c++ 高精度模板

来源：互联网发布：数码宝贝2知乎编辑：程序博客网时间：2024/06/05 16:25
一个高精度模板, 我一直用它, 很好用.
#include <stdio.h>#include <string.h> #include <stdlib.h> #include <math.h>#include <assert.h>  #include <ctype.h> #include <map>#include <string>#include <set>#include <bitset>#include <utility>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <iostream>#include <fstream>#include <list>using  namespace  std;      const  int MAXL = 1000;  //位数！！      struct  BigNum      {          int  num[MAXL];          int  len;      };      //高精度比较 a > b return 1, a == b return 0; a < b return -1;      int  Comp(BigNum &a, BigNum &b)      {          int  i;          if(a.len != b.len) return (a.len > b.len) ? 1 : -1;          for(i = a.len-1; i >= 0; i--)              if(a.num[i] != b.num[i]) return  (a.num[i] > b.num[i]) ? 1 : -1;          return  0;      }      //高精度加法      BigNum  Add(BigNum &a, BigNum &b)      {          BigNum c;          int  i, len;          len = (a.len > b.len) ? a.len : b.len;          memset(c.num, 0, sizeof(c.num));          for(i = 0; i < len; i++)          {              c.num[i] += (a.num[i]+b.num[i]);              if(c.num[i] >= 10)              {                  c.num[i+1]++;                  c.num[i] -= 10;              }          }          if(c.num[len])  len++;          c.len = len;          return  c;      }      //高精度减法,保证a >= b      BigNum Sub(BigNum &a, BigNum &b)      {          BigNum  c;          int  i, len;          len = (a.len > b.len) ? a.len : b.len;          memset(c.num, 0, sizeof(c.num));          for(i = 0; i < len; i++)          {              c.num[i] += (a.num[i]-b.num[i]);              if(c.num[i] < 0)              {                  c.num[i] += 10;                  c.num[i+1]--;              }          }          while(c.num[len] == 0 && len > 1)  len--;          c.len = len;          return  c;      }      //高精度乘以低精度，当b很大时可能会发生溢出int范围,具体情况具体分析      //如果b很大可以考虑把b看成高精度      BigNum Mul1(BigNum &a, int  &b)      {          BigNum c;          int  i, len;          len = a.len;          memset(c.num, 0, sizeof(c.num));          //乘以0，直接返回0          if(b == 0)           {              c.len = 1;              return  c;          }          for(i = 0; i < len; i++)          {              c.num[i] += (a.num[i]*b);              if(c.num[i] >= 10)              {                  c.num[i+1] = c.num[i]/10;                  c.num[i] %= 10;              }          }          while(c.num[len] > 0)          {              c.num[len+1] = c.num[len]/10;              c.num[len++] %= 10;          }          c.len = len;           return  c;      }      //高精度乘以高精度,注意要及时进位，否则肯能会引起溢出,但这样会增加算法的复杂度，      //如果确定不会发生溢出, 可以将里面的while改成if      BigNum  Mul2(BigNum &a, BigNum &b)      {          int i, j, len = 0;          BigNum  c;          memset(c.num, 0, sizeof(c.num));          for(i = 0; i < a.len; i++) {        for(j = 0; j < b.len; j++)              {                  c.num[i+j] += (a.num[i]*b.num[j]);                  if(c.num[i+j] >= 10)                  {                      c.num[i+j+1] += c.num[i+j]/10;                      c.num[i+j] %= 10;                  }              } }    len = a.len+b.len-1;          while(c.num[len-1] == 0 && len > 1)  len--;          if(c.num[len])  len++;          c.len = len;          return  c;      }      //高精度除以低精度,除的结果为c, 余数为f      void Div1(BigNum &a, int &b, BigNum &c, int &f)      {          int  i, len = a.len;          memset(c.num, 0, sizeof(c.num));          f = 0;          for(i = a.len-1; i >= 0; i--)          {              f = f*10+a.num[i];              c.num[i] = f/b;              f %= b;          }          while(len > 1 && c.num[len-1] == 0)  len--;          c.len = len;      }      //高精度*10      void  Mul10(BigNum &a)      {          int  i, len = a.len;          for(i = len; i >= 1; i--)              a.num[i] = a.num[i-1];          a.num[i] = 0;          len++;          //if a == 0          while(len > 1 && a.num[len-1] == 0)  len--;      }      //高精度除以高精度，除的结果为c，余数为f      void Div2(BigNum &a, BigNum &b, BigNum &c, BigNum &f)      {          int  i, len = a.len;          memset(c.num, 0, sizeof(c.num));          memset(f.num, 0, sizeof(f.num));          f.len = 1;          for(i = len-1;i >= 0;i--)          {              Mul10(f);              //余数每次乘10              f.num[0] = a.num[i];              //然后余数加上下一位              ///利用减法替换除法              while(Comp(f, b) >= 0)              {            f = Sub(f, b);                  c.num[i]++;              }          }          while(len > 1 && c.num[len-1] == 0)  len--;          c.len = len;      }   void  print(BigNum &a)   //输出大数   {          int  i;          for(i = a.len-1; i >= 0; i--)              printf("%d", a.num[i]);          puts("");      }      //将字符串转为大数存在BigNum结构体里面      BigNum ToNum(char *s)      {          int i, j;          BigNum  a;          a.len = strlen(s);          for(i = 0, j = a.len-1; s[i] != '\0'; i++, j--)              a.num[i] = s[j]-'0';          return  a;      }      void Init(BigNum &a, char *s, int &tag)   //将字符串转化为大数{       int  i = 0, j = strlen(s);     if(s[0] == '-') {  j--;  i++;  tag *= -1; }    a.len = j;    for(; s[i] != '\0'; i++, j--)        a.num[j-1] = s[i]-'0';}   int main(void)      {          BigNum a, b;       char  s1[100], s2[100];       while(scanf("%s %s", s1, s2) != EOF)       {           int tag = 1;           Init(a, s1, tag);    //将字符串转化为大数        Init(b, s2, tag);           a = Mul2(a, b);           if(a.len == 1 && a.num[0] == 0)           {               puts("0");           }           else            {               if(tag < 0) putchar('-');               print(a);           }       }       return 0;   }
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