bign（高精度）大整数输入输出及运算

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<string>
  5 #include<algorithm>
  6 #include<vector>
  7 #include<queue>
  8 #include<stack>
  9 #include<map>
 10 #include<set>
 11 #include<cmath>
 12 using namespace std;
 13 #define ll long long
 14 #define ull unsigned long long
 15 #define ldb long double
 16 #define db double
 17 #define fi first
 18 #define se second
 19 #define INF 0x3f3f3f3f
 20 #define endl "\n"
 21 #define rush() int T;cin>>T;while(T--)
 22 #define mem(a,b) memset((a),(b),sizeof(a))
 23 const db pi = acos((db)-1);
 24 const ll MAXN = 5005;
 25 const ll mod = 1e9+7; 
 26 
 27 struct bign  
 28 {  
 29     int len, s[MAXN];  
 30     bign ()  //初始化
 31     {  
 32         memset(s, 0, sizeof(s));  
 33         len = 1;  
 34     }  
 35     bign (int num) { *this = num; }  
 36     bign (const char *num) { *this = num; }  //让this指针指向当前字符串
 37     bign operator = (const int num)  
 38     {  
 39         char s[MAXN];  
 40         sprintf(s, "%d", num);  //sprintf函数将整型映到字符串中
 41         *this = s;  
 42         return *this;  //再将字符串转到下面字符串转化的函数中
 43     }  
 44     bign operator = (const char *num)  
 45     {  
 46         for(int i = 0; num[i] == ‘0‘; num++) ;  //去前导0  
 47         len = strlen(num);  
 48         for(int i = 0; i < len; i++) s[i] = num[len-i-1] - ‘0‘; //反着存
 49         return *this;
 50     }  
 51     bign operator + (const bign &b) const //对应位相加，最为简单
 52     {  
 53         bign c;  
 54         c.len = 0;  
 55         for(int i = 0, g = 0; g || i < max(len, b.len); i++)  
 56         {  
 57             int x = g;  
 58             if(i < len) x += s[i];  
 59             if(i < b.len) x += b.s[i];  
 60             c.s[c.len++] = x % 10;  //关于加法进位
 61             g = x / 10;  
 62         }  
 63         return c;  
 64     }  
 65     bign operator += (const bign &b)  //如上文所说，此类运算符皆如此重载
 66     {  
 67         *this = *this + b;  
 68         return *this;  
 69     }  
 70     void clean()  //由于接下来的运算不能确定结果的长度，先大而估之然后再查
 71     {  
 72         while(len > 1 && !s[len-1]) len--;  //首位部分‘0’故删除该部分长度
 73     }   
 74     bign operator * (const bign &b) //乘法重载在于列竖式，再将竖式中的数转为抽象，即可看出运算法则。
 75     {  
 76         bign c;  
 77         c.len = len + b.len;  
 78         for(int i = 0; i < len; i++)  
 79         {  
 80             for(int j = 0; j < b.len; j++)  
 81             {  
 82                 c.s[i+j] += s[i] * b.s[j];//不妨列个竖式看一看
 83             }  
 84         }  
 85         for(int i = 0; i < c.len; i++) //关于进位，与加法意同 
 86         {  
 87             c.s[i+1] += c.s[i]/10;
 88             c.s[i] %= 10;
 89         }  
 90         c.clean();  //我们估的位数是a+b的长度和，但可能比它小（1*1 = 1）
 91         return c;  
 92     }  
 93     bign operator *= (const bign &b)  
 94     {  
 95         *this = *this * b;  
 96         return *this;  
 97     }  
 98     bign operator - (const bign &b)  //对应位相减，加法的进位改为借1
 99     {  //不考虑负数
100         bign c;  
101         c.len = 0;  
102         for(int i = 0, g = 0; i < len; i++)  
103         {  
104             int x = s[i] - g;  
105             if(i < b.len) x -= b.s[i];  //可能长度不等
106             if(x >= 0) g = 0;  //是否向上移位借1
107             else  
108             {  
109                 g = 1;  
110                 x += 10;  
111             }  
112             c.s[c.len++] = x;  
113         }  
114         c.clean();  
115         return c;   
116     }  
117     bign operator -= (const bign &b)  
118     {  
119         *this = *this - b;  
120         return *this;  
121     }  
122     bign operator / (const bign &b)  //运用除是减的本质，不停地减，直到小于被减数
123     {  
124         bign c, f = 0; //可能会在使用减法时出现高精度运算 
125         for(int i = len-1; i >= 0; i--)  //正常顺序，从最高位开始
126         {  
127             f = f*10;  //上面位的剩余到下一位*10
128             f.s[0] = s[i];  //加上当前位
129             while(f >= b)  
130             {  
131                 f -= b;  
132                 c.s[i]++;  
133             }  
134         }  
135         c.len = len;  //估最长位
136         c.clean();  
137         return c;  
138     }  
139     bign operator /= (const bign &b)  
140     {  
141         *this  = *this / b;  
142         return *this;  
143     }  
144     bign operator % (const bign &b)  //取模就是除完剩下的
145     {  
146         bign r = *this / b;  
147         r = *this - r*b;  
148         r.clean();
149         return r;  
150     }  
151     bign operator %= (const bign &b)  
152     {  
153         *this = *this % b;  
154         return *this;  
155     }  
156     bool operator < (const bign &b) //字符串比较原理 
157     {  
158         if(len != b.len) return len < b.len;  
159         for(int i = len-1; i != -1; i--)  
160         {  
161             if(s[i] != b.s[i]) return s[i] < b.s[i];  
162         }  
163         return false;  
164     }  
165     bool operator > (const bign &b)  //同理
166     {  
167         if(len != b.len) return len > b.len;  
168         for(int i = len-1; i != -1; i--)  
169         {  
170             if(s[i] != b.s[i]) return s[i] > b.s[i];  
171         }  
172         return false;  
173     }  
174     bool operator == (const bign &b)  
175     {  
176         return !(*this > b) && !(*this < b);  
177     }  
178     bool operator != (const bign &b)  
179     {  
180         return !(*this == b);  
181     }  
182     bool operator <= (const bign &b)  
183     {  
184         return *this < b || *this == b;  
185     }  
186     bool operator >= (const bign &b)  
187     {  
188         return *this > b || *this == b;  
189     }  
190     string str() const  //将结果转化为字符串（用于输出）
191     {  
192         string res = "";  
193         for(int i = 0; i < len; i++) res = char(s[i]+‘0‘)+res;  
194         return res;  
195     }  
196 };  
197 
198 istream& operator >> (istream &in, bign &x) //重载输入流 
199 {  
200     string s;  
201     in >> s;  
202     x = s.c_str();  //string转化为char[]
203     return in;  
204 }  
205 
206 ostream& operator << (ostream &out, const bign &x)  //重载输出流
207 {  
208     out << x.str();  
209     return out;  
210 }
211 
212 int main()
213 {
214     bign a,b;
215     
216     cin>>a>>b;
217     cout<<a*b<<endl;
218     
219     return 0; 
220 }