Data-Structure/模板/其他/高精度.cpp

#include <iostream>
#include <vector>
#include <string>
#include <iomanip>
#include <algorithm>
using namespace std;

/*
 * 高精度整数 BigInteger（正数）
 * 存储方式：vector<int> d，低位在前（little-endian）
 * 每位存储 BASE = 10^4 = 10000 的一个“数字块”，即四位十进制
 * 这样可以大幅提升运算效率
 */

struct BigInteger {
    static const int BASE = 10000;    // 一个“数字块”的进制
    static const int WIDTH = 4;       // 每个块的十进制位数（用于输出补零）

    vector<int> d;   // 数字块存储，低位在 d[0]

    /* -------- 构造函数部分 -------- */

    BigInteger(long long num = 0) { *this = num; }

    BigInteger& operator = (long long num) {
        d.clear();
        do {
            d.push_back(num % BASE);
            num /= BASE;
        } while (num > 0);
        return *this;
    }

    BigInteger(const string& s) { *this = s; }

    BigInteger& operator = (const string& s) {
        d.clear();
        int len = s.size();
        // 每 WIDTH 位作为一个数字块从后往前解析
        for (int i = len; i > 0; i -= WIDTH) {
            int x = 0;
            int l = max(0, i - WIDTH);
            for (int j = l; j < i; j++) x = x * 10 + (s[j] - '0');
            d.push_back(x);
        }
        trim();
        return *this;
    }

    /* -------- 删除高位的前导零 -------- */
    void trim() {
        while (d.size() > 1 && d.back() == 0) d.pop_back();
    }

    /* -------- 比较运算符 -------- */

    bool operator < (const BigInteger& b) const {
        if (d.size() != b.d.size()) return d.size() < b.d.size();
        for (int i = d.size() - 1; i >= 0; i--) {
            if (d[i] != b.d[i]) return d[i] < b.d[i];
        }
        return false; // 相等则不小于
    }

    bool operator > (const BigInteger& b) const { return b < *this; }
    bool operator <= (const BigInteger& b) const { return !(b < *this); }
    bool operator >= (const BigInteger& b) const { return !(*this < b); }
    bool operator == (const BigInteger& b) const { return d == b.d; }
    bool operator != (const BigInteger& b) const { return !(*this == b); }

    /* -------- 高精度加法 -------- */

    BigInteger operator + (const BigInteger& b) const {
        BigInteger c;
        c.d.clear();
        int carry = 0;
        for (size_t i = 0; i < d.size() || i < b.d.size() || carry; i++) {
            int x = carry;
            if (i < d.size()) x += d[i];
            if (i < b.d.size()) x += b.d[i];
            c.d.push_back(x % BASE);
            carry = x / BASE;
        }
        return c;
    }

    /* -------- 高精度减法（默认 *this >= b） -------- */

    BigInteger operator - (const BigInteger& b) const {
        BigInteger c;
        c.d.clear();
        int borrow = 0;
        for (size_t i = 0; i < d.size(); i++) {
            int x = d[i] - borrow - (i < b.d.size() ? b.d[i] : 0);
            if (x < 0) { x += BASE; borrow = 1; }
            else borrow = 0;
            c.d.push_back(x);
        }
        c.trim();
        return c;
    }

    /* -------- 高精度 * int -------- */

    BigInteger operator * (int m) const {
        BigInteger c;
        long long carry = 0;
        for (size_t i = 0; i < d.size() || carry; i++) {
            long long x = carry + (long long)(i < d.size() ? d[i] : 0) * m;
            c.d.push_back(x % BASE);
            carry = x / BASE;
        }
        c.trim();
        return c;
    }

    /* -------- 高精度乘法 O(n^2) -------- */

    BigInteger operator * (const BigInteger& b) const {
        BigInteger c;
        c.d.assign(d.size() + b.d.size(), 0);

        for (size_t i = 0; i < d.size(); i++) {
            long long carry = 0;
            for (size_t j = 0; j < b.d.size() || carry; j++) {
                long long cur = c.d[i+j] +
                                (long long)d[i] * (j < b.d.size() ? b.d[j] : 0) +
                                carry;
                c.d[i+j] = cur % BASE;
                carry = cur / BASE;
            }
        }
        c.trim();
        return c;
    }

    /* -------- 高精度 / int，返回商，并把余数存入 r -------- */

    BigInteger operator / (int m) const {
        BigInteger c;
        c.d.resize(d.size());
        long long r = 0;

        for (int i = d.size() - 1; i >= 0; i--) {
            long long x = d[i] + r * BASE;
            c.d[i] = x / m;
            r = x % m;
        }
        c.trim();
        return c;
    }

    int operator % (int m) const {
        long long r = 0;
        for (int i = d.size() - 1; i >= 0; i--) {
            long long x = d[i] + r * BASE;
            r = x % m;
        }
        return r;
    }

    /* -------- 输入输出重载 -------- */

    friend ostream& operator << (ostream& os, const BigInteger& x) {
        os << x.d.back();
        for (int i = x.d.size() - 2; i >= 0; i--) {
            os << setw(WIDTH) << setfill('0') << x.d[i];
        }
        return os;
    }

    friend istream& operator >> (istream& is, BigInteger& x) {
        string s;
        is >> s;
        x = s;
        return is;
    }
};