大数模板

acm竞赛版

#include<iostream>
#include<string>
#include<iomanip>
#include<algorithm>
using namespace std;

#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum
{
private:
    int a[500];    //可以控制大数的位数
    int len;       //大数长度
public:
    BigNum()
    {
        len = 1;    //构造函数
        memset(a,0,sizeof(a));
    }
    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;
    else
        t.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];
        }
        else
            t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while(t1.a[t1.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;
        else
            return false;
    }
    else
        return 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(void)
{
    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)==1 && n!=-1)
    {
        x[n].print();
    }
}

高精度版

const  int MAXL = 500;      
struct  BigNum      
{      
    int  num[MAXL];      
    int  len;      
};      
     
//高精度比较 a > b return 1, a == b return 0; a < b return -1;      
int  Comp(BigNum &a, BigNum &b)      
{      
    int  i;      
    if(a.len != b.len) return (a.len > b.len) ? 1 : -1;      
    for(i = a.len-1; i >= 0; i--)      
        if(a.num[i] != b.num[i]) return  (a.num[i] > b.num[i]) ? 1 : -1;      
    return  0;      
}      
     
//高精度加法      
BigNum  Add(BigNum &a, BigNum &b)      
{      
    BigNum c;      
    int  i, len;      
    len = (a.len > b.len) ? a.len : b.len;      
    memset(c.num, 0, sizeof(c.num));      
    for(i = 0; i < len; i++)      
    {      
        c.num[i] += (a.num[i]+b.num[i]);      
        if(c.num[i] >= 10)      
        {      
            c.num[i+1]++;      
            c.num[i] -= 10;      
        }      
    }      
    if(c.num[len])
		len++;      
    c.len = len;      
    return  c;      
}      
//高精度减法,保证a >= b      
BigNum Sub(BigNum &a, BigNum &b)      
{      
    BigNum  c;      
    int  i, len;      
    len = (a.len > b.len) ? a.len : b.len;      
    memset(c.num, 0, sizeof(c.num));      
    for(i = 0; i < len; i++)      
    {      
        c.num[i] += (a.num[i]-b.num[i]);      
        if(c.num[i] < 0)      
        {      
            c.num[i] += 10;      
            c.num[i+1]--;      
        }      
    }      
    while(c.num[len] == 0 && len > 1)
		len--;      
    c.len = len;      
    return  c;      
}      
//高精度乘以低精度，当b很大时可能会发生溢出int范围,具体情况具体分析      
//如果b很大可以考虑把b看成高精度      
BigNum Mul1(BigNum &a, int  &b)      
{      
    BigNum c;      
    int  i, len;      
    len = a.len;      
    memset(c.num, 0, sizeof(c.num));      
    //乘以0，直接返回0      
    if(b == 0)       
    {      
        c.len = 1;      
        return  c;      
    }      
    for(i = 0; i < len; i++)      
    {      
        c.num[i] += (a.num[i]*b);      
        if(c.num[i] >= 10)      
        {      
            c.num[i+1] = c.num[i]/10;      
            c.num[i] %= 10;      
        }      
    }      
    while(c.num[len] > 0)      
    {      
        c.num[len+1] = c.num[len]/10;      
        c.num[len++] %= 10;      
    }      
    c.len = len;       
    return  c;      
}      
     
//高精度乘以高精度,注意要及时进位，否则肯能会引起溢出,但这样会增加算法的复杂度，      
//如果确定不会发生溢出, 可以将里面的while改成if      
BigNum  Mul2(BigNum &a, BigNum &b)      
{      
    int i, j, len = 0;      
    BigNum  c;      
    memset(c.num, 0, sizeof(c.num));      
    for(i = 0; i < a.len; i++)
	{
        for(j = 0; j < b.len; j++)      
        {      
            c.num[i+j] += (a.num[i]*b.num[j]);      
            if(c.num[i+j] >= 10)      
            {      
                c.num[i+j+1] += c.num[i+j]/10;      
                c.num[i+j] %= 10;      
            }      
        }
	}
    len = a.len+b.len-1;      
    while(c.num[len-1] == 0 && len > 1)
		len--;      
    if(c.num[len])
		len++;      
    c.len = len;      
    return  c;      
}      
     
//高精度除以低精度,除的结果为c, 余数为f      
void Div1(BigNum &a, int &b, BigNum &c, int &f)      
{      
    int  i, len = a.len;      
    memset(c.num, 0, sizeof(c.num));      
    f = 0;      
    for(i = a.len-1; i >= 0; i--)      
    {      
        f = f*10+a.num[i];      
        c.num[i] = f/b;      
        f %= b;      
    }      
    while(len > 1 && c.num[len-1] == 0)
		len--;      
    c.len = len;      
}      
//高精度*10      
void  Mul10(BigNum &a)      
{      
    int  i, len = a.len;      
    for(i = len; i >= 1; i--)      
        a.num[i] = a.num[i-1];      
    a.num[i] = 0;      
    len++;      
    //if a == 0      
    while(len > 1 && a.num[len-1] == 0)
		len--;      
}      
     
//高精度除以高精度，除的结果为c，余数为f      
void Div2(BigNum &a, BigNum &b, BigNum &c, BigNum &f)      
{      
    int  i, len = a.len;      
    memset(c.num, 0, sizeof(c.num));      
    memset(f.num, 0, sizeof(f.num));      
    f.len = 1;      
    for(i = len-1;i >= 0;i--)      
    {      
        Mul10(f);      
        //余数每次乘10      
        f.num[0] = a.num[i];      
        //然后余数加上下一位      
        ///利用减法替换除法      
        while(Comp(f, b) >= 0)      
        {
            f = Sub(f, b);      
            c.num[i]++;      
        }      
    }      
    while(len > 1 && c.num[len-1] == 0)
		len--;      
    c.len = len;      
}   
void  print(BigNum &a)   //输出大数   
{      
    int  i;      
    for(i = a.len-1; i >= 0; i--)      
        printf("%d", a.num[i]);      
    puts("");      
}      
//将字符串转为大数存在BigNum结构体里面      
BigNum ToNum(char *s)      
{      
    int i, j;      
    BigNum  a;      
    a.len = strlen(s);      
    for(i = 0, j = a.len-1; s[i] != '\0'; i++, j--)      
        a.num[i] = s[j]-'0';      
    return  a;      
}      
     
void Init(BigNum &a, char *s, int &tag)   //将字符串转化为大数
{   
    int  i = 0, j = strlen(s); 
    if(s[0] == '-')
	{
		j--;
		i++;
		tag *= -1;
	}
    a.len = j;
    for(; s[i] != '\0'; i++, j--)
        a.num[j-1] = s[i]-'0';
}   
  
int main(void)      
{      
    BigNum a, b;   
    char  s1[100], s2[100];   
    while(scanf("%s %s", s1, s2) != EOF)   
    {   
        int tag = 1;   
        Init(a, s1, tag);    //将字符串转化为大数
        Init(b, s2, tag);   
        a = Mul2(a, b);   
        if(a.len == 1 && a.num[0] == 0)   
        {   
            puts("0");   
        }   
        else    
        {   
            if(tag < 0) putchar('-');   
            print(a);   
        }   
    }   
    return 0;   
}