C++大数模板

转自http://blog.csdn.net/hackbuteer1/article/details/6595881

分别使用C++中的运算符重载的方法来实现大数之间的数学运算，包括加法、减法、乘法、除法、n次方、取模、大小比较、赋值以及输入流、输出流的重载。。

并且使用这个大数模板，顺利AC了HDOJ上的1134这个题目的Catalan数计数问题。。

http://acm.hdu.edu.cn/showproblem.php?pid=1134

大数模板的代码如下：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;#define MAXN 9999#define MAXSIZE 10#define DLEN 4class BigNum{private:int a[500];    //可以控制大数的位数int len;       //大数长度public:BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数BigNum(const int);       //将一个int类型的变量转化为大数BigNum(const char*);     //将一个字符串类型的变量转化为大数BigNum(const BigNum &);  //拷贝构造函数BigNum &operator=(const BigNum &);   //重载赋值运算符，大数之间进行赋值运算friend istream& operator>>(istream&,  BigNum&);   //重载输入运算符friend ostream& operator<<(ostream&,  BigNum&);   //重载输出运算符BigNum operator+(const BigNum &) const;   //重载加法运算符，两个大数之间的相加运算BigNum operator-(const BigNum &) const;   //重载减法运算符，两个大数之间的相减运算BigNum operator*(const BigNum &) const;   //重载乘法运算符，两个大数之间的相乘运算BigNum operator/(const int   &) const;    //重载除法运算符，大数对一个整数进行相除运算BigNum operator^(const int  &) const;    //大数的n次方运算int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较void print();       //输出大数};BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数{int c,d = b;len = 0;memset(a,0,sizeof(a));while(d > MAXN){c = d - (d / (MAXN + 1)) * (MAXN + 1);d = d / (MAXN + 1);a[len++] = c;}a[len++] = d;}BigNum::BigNum(const char*s)     //将一个字符串类型的变量转化为大数{int t,k,index,l,i;memset(a,0,sizeof(a));l=strlen(s);len=l/DLEN;if(l%DLEN)len++;index=0;for(i=l-1;i>=0;i-=DLEN){t=0;k=i-DLEN+1;if(k<0)k=0;for(int j=k;j<=i;j++)t=t*10+s[j]-'0';a[index++]=t;}}BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数{int i;memset(a,0,sizeof(a));for(i = 0 ; i < len ; i++)a[i] = T.a[i];}BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符，大数之间进行赋值运算{int i;len = n.len;memset(a,0,sizeof(a));for(i = 0 ; i < len ; i++)a[i] = n.a[i];return *this;}istream& operator>>(istream & in,  BigNum & b)   //重载输入运算符{char ch[MAXSIZE*4];int i = -1;in>>ch;int l=strlen(ch);int count=0,sum=0;for(i=l-1;i>=0;){sum = 0;int t=1;for(int j=0;j<4&&i>=0;j++,i--,t*=10){sum+=(ch[i]-'0')*t;}b.a[count]=sum;count++;}b.len =count++;return in;}ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符{int i;cout << b.a[b.len - 1];for(i = b.len - 2 ; i >= 0 ; i--){cout.width(DLEN);cout.fill('0');cout << b.a[i];}return out;}BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算{BigNum t(*this);int i,big;      //位数big = T.len > len ? T.len : len;for(i = 0 ; i < big ; i++){t.a[i] +=T.a[i];if(t.a[i] > MAXN){t.a[i + 1]++;t.a[i] -=MAXN+1;}}if(t.a[big] != 0)t.len = big + 1;elset.len = big;return t;}BigNum BigNum::operator-(const BigNum & T) const   //两个大数之间的相减运算{int i,j,big;bool flag;BigNum t1,t2;if(*this>T){t1=*this;t2=T;flag=0;}else{t1=T;t2=*this;flag=1;}big=t1.len;for(i = 0 ; i < big ; i++){if(t1.a[i] < t2.a[i]){j = i + 1;while(t1.a[j] == 0)j++;t1.a[j--]--;while(j > i)t1.a[j--] += MAXN;t1.a[i] += MAXN + 1 - t2.a[i];}elset1.a[i] -= t2.a[i];}t1.len = big;while(t1.a[len - 1] == 0 && t1.len > 1){t1.len--;big--;}if(flag)t1.a[big-1]=0-t1.a[big-1];return t1;}BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算{BigNum ret;int i,j,up;int temp,temp1;for(i = 0 ; i < len ; i++){up = 0;for(j = 0 ; j < T.len ; j++){temp = a[i] * T.a[j] + ret.a[i + j] + up;if(temp > MAXN){temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);up = temp / (MAXN + 1);ret.a[i + j] = temp1;}else{up = 0;ret.a[i + j] = temp;}}if(up != 0)ret.a[i + j] = up;}ret.len = i + j;while(ret.a[ret.len - 1] == 0 && ret.len > 1)ret.len--;return ret;}BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算{BigNum ret;int i,down = 0;for(i = len - 1 ; i >= 0 ; i--){ret.a[i] = (a[i] + down * (MAXN + 1)) / b;down = a[i] + down * (MAXN + 1) - ret.a[i] * b;}ret.len = len;while(ret.a[ret.len - 1] == 0 && ret.len > 1)ret.len--;return ret;}int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算{int i,d=0;for (i = len-1; i>=0; i--){d = ((d * (MAXN+1))% b + a[i])% b;}return d;}BigNum BigNum::operator^(const int & n) const    //大数的n次方运算{BigNum t,ret(1);int i;if(n<0)exit(-1);if(n==0)return 1;if(n==1)return *this;int m=n;while(m>1){t=*this;for( i=1;i<<1<=m;i<<=1){t=t*t;}m-=i;ret=ret*t;if(m==1)ret=ret*(*this);}return ret;}bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较{int ln;if(len > T.len)return true;else if(len == T.len){ln = len - 1;while(a[ln] == T.a[ln] && ln >= 0)ln--;if(ln >= 0 && a[ln] > T.a[ln])return true;elsereturn false;}elsereturn false;}bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较{BigNum b(t);return *this>b;}void BigNum::print()    //输出大数{int i;cout << a[len - 1];for(i = len - 2 ; i >= 0 ; i--){cout.width(DLEN);cout.fill('0');cout << a[i];}cout << endl;}int main(){int i,n;BigNum x[101];      //定义大数的对象数组x[0]=1;for(i=1;i<101;i++)x[i]=x[i-1]*(4*i-2)/(i+1);while(~scanf("%d",&n) && n!=-1){x[n].print();}return 0;}