【原创】高精度(压位储存)模板

无聊写了个高精度模板玩玩......

(以前有些地方写错了TAT)

  1 /*

  2 唐代李白

  3 《江夏别宋之悌》

  4 楚水清若空，遥将碧海通。人分千里外，兴在一杯中。

  5 谷鸟吟晴日，江猿啸晚风。平生不下泪，于此泣无穷.

  6 */

  7 #include <iostream>

  8 #include <cstdio>

  9 #include <algorithm>

 10 #include <cstring>

 11 #include <vector>

 12 #include <utility>

 13 #include <iomanip>

 14 #include <string>

 15 #include <cmath>

 16 #include <queue>

 17 #include <assert.h>

 18 #include <map>

 19 #include <ctime>

 20 #include <cstdlib>

 21 #include <stack>

 22 #include <set> 

 23 #define LOCAL

 24 const int INF = 0x7fffffff;

 25 const int MAXN = 100000  + 10;

 26 const int maxnode = 20000 * 2 + 200000 * 20;

 27 const int MAXM = 50000 + 10;

 28 const int MAX = 100;

 29 using namespace std;

 30 struct hp{

 31        int num[MAX];

 32        ////hp(){num[0] = 0;}

 33        hp & operator = (const char *str);

 34        hp & operator = (int b);

 35        

 36        //最好先比较一下再大的减小的 

 37        hp operator + (const hp &)const;

 38        hp operator * (const hp &)const;

 39        hp operator - (const hp &)const;

 40        hp operator / (const hp &)const;

 41        

 42        bool operator < (const hp &)const;

 43        bool operator > (const hp &)const;

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

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

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

 47 };

 48 bool hp::operator < (const hp &b)const{

 49      if (num[0] < b.num[0]) return 1;

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

 51          if (num[i] < b.num[i]) return 1;

 52          else if (num[i] > b.num[i]) return 0;

 53      }

 54      return 0;

 55 }

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

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

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

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

 60 //注意d为余数 

 61 hp hp::operator / (const hp &b)const{

 62     hp d, c;

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

 64     memset(d.num, 0, sizeof(d));

 65     c.num[0] = num[0] + b.num[0] + 1;//和乘法的长度是一样的,长除法..

 66     d.num[0] = 0;

 67     

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

 69         //这个是腾出一个d.num[1]的空间 

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

 71         d.num[0]++;

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

 73         

 74         int l = 0, r = 9999, mid;//枚举除数

 75         while (l < r){//不同的写法当然可以 

 76               int mid = (l + r) >> 1;

 77               hp tmp; 

 78               tmp = mid;tmp = tmp * b;

 79               //b乘以乘数当然不能大于余数 

 80               if (tmp <= d) l = mid + 1;

 81               else r  = mid;

 82         } 

 83         c.num[i] = r - 1;

 84         hp tmp; tmp = r - 1;//r - 1为得到答案 

 85         tmp = tmp * b; 

 86         d = d - tmp;

 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 int b){

 92    char str[MAX];

 93    sprintf(str, "%d", b);

 94    return *this = str; 

 95 }

 96 hp & hp::operator = (const char *str){

 97    int len = strlen(str), k = 1;

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

 99    num[0] = 1;

100    for (int i = len - 1 ; i >= 0; i--){

101        if (k == 10000) {k = 1; num[0]++;}

102        num[num[0]] += k * (str[i] - '0');

103        k *= 10;

104    }

105    return *this;

106 }

107 hp hp::operator + (const hp &b)const{

108    hp c;

109    memset(c.num, 0, sizeof(c.num[0]));

110    c.num[0] = max(num[0], b.num[0]);

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

112        c.num[i] += num[i] + b.num[i];

113        c.num[i + 1] += c.num[i] / 10000;

114        c.num[i] %= 10000;

115    }

116    if (c.num[c.num[0] + 1] > 0) c.num[0]++;

117    return c;

118 }

119 hp hp::operator * (const hp &b)const{

120    hp c;

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

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

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

124    for (int j = 1; j <= b.num[0]; j++){

125        c.num[i + j - 1] += num[i] * b.num[j];

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

127        c.num[i + j - 1] %= 10000;

128    }

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

130    return c;

131 }

132 hp hp::operator - (const hp &b)const{

133    hp c;

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

135    c.num[0] = max(num[0], b.num[0]);

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

137        c.num[i] += num[i] - b.num[i];

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

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

140           c.num[i] += 10000; 

141        }

142    }

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

144    return c;

145 }

146 void print(hp c){

147      printf("%d", c.num[c.num[0]]);

148      for (int i = c.num[0] - 1; i >= 1; i--) printf("%04d", c.num[i]);

149      return;

150 }

151 

152 int main(){

153    #ifdef LOCAL

154    freopen("data.txt", "r", stdin);

155    freopen("out.txt", "w", stdout);

156    #endif

157    hp c;

158    c = 0;

159    hp a; 

160    a = 20300;

161    hp b;

162    //print(a);

163    //print(c);

164    b = c / a; 

165    print(b);

166    //printf("%d", (a == c));

167    return 0; 

168 }

View Code