数int max(int a, int b){return a > b ? a : b;}struct bign{int len, s[maxn];                  //len记录数字的位数，s存储数字s[0]，s[1],s[2]。分别是个位十位百位（以此类推）bign() { memset(s, 0, sizeof(s)); len = 1; }bign operator=(const char * num)   //将字符串赋值给大数类型{len = strlen(num);for (int i = 0; i < len; i++)s[i] = num[len - i - 1] - '0';int i = len - 1;while (i > 0 && s[i] == 0) i--;  // 验证数字的真正大小，避免“0000010”的这样情况len = i + 1;return *this;}bign operator=(int num)           //将int类型赋值给大数类型{char s[maxn];sprintf(s, "%d", num);*this = s;return *this;}bign operator=(const bign & b)   //大数类型的互相赋值{len = b.len;for (int i = 0; i < len; i++)s[i] = b.s[i];return *this;}bign(int num) { *this = num; }   //将int转换为大数类型bign(const char * num) { *this = num; }  //同上string str() const               //将大数类型（副本）转换为string{string res = "";for (int i = 0; i < len; i++)res = (char)(s[i] + '0') + res;if (res == "")res == "0";return res;}bign operator+(const bign & b) const  {bign c;c.len = 0;for (int i = 0, g = 0; g || i < max(len, b.len); i++){int x = g;if (i < len)x += s[i];if (i < b.len)x += b.s[i];c.s[c.len++] = x % 10;g = x / 10;}return c;}bign operator-(const bign & b) const{bign c;bign temp;temp = *this;c.len = 0;for (int i = 0; i < max(temp.len, b.len); i++){int x = 0;if (i < temp.len)x += temp.s[i];if (i < b.len)x -= b.s[i];if (x >= 0)c.s[c.len++] = x;else{c.s[c.len++] = x + 10;temp.s[i + 1]--;}}int i = c.len - 1;while (i > 0 && c.s[i] == 0) i--;       //确定数字的真正位数c.len = i + 1;return c;}bign operator*(const bign & b) const{bign c;for (int i = 0; i < len; i++){for (int j = 0; j < b.len; j++){c.s[i + j] += s[i] * b.s[j];}}for (int i = 0; i < len + b.len - 1; i++){c.s[i + 1] += c.s[i] / 10;c.s[i] = c.s[i] % 10;}for (int i = len + b.len + 2; i > 0; i--)if (c.s[i] != 0){c.len = i + 1;break;}return c;}    //定义逻辑运算符bool operator<(const bign & b) const{if (len != b.len)return len < b.len;for (int i = len; i > 0; i--)if (s[i] != b.s[i])return s[i] < b.s[i];}bool operator>(const bign & b) const{return b < *this;}bool operator<=(const bign & b) const{return !(b < *this);}bool operator>=(const bign & b) const{return !(*this < b);}bool operator==(const bign & b) const{if (*this < b)return false;else if (*this > b)return false;elsereturn true;}bool operator!=(const bign & b) const{return !(*this == b);}//除法（整除）bign operator/(const bign & b) const{    bign temp, result;    temp = *this;    result = 0;    while (temp > b)        {            temp = temp - b;            result = result + 1;        }    return result;}//取余数bign operator%(const bign & b) const{    bign result;    result = *this;    while (result > b)            result = result - b;        return result;}};istream & operator>>(istream & in, bign & x){string s;in >> s;x = s.c_str();return in;}ostream & operator<<(ostream & out, const bign & x){out << x.str();return out;}int main(){bign a, b, c;cin >> a >> b;    cout << a + b << endl;    cout << a * b << endl;    cout << a - b << endl;    cout << a / b << endl;    cout << a % b << endl;return 0;}

一直觉得除法的效率很低，但是其他方法不会啊2333333

以后有时间再优化一下吧。。

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