高精度加、减、乘模板

今天早上写了一道高精度的题，一口气敲完了高精度加、减、乘。现在把模板放在这备用着 …… 需要者自取。

#include <cstring>
#include <iostream>
#include <vector>

struct BigInt {
    std::vector<char> v;

    BigInt() {
        *this = 0;
    }

    BigInt(int x) {
        *this = x;
    }

    BigInt &operator=(int x) {
        v.clear();
        do v.push_back(x % 10); while (x /= 10);
        return *this;
    }

    BigInt &operator=(const BigInt &x) {
        v.resize(x.v.size());
        memcpy(const_cast<char *>(v.data()), x.v.data(), 
x.v.size() * sizeof(char));
        return *this;
    }
};

std::ostream &operator<<(std::ostream &out, const BigInt &x) {
    for (int i = x.v.size() - 1; i >= 0; i--) out << (char)(x.v[i] 
+ '0');
    return out;
}

BigInt operator+(const BigInt &a, const BigInt &b) {
    BigInt result;
    result.v.clear();
    bool flag = false;
    for (int i = 0; i < (int)std::max(a.v.size(), b.v.size()); 
i++) {
        int tmp = 0;
        if (i < (int)a.v.size()) tmp += a.v[i];
        if (i < (int)b.v.size()) tmp += b.v[i];
        if (flag) tmp++, flag = false;
        if (tmp >= 10) tmp -= 10, flag = true;
        result.v.push_back(tmp);
    }
    if (flag) result.v.push_back(1);

    return result;
}

BigInt &operator+=(BigInt &a, const BigInt &b) {
    return a = a + b;
}

BigInt operator-(const BigInt &a, const BigInt &b) {
    BigInt result;
    result.v.clear();
    bool flag = false;
    for (int i = 0; i < (int)a.v.size(); i++) {
        int tmp = a.v[i];
        if (i < (int)b.v.size()) tmp -= b.v[i];
        if (flag) tmp--, flag = false;
        if (tmp < 0) tmp += 10, flag = true;
        result.v.push_back(tmp);
    }

    int size = result.v.size();
    while (size > 1 && result.v[size - 1] == 0) size--;
    result.v.resize(size);

    return result;
}

BigInt operator*(const BigInt &a, const BigInt &b) {
    BigInt result;
    result.v.resize(a.v.size() + b.v.size());
    for (int i = 0; i < (int)a.v.size(); i++) {
        for (int j = 0; j < (int)b.v.size(); j++){
            result.v[i + j] += a.v[i] * b.v[j];
            result.v[i + j + 1] += result.v[i + j] / 10;
            result.v[i + j] %= 10;
        }
    }

    int size = result.v.size();
    while (size > 1 && result.v[size - 1] == 0) size--;
    result.v.resize(size);

    return result;
}