完全高精度

Posted 2022-07-20 cadcadcad

from kuangbin

包含了 + - * / ^ % = > >> << == 运算符的重载，可用int char* BigNum调用构造函数

#define MAXN 9999
#define MAXSIZE 100000
#define DLEN 4
class BigNum

public:
    int a[MAXSIZE];
    int len;
public:
    BigNum()len = 1;memset(a,0,sizeof(a));
    BigNum(const int);
    BigNum(const char*);
    BigNum(const BigNum &);
    BigNum &operator=(const BigNum &);
    friend istream& operator>>(istream&,  BigNum&);
    friend ostream& operator<<(ostream&,  BigNum&);
    BigNum operator+(const BigNum &) const;
    BigNum operator-(const BigNum &) const;
    BigNum operator*(const BigNum &) const;
    BigNum operator/(const int  &) const;
    BigNum operator^(const int  &) const;
    int    operator%(const int  &) const;
    bool   operator>(const BigNum & T)const;
    bool   operator==(const BigNum & T)const;
    bool   operator==(const int & t)const;
    bool   operator>(const int & t)const;
;
istream& operator>>(istream & in,  BigNum & b)

    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)

    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(const int b)

    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)

    int t,k,index,l;
    memset(a,0,sizeof(a));
    l=strlen(s);
    len=l/DLEN;
    if (l%DLEN)len++;
    index=0;
    for (int 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)

    int i;
    memset(a,0,sizeof(a));
    for (i = 0 ; i < len ; i++)  a[i] = T.a[i];

BigNum & BigNum::operator=(const BigNum & n)

    len = n.len;
    memset(a,0,sizeof(a));
    for (int i = 0 ; i < len ; i++)
        a[i] = n.a[i];
    return *this;

BigNum BigNum::operator+(const BigNum & T) const

    BigNum t(*this);
    int i,big;
    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

    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[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

    BigNum ret;
    int i,j,up,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

    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

    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

    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

    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--;
        return ln>=0 && a[ln]>T.a[ln];
    
    else return false;


bool BigNum::operator==(const BigNum & T) const

    int ln;
    if (len != T.len) return false;
    else
    
        ln = len - 1;
        while (a[ln] == T.a[ln] && ln-- );
        return ln<0;
    


bool BigNum::operator >(const int & t) const

    BigNum b(t);
    return *this>b;


bool BigNum::operator==(const int & t) const

    BigNum b(t);
    return *this==b;