【原创】高精度(压位储存)模板

无聊写了个高精度模板玩玩......

  1 /*

  2 高精度(压位储存)

  3 */

  4 #include <cstdlib>

  5 #include <iostream>

  6 #include <cstdio>

  7 #include <cstring>

  8 #include <cmath>

  9 const int MAX=10005;//最长长度

 10 using namespace std;

 11 //高精度结构体,先声明后实现

 12 struct hp

 13 {

 14        //注意，这里用num[0]来保存位数

 15        int num[MAX];

 16

 17        //-------赋值与初始化-------//

 18        hp() {}

 19        hp & operator =(const char*);//从字符串赋值

 20        hp & operator =(int);//从整型赋值

 21        hp (int);

 22

 23        //-------普通运算符-------//

 24        hp operator + (const hp &) const;

 25        hp operator - (const hp &) const;

 26        hp operator * (const hp &) const;

 27        hp operator / (const hp &) const;//注意后面两个的实现

 28        hp operator % (const hp &) const;

 29

 30        //-------比较运算符-------//

 31        bool operator > (const hp &) const;

 32        bool operator < (const hp &) const;

 33        bool operator == (const hp &) const;

 34        bool operator >= (const hp &) const;

 35        bool operator <= (const hp &) const;

 36

 37 }shu;

 38 //-------实现-------//

 39 hp & hp::operator =(const char* str)

 40 {

 41      memset(num,0,sizeof(num));

 42      int len=strlen(str),j=1,k=1;

 43      for (int i=1;i<=len;i++)//注意从1开始循环

 44      {

 45          if (k==10000) j++,k=1;

 46          num[j]+=k*(str[len-i]-‘0‘);

 47          k*=10;

 48      }

 49      num[0]=j;

 50      return *this;

 51 }

 52 hp & hp::operator =(int a)

 53 {

 54      char s[MAX];

 55      sprintf(s,"%d",a);

 56      return *this=s;

 57 }

 58 hp::hp(int n){*this=n;}

 59 hp hp::operator +(const hp & t)const

 60 {

 61      hp c;

 62      memset(c.num,0,sizeof(c.num));

 63      c.num[0]=max(num[0],t.num[0]);

 64      for (int i=1;i<=c.num[0];i++)

 65      {

 66          c.num[i]=num[i]+t.num[i];

 67          if (c.num[i]>=10000){

 68              c.num[i+1]++;

 69              c.num[i]-=10000;

 70          }

 71      }

 72      while (c.num[c.num[0]+1]>0) c.num[0]++;

 73      return c;

 74 }

 75 hp hp::operator -(const hp & t)const

 76 {

 77      hp c;

 78      memset(c.num,0,sizeof(c.num));

 79      c.num[0]=num[0];

 80      for (int i=1;i<=c.num[0];i++)

 81      {

 82          c.num[i]+=num[i]-t.num[i];

 83          if (c.num[i]<0){

 84              c.num[i+1]--;

 85              c.num[i]+=10000;

 86          }

 87      }

 88      while (c.num[c.num[0]]==0 && c.num[0]>1) c.num[0]--;

 89      return c;

 90 }

 91 hp hp::operator *(const hp & t)const

 92 {

 93      hp c;

 94      memset(c.num,0,sizeof(c.num));

 95      c.num[0]=num[0]+t.num[0]+1;//位数

 96      for (int i=1;i<=num[0];i++)

 97      for (int j=1;j<=t.num[0];j++)

 98      {

 99          c.num[i+j-1]=num[i]*t.num[j];

100          c.num[i+j]+=c.num[i+j-1]/10000;

101          c.num[i+j]%=10000;

102      }

103      while (c.num[c.num[0]]==0 && c.num[0]>1) c.num[0]--;

104      return c;

105 }

106 bool hp::operator > (const hp &b) const

107 {

108     if (num[0]!=b.num[0]) return num[0]>b.num[0];

109     for (int i=num[0];i>=1;i--)

110         if (num[i]!=b.num[i])

111             return (num[i]>b.num[i]);

112     return false;

113 }

114 bool hp::operator < (const hp &b) const {return b>*this;}

115 bool hp::operator <= (const hp &b) const {return !(*this>b);}

116 bool hp::operator >= (const hp &b) const {return !(b>*this);}

117 bool hp::operator == (const hp &b) const {return !(b>*this)&&!(*this>b);}

118 hp hp::operator /(const hp & b)const//二分除法

119 {

120     hp c, d;

121     c.num[0]=num[0]+b.num[0]+1;

122     d.num[0]=0;//d用来存储余数

123     for (int i=num[0];i>=1;i--)

124     {

125         memmove(d.num+2,d.num+1,sizeof(d.num)-sizeof(int)*2);

126         d.num[0]++;

127         d.num[1]=num[i];

128

129         int left=0,right=9999,mid;

130         while (left<right)

131         {

132             mid =(left+right)/2;

133             if (b*hp(mid)<=d) left=mid+1;

134             else right=mid;

135         }

136         c.num[i]=right-1;

137         d=d-b*hp(right-1);

138     }

139     while (c.num[c.num[0]]==0&&c.num[0]>1) c.num[0]--;

140     return c;

141 }