BigInt 的使用！

今天学长讲的卡特兰数真的是卡的一批，整个全是高精的题，这时我就使用重载运算符，然后一下午就过去了

首先来看一波水题（也就卡了2小时）

.

A. 网格

内存限制：512 MiB 时间限制：1000 ms 标准输入输出

 

 

题目描述

原题来自：BZOJ 3907

某城市的街道呈网格状，左下角坐标为 A(0,0)A(0, 0)A(0,0)，右上角坐标为 B(n,m)B(n, m)B(n,m)，其中 n≥mn \ge mn≥m。现在从 A(0,0)A(0, 0)A(0,0) 点出发，只能沿着街道向正右方或者正上方行走，且不能经过图示中直线左上方的点，即任何途径的点 (x,y)(x, y)(x,y) 都要满足 x≥yx \ge yx≥y，请问在这些前提下，到达 B(n,m)B(n, m)B(n,m) 有多少种走法。

这道题是把对角线上方的对角线翻折，然后，自己一看就是C（n+m,n）-C(n+m,n-1),然后就是高精部分，我就不说了;

先看一下我一开始打的代码：

前方高能！

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<cstdlib>
 #include<ctime>
 using namespace std;
 #define maxn 10000
 #define base 10000
 //可能重载运算符码量稍大qwq,而且巨难调试
 struct Bigint
 {
     int c[maxn],len,sign;
     Bigint(){memset(c,,sizeof(c)),len=,sign=;}
     void Zero()
     {
         while(len>&&c[len]==)len--;
         if(len==&&c[len]==)sign=;
     }
     void Write(char *s)
     {
         int k=,l=strlen(s);
         for(int i=l-;i>=;i--)
         {
             c[len]+=(s[i]-'')*k;
             k*=;
             if(k==base)
                 k=,len++;
         }
     }
     void Read()
     {
         char s[maxn]={};
         scanf("%s",s);
         Write(s);
     }
     void Print()
     {
         if(sign)printf("-");
         printf("%d",c[len]);
         for(int i=len-;i>=;i--)printf("%d",c[i]);
         printf("\n");
     }
     bool operator < (const Bigint &a)const
     {
         if(len!=a.len)return len<a.len;
         for(int i=len;i>=;i--)
             if(c[i]!=a.c[i])return c[i]<a.c[i];
         return ;
     }
     bool operator > (const Bigint &a)const
     {
         return a<*this;
     }
     Bigint operator = (int a)
     {
         char s[];
         sprintf(s,"%d",a);      //int ->string
         Write(s);
         return *this;
     }
     Bigint operator + (const Bigint &a)
     {
         Bigint r;
         r.len=max(len,a.len)+;
         for(int i=;i<=r.len;i++)
         {
             r.c[i]+=c[i]+a.c[i];
             r.c[i+]+=r.c[i]/base;
             r.c[i]%=base;
         }
         r.Zero();
         return r;
     }
     Bigint operator + (const int &a)
     {
         Bigint b;b=a;
         return *this+b;
     }
     Bigint operator - (const Bigint &a)
     {
         Bigint b,c;
         b=*this;
         c=a;
         if(c>b)
         {
             swap(b,c);
             b.sign=;
         }
         for(int i=;i<=b.len;i++)
         {
             b.c[i]=b.c[i]-c.c[i];
             if(b.c[i]<)
             {
                 b.c[i]+=base;
                 b.c[i+]--;
             }
         }
         b.Zero();
         return b;
     }
     Bigint operator - (const int &a)
     {
         Bigint b;b=a;return *this-b;
     }
     Bigint operator * (const Bigint &a)
     {
         Bigint r;
         r.len=len+a.len+;
         for(int i=;i<=len;i++)
         {
             for(int j=;j<=a.len;j++)
                 r.c[i+j-]+=c[i]*a.c[j];
         }
         for(int i=;i<=r.len;i++)
         {
             r.c[i+]+=r.c[i]/base;
             r.c[i]%=base;
         }
         r.Zero();
         return r;
     }
     Bigint operator * (const int &a)
     {
         Bigint b;b=a;
         return *this*b;
     }
     Bigint operator / (const Bigint &b)
     {
         Bigint r,t,a;
         a=b;
         r.len=len;
         for(int i=len;i>=;i--)
         {
             t=t*base+c[i];
             int div,ll=,rr=base;
             while(ll<=rr)
             {
                 int mid=(ll+rr)/;
                 Bigint k=a*mid;
                 if(k>t)rr=mid-;
                 else
                 {
                     ll=mid+;
                     div=mid;
                 }
             }
             r.c[i]=div;
             t=t-a*div;
         }
         r.Zero();
         return r;
     }
     Bigint operator / (const int &a)
     {
         Bigint b;b=a;
         return *this/b;
     }
 };
 int main()
 {
     //freopen("cd.txt","r",stdin);
     Bigint a,b,c;
     int n,m;
     scanf("%d%d",&n,&m);
     a=,b=;
     for(int i=m+;i<=n+m;i++)
         a=a*i;
     if(m+-n>)
     a=a*(m+-n);
     for(int i=;i<=n;i++)
         b=b*i;
     c=a/b;
     c.Print();
     //cout<<clock()<<endl;
     return ;
 } 

