高精度模板(含大数开平方)

/*    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;}