高精度模板

大数模板（进制优化）：

/*
    高精度模版
*/

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

const int numlen = 2005; // 需要的位数
const int numbit = 4;   // 数组一位表示的整数
const int addbit = 10000;//进位数
const int maxn = numlen/numbit + 10;   // 数组要开的位数

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) {
        int clen = strlen(num);
        while(clen > 1 && num[0] == '0') num++, clen--;
        len = 0;
        for(int i = clen-1;i >= 0;i -= numbit) {
            int top = min(numbit, i+1), mul = 1;
            s[len] = 0;
            for(int j = 0;j < top; j++) {
                s[len] += (num[i-j]-'0')*mul;
                mul *= 10;
            }
            len++;
        }
        deal();
        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%addbit;
            add = now/addbit;
        }
        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 += addbit;
            }
            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++) {
            int pre = 0;
            for(int j = 0;j < a.len; j++) {
                int now = s[i]*a.s[j] + pre;
                pre = 0;
                ret.s[i+j] += now;
                if(ret.s[i+j] >= addbit) {
                    pre = ret.s[i+j]/addbit;
                    ret.s[i+j] -= pre*addbit;
                }
            }
            if(pre) ret.s[i+a.len] = pre;
        }
        ret.deal();
        return ret;
    }

    //乘以小数，直接乘快点  ***********注意计算过程可能会爆int
    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%addbit;
            bb = now/addbit;
        }
        while(bb) {
            ret.s[ret.len++] = bb % addbit;
            bb /= addbit;
        }
        ret.deal();
        return ret;
    }
    // 除以一个小整数     ***********注意计算过程可能会爆int
    bign operator / (const int a) const {
        bign ret;
        ret.len = len;
        int pre = 0;
        for(int i = len-1;i >= 0; i--) {
            ret.s[i] = (s[i] + pre*addbit)/a;
            pre = s[i] + pre*addbit - a*ret.s[i];
        }
        ret.deal();
        return ret;
    }


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

    void print() {
        printf("%d", s[len-1]);
        for(int i = len-2;i >= 0; i--) {
            printf("%04d", s[i]);
        }
        puts("");
    }

    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) {
    printf("%d", x.s[x.len-1]);
    for(int i = x.len-2;i >= 0; i--)    printf("%04d", x.s[i]);
    return out;
}

bign a, b;//a, b是两个大数

int main(){
    while(1){
        cin >> a >> b;
        cout << a+b << endl;
        cout << a-b << endl;
        cout << a*b << endl;
    }
    return 0;
}

大数模板（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;
}
 bign a, b;

 int main(){
     while(1){
         cin >> a >> b;
         cout << a + b << endl;
         cout << a - b << endl;
         cout << a * b << endl;
         cout << Sqrt(a) << endl;
     }
     return 0;
 }