关于高精度

　　　　　　　　　　　　　　　　　关于高精度

虽然我接触过的大多数模拟赛或者考试一般都是int类型或者是long long类型，但是高精度还是很重要的，整理一下也方便些，以便不时之需。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 410;
struct bign {
    int len, s[MAXN];
    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 t[MAXN];
        sprintf(t, "%d", num);
        *this = t;
        return *this;
    }
    bign operator = (const char *num) {
        for(int i = 0; num[i] == '0'; num++);
        len = strlen(num);
        for(int i = 0; i < len; i++) s[i] = num[len - i - 1] - '0';
        return *this;
    }
    bign operator + (const bign &b) {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; g || i < max(len, b.len); i++) {
            int x = g;
            if(i < len) x += s[i];
            if(i < b.len) x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;  
        }
        return c;
    }
    bign operator += (const bign &b) {
        *this = *this + b;
        return *this;
    }
    void clean() {
        while(len > 1 && !s[len - 1]) len--; 
    }
    bign operator * (const bign &b) {
        bign c;
        c.len = len + b.len;
        for(int i = 0; i < len; i++)
            for(int j = 0; j < b.len; j++)
                c.s[i + j] += s[i] * b.s[j];
        for(int i = 0; i < c.len; i++) {
            c.s[i + 1] = c.s[i] / 10;
            c.s[i] %= 10;
        }
        c.clean();
        return c;
    }
    bign operator *= (const bign &b) {
        *this = *this * b;
        return *this;
    }
    bign operator - (const bign &b) {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; i < len; i++) {
            int x = s[i] - g;
            if(i < b.len) x -= b.s[i];
            if(x >= 0) g = 0;
            else {
                g = 1;
                x += 10;
            }
            c.s[c.len++] = x;
        }
        c.clean();
        return c;
    }
    bign operator -= (const bign &b) {
        *this = *this - b;
        return *this;
    }
    bign operator / (const bign &b) {
        bign c, f = 0;
        for(int i = len - 1; i >= 0; i--) {
            f = f * 10;
            f.s[0] = s[i];
            while(f >= b) {
                f -= b;
                c.s[i]++;
            }
        }
        c.len = len;
        c.clean();
        return c;
    }
    bign operator /= (const bign &b) {
        *this = *this / b;
        return *this;
    }
    bign operator % (const bign &b) {
        bign r = *this / b;
        r = *this - r * b;
        return r;
    }
    bign operator %= (const bign &b) {
        *this = *this % b;
        return *this;
    }
    bool operator < (const bign &b) {
        if(len != b.len) return len < b.len;
        for(int i = len - 1; i >= 0; i--)
            if(s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }
    bool operator > (const bign &b) {
        if(len != b.len) return len > b.len;
        for(int i = len - 1; i >= 0; i--)
            if(s[i] != b.s[i]) return s[i] > b.s[i];
        return false;
    }
    bool operator == (const bign &b) {
        return !(*this > b) && !(*this < b);
    }
    bool operator >= (const bign &b) {
        return *this > b || *this == b;
    }
    bool operator <= (const bign &b) {
        return *this < b || *this == b;
    }
    bool operator != (const bign &b) {
        return *this > b || *this < b;
    }
    string str() const {
        string ret = "";
        for(int i = len - 1; i >= 0; i--) ret += char(s[i] + '0');
        return ret;
    }
};
istream& operator >> (istream &in, bign &x) {
    string s;
    in >> s;
    x = s.c_str();
    return in;
}
ostream& operator << (ostream &out, bign &x) {
    out << x.str();
    return out;
}
int main() {
    bign a, b, c, d;//可加
    cin >> a >> b >> d;//可改
    c = a - b + d;//可改
    cout << c;
    return 0;
}

哪怕遇到你的的可能是十三亿分之一，但我仍旧不愿放弃追寻你的足迹。