高精度模板

来自杰哥的模板

/*

    高精度模版 含大数开平方

*/



#include <stdio.h>

#include <math.h>

#include <string.h>

#include <iostream>

#include <string>

#include <algorithm>

using namespace std;



const int numlen = 105; // 位数



int max(int a, int b) { return a>b?a:b; }

struct bign {

    int len, s[numlen];

    bign() {

        memset(s, 0, sizeof(s));

        len = 1;

    }

    bign(int num) { *this = num; }

    bign(const char *num) { *this = num; }

    bign operator = (const int num) {

        char s[numlen];

        sprintf(s, "%d", num);

        *this = s;

        return *this;

    }

    bign operator = (const char *num) {

        len = strlen(num);

        while(len > 1 && num[0] == '0') num++, len--;

        for(int i = 0;i < len; i++) s[i] = num[len-i-1] - '0';

        return *this;

    }



    void deal() {

        while(len > 1 && !s[len-1]) len--;

    }



    bign operator + (const bign &a) const {

        bign ret;

        ret.len = 0;

        int top = max(len, a.len) , add = 0;

        for(int i = 0;add || i < top; i++) {

            int now = add;

            if(i < len) now += s[i];

            if(i < a.len)   now += a.s[i];

            ret.s[ret.len++] = now%10;

            add = now/10;

        }

        return ret;

    }

    bign operator - (const bign &a) const {

        bign ret;

        ret.len = 0;

        int cal = 0;

        for(int i = 0;i < len; i++) {

            int now = s[i] - cal;

            if(i < a.len)   now -= a.s[i];

            if(now >= 0)    cal = 0;

            else {

                cal = 1; now += 10;

            }

            ret.s[ret.len++] = now;

        }

        ret.deal();

        return ret;

    }

    bign operator * (const bign &a) const {

        bign ret;

        ret.len = len + a.len;

        for(int i = 0;i < len; i++) {

            for(int j = 0;j < a.len; j++)

                ret.s[i+j] += s[i]*a.s[j];

        }

        for(int i = 0;i < ret.len; i++) {

            ret.s[i+1] += ret.s[i]/10;

            ret.s[i] %= 10;

        }

        ret.deal();

        return ret;

    }



    //乘以小数，直接乘快点

    bign operator * (const int num) {

        bign ret;

        ret.len = 0;

        int bb = 0;

        for(int i = 0;i < len; i++) {

            int now = bb + s[i]*num;

            ret.s[ret.len++] = now%10;

            bb = now/10;

        }

        while(bb) {

            ret.s[ret.len++] = bb % 10;

            bb /= 10;

        }

        ret.deal();

        return ret;

    }



    bign operator / (const bign &a) const {

        bign ret, cur = 0;

        ret.len = len;

        for(int i = len-1;i >= 0; i--) {

            cur = cur*10;

            cur.s[0] = s[i];

            while(cur >= a) {

                cur -= a;

                ret.s[i]++;

            }

        }

        ret.deal();

        return ret;

    }



    bign operator % (const bign &a) const {

        bign b = *this / a;

        return *this - b*a;

    }



    bign operator += (const bign &a) { *this = *this + a; return *this; }

    bign operator -= (const bign &a) { *this = *this - a; return *this; }

    bign operator *= (const bign &a) { *this = *this * a; return *this; }

    bign operator /= (const bign &a) { *this = *this / a; return *this; }

    bign operator %= (const bign &a) { *this = *this % a; return *this; }



    bool operator < (const bign &a) const {

        if(len != a.len)    return len < a.len;

        for(int i = len-1;i >= 0; i--) if(s[i] != a.s[i])

            return s[i] < a.s[i];

        return false;

    }

    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 !(*this < a); }

    bool operator == (const bign &a) const { return !(*this > a || *this < a); }

    bool operator != (const bign &a) const { return *this > a || *this < a; }



    string str() const {

        string ret = "";

        for(int i = 0;i < len; i++) ret = char(s[i] + '0') + ret;

        return ret;

    }

};

istream& operator >> (istream &in, bign &x) {

    string s;

    in >> s;

    x = s.c_str();

    return in;

}

ostream& operator << (ostream &out, const bign &x) {

    out << x.str();

    return out;

}

// 大数开平方

bign Sqrt(bign x) {

    int a[numlen/2];

    int top = 0;

    for(int i = 0;i < x.len; i += 2) {

        if(i == x.len-1) {

            a[top++] = x.s[i];

        }

        else

            a[top++] = x.s[i] + x.s[i+1]*10;

    }

    bign ret = (int)sqrt((double)a[top-1]);

    int xx = (int)sqrt((double)a[top-1]);

    bign pre = a[top-1] - xx*xx;

    bign cc;

    for(int i = top-2;i >= 0; i--) {

        pre = pre*100 + a[i];

        cc = ret*20;

        for(int j = 9;j >= 0; j--) {

            bign now = (cc + j)*j;

            if(now <= pre) {

                ret = ret*10 + j;

                pre -= now;

                break;

            }

        }

    }

    return ret;

}