{模板}高精度套装

来源：互联网发布：ping端口号指令编辑：程序博客网时间：2024/06/02 12:55
绝大部分来自@代号4101的这篇博客
随更新可能部分描述失效
第一个轮子~
现支持
1. 高精度加减乘除及 mod 运算
2. 单精度除法
3. 小范围下按位右移
4. 高精度直接输入输出
5. 字符串 / Long long 转高精度格式
6. 自动压位及去前缀0还有一大堆有的没的玩意儿
2017.8.6. Updata
1. 支持较小范围下右移运算 ( 卡常 )
2. 支持单精度除法
#include <algorithm> // max  #include <cassert>   // assert  #include <cstdio>    // printf,sprintf  #include <cstring>   // strlen  #include <iostream>  // cin,cout  #include <string>    // string类  #include <vector>    // vector类  using namespace std;  struct bign {      typedef unsigned long long ULL;      static const int BASE = 100000000;      static const int WIDTH = 8;      static const int TOTDIT = 10000;//最多总共10000位     vector<int> s;      void write() {        printf("%d", s.back());        for (int i = s.size() - 2;i >= 0;-- i)        {            char buf[WIDTH + 2];            sprintf(buf,"%08d", s[i]);            for (int j = 0;j < strlen(buf);++ j) printf("%c", buf[j]);        }    }    void read() {        char buf[TOTDIT];//位数 最高位        scanf("%s", buf + 1);        s.clear();        int len = (strlen(buf + 1) - 1) / WIDTH + 1;        int num = 0,tmp = 0;        for (int i = len - 1;i >= 1;-- i)        {            int en = strlen(buf + 1) - tmp,st = en - WIDTH + 1;            tmp += WIDTH;            num = 0;            for (int i = st;i <= en;++ i)                num = num * 10 + buf[i] - '0';              s.push_back(num);        }          if (strlen(buf + 1) - tmp > 0)        {            num = 0;            int en = strlen(buf + 1) - tmp,st = 1;            for (int i = st;i <= en;++ i)                num = num * 10 + buf[i] - '0';              s.push_back(num);        }     }    bign& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;}      bign(ULL num = 0) {*this = num;}      bign(string s) {*this = s;}      bign& operator = (long long num) {          s.clear();          do {              s.push_back(num % BASE);              num /= BASE;          } while (num > 0);          return *this;      }      bign& operator = (const string& str) {          s.clear();          int x, len = (str.length() - 1) / WIDTH + 1;          for (int i = 0; i < len; i++) {              int end = str.length() - i*WIDTH;              int start = max(0, end - WIDTH);              sscanf(str.substr(start,end-start).c_str(), "%d", &x);              s.push_back(x);          }          return (*this).clean();      }      bign operator + (const bign& b) const {          bign c; c.s.clear();          for (int i = 0, g = 0; ; i++) {              if (g == 0 && i >= s.size() && i >= b.s.size()) break;              int x = g;              if (i < s.size()) x += s[i];              if (i < b.s.size()) x += b.s[i];              c.s.push_back(x % BASE);              g = x / BASE;          }          return c;      }      bign operator - (const bign& b) const {          assert(b <= *this); // 减数不能大于被减数          bign c; c.s.clear();          for (int i = 0, g = 0; ; i++) {              if (g == 0 && i >= s.size() && i >= b.s.size()) break;              int x = s[i] + g;              if (i < b.s.size()) x -= b.s[i];              if (x < 0) {g = -1; x += BASE;} else g = 0;              c.s.push_back(x);          }          return c.clean();      }      bign operator * (const bign& b) const {          int i, j; ULL g;          vector<ULL> v(s.size()+b.s.size(), 0);          bign c; c.s.clear();          for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=ULL(s[i])*b.s[j];          for (i = 0, g = 0; ; i++) {              if (g ==0 && i >= v.size()) break;              ULL x = v[i] + g;              c.s.push_back(x % BASE);              g = x / BASE;          }          return c.clean();      }     bign operator >> (const int& b) const {        assert(b > 0);          ll p = (1 << b) - 1,q = p + 1;        bign c = *this;        ll m = 0;//余数        for (int i = s.size()-1; i >= 0; i--) {            m = m*BASE + s[i];            c.s[i] = (1ll * m) >> b;            m &= p;            }        return c.clean();    }    bign operator / (const int& b) const {        assert(b > 0);  // 除数必须大于0        bign c = *this;        ll m = 0;//余数        for (int i = s.size()-1; i >= 0; i--) {            m = 1ll * m * BASE + s[i];            c.s[i] = 1ll * m / b;            m %= b;            }        return c.clean();    }    bign operator / (const bign& b) const {          assert(b > 0);  // 除数必须大于0          bign c = *this;       // 商:主要是让c.s和(*this).s的vector一样大          bign m;               // 余数:初始化为0          for (int i = s.size()-1; i >= 0; i--) {              m = m*BASE + s[i];              c.s[i] = bsearch(b, m);              m -= b*c.s[i];              }          return c.clean();      }      bign operator % (const bign& b) const { //方法与除法相同          bign c = *this;          bign m;          for (int i = s.size()-1; i >= 0; i--) {              m = m*BASE + s[i];              c.s[i] = bsearch(b, m);              m -= b*c.s[i];          }          return m;      }      // 二分法找出满足bx<=m的最大的x      int bsearch(const bign& b, const bign& m) const{          int L = 0, R = BASE-1, x;          while (1) {              x = (L+R)>>1;              if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;}              else R = x;          }      }      bign& operator += (const bign& b) {*this = *this + b; return *this;}      bign& operator -= (const bign& b) {*this = *this - b; return *this;}      bign& operator *= (const bign& b) {*this = *this * b; return *this;}      bign& operator /= (const bign& b) {*this = *this / b; return *this;}      bign& operator %= (const bign& b) {*this = *this % b; return *this;}      bign& operator /= (const int& b) {*this = *this / b; return *this;}    bign& operator >>= (const int& b) {*this = *this >> b; return *this;}    bool operator < (const bign& b) const {          if (s.size() != b.s.size()) return s.size() < b.s.size();          for (int i = s.size()-1; i >= 0; i--)              if (s[i] != b.s[i]) return s[i] < b.s[i];          return false;      }      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{return b < *this || *this < b;}      bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);}  };
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