高能代码

然后，我就直接站一下，我今天一下午顺便集齐的BigInt操作：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 #define maxn 2000
 #define base 10000

 struct Bign
 {
     int c[maxn],len,sign;
     //初始化
     Bign(){memset(c,,sizeof(c)),len = ,sign = ;}
     //高位清零
     void Zero()
     {
         while(len >  && c[len] == )len--;
         if(len ==  && c[len] == )sign = ;
     }
     //压位读入
     void Write(char *s)
     {
         int k = ,l = strlen(s);
         for(int i = l - ;i >= ;i--)
         {
             c[len] += (s[i] - '') * k;
             k *= ;
             if(k == base)
             {
                 k = ;
                 len++;
             }
         }
     }
     void Read()
     {
         char s[maxn] = {};
         scanf("%s",s);
         Write(s);
     }
     //输出
     void Print()
     {
         if(sign)printf("-");
         printf("%d",c[len]);
         for(int i = len - ;i >= ;i--)printf("%04d",c[i]);
         printf("\n");
     }
     //重载 = 运算符，将低精赋值给高精
     Bign operator = (int a)
     {
         char s[];
         sprintf(s,"%d",a);
         Write(s);
         return *this;//this只能用于成员函数，表示当前对象的地址
     }
     //重载 < 运算符
     bool operator < (const Bign &a)const
     {
         if(len != a.len)return len < a.len;
         for(int i = len;i >= ;i--)
         {
             if(c[i] != a.c[i])return c[i] < a.c[i];
         }
         return ;
     }
     bool operator > (const Bign &a)const
     {
         return a < *this;
     }
     bool operator <= (const Bign &a)const
     {
         return !(a < *this);
     }
     bool operator >= (const Bign &a)const
     {
         return !(*this < a);
     }
     bool operator != (const Bign &a)const
     {
         return a < *this || *this < a;
     }
     bool operator == (const Bign &a)const
     {
         return !(a < *this) && !(*this < a);
     }
     bool operator == (const int &a)const
     {
         Bign b;b = a;
         return *this == b;
     }

     //重载 + 运算符
     Bign operator + (const Bign &a)
     {
         Bign r;
         r.len = max(len,a.len) + ;
         for(int i = ;i <= r.len;i++)
         {
             r.c[i] += c[i] + a.c[i];
             r.c[i + ] += r.c[i] / base;
             r.c[i] %= base;
         }
         r.Zero();
         return r;
     }
     Bign operator + (const int &a)
     {
         Bign b;b = a;
         return *this + b;
     }
     //重载 - 运算符
     Bign operator - (const Bign &a)
     {
         Bign b,c;// b - c
         b = *this;
         c = a;
         if(c > b)
         {
             swap(b,c);
             b.sign = ;
         }
         for(int i = ;i <= b.len;i++)
         {
             b.c[i] -= c.c[i];
             if(b.c[i] < )
             {
                 b.c[i] += base;
                 b.c[i + ]--;
             }
         }
         b.Zero();
         return b;
     }
     Bign operator - (const int &a)
     {
         Bign b;b = a;
         return *this - b;
     }
     //重载 * 运算符
     Bign operator * (const Bign &a)
     {
         Bign r;
         r.len = len + a.len + ;
         for(int i = ;i <= len;i++)
         {
             for(int j = ;j <= a.len;j++)
             {
                 r.c[i + j - ] += c[i] * a.c[j];
             }
         }
         for(int i = ;i <= r.len;i++)
         {
             r.c[i + ] += r.c[i] / base;
             r.c[i] %= base;
         }
         r.Zero();
         return r;
     }
     Bign operator * (const int &a)
     {
         Bign b;b = a;
         return *this * b;
     }
     //重载 / 运算符
     Bign operator / (const Bign &b)
     {
         Bign r,t,a;
         a = b;
         //if(a == 0)return r;
         r.len = len;
         for(int i = len;i >= ;i--)
         {
             t = t * base + c[i];
             int div,ll = ,rr = base;
             while(ll <= rr)
             {
                 int mid = (ll + rr) / ;
                 Bign k = a * mid;
                 if(k > t)rr = mid - ;
                 else
                 {
                     ll = mid + ;
                     div = mid;
                 }
             }
             r.c[i] = div;
             t = t - a * div;
         }
         r.Zero();
         return r;
     }
     Bign operator / (const int &a)
     {
         Bign b;b = a;
         return *this / b;
     }
     //重载 % 运算符
     Bign operator % (const Bign &a)
     {
         return *this - *this / a * a;
     }
     Bign operator % (const int &a)
     {
         Bign b;b = a;
         return *this % b;
     }
 };

 int main()
 {
     freopen("Bign.in","r",stdin);
     freopen("Bign.out","w",stdout);
     Bign a,b,c,d,e,f,g;
     a.Read();
     b.Read();
     c = a + b;
     c.Print();
     d = a - b;
     d.Print();
     e = a * b;
     e.Print();
     f = a / b;
     f.Print();
     g = a % b;
     g.Print();
     return ;
 } 

BigInt

然后第二题也是卡特兰数的水题，但是.............我又卡了2个小时

重载运算符，确实很好用，但是不能拿重载之前的低精的复杂度来考虑，稍有不慎，重载运算符的复杂度就能高的惊人！