A+B problem最简单不会报错题解

废话不多说，直接上代码，代码不难，就是A+B，但是避免意外用个高精度，在避免出错的同时装个B，浅浅的检查亿遍
#include <iostream>
#include <string>
#include <vector>
#include <stdexcept>
#include <algorithm>
#include <cctype>
using namespace std;

class BN {
private:
vector<int> d;
bool neg;

void tlz() {
    while (d.size() > 1 && d.back() == 0) d.pop_back();
    if (d.size() == 1 && d[0] == 0) neg = false;
}

public:
BN(const string& s) {
if (s.empty()) throw invalid_argument("");
neg = (s[0] == '-');
d.clear();

    for (int i = s.size() - 1; i >= (neg ? 1 : 0); --i) {
        if (!isdigit(s[i])) throw invalid_argument("");
        d.push_back(s[i] - '0');
    }
    
    tlz();
}

BN operator+(const BN& o) const {
    if (!neg && !o.neg) return add(o);
    else if (neg && o.neg) { BN r = add(o); r.neg = true; return r; }
    else if (neg && !o.neg) { BN a = *this; a.neg = false; return o.sub(a); }
    else { BN a = o; a.neg = false; return sub(a); }
}

BN add(const BN& o) const {
    BN r = *this;
    int c = 0;
    
    for (size_t i = 0; i < max(r.d.size(), o.d.size()) || c; ++i) {
        if (i == r.d.size()) r.d.push_back(0);
        r.d[i] += c + (i < o.d.size() ? o.d[i] : 0);
        c = r.d[i] / 10;
        r.d[i] %= 10;
    }
    
    return r;
}

BN sub(const BN& o) const {
    if (cmp(o) < 0) { BN r = o.sub(*this); r.neg = true; return r; }
    
    BN r = *this;
    int b = 0;
    
    for (size_t i = 0; i < o.d.size() || b; ++i) {
        r.d[i] -= b + (i < o.d.size() ? o.d[i] : 0);
        b = r.d[i] < 0;
        if (b) r.d[i] += 10;
    }
    
    r.tlz();
    return r;
}

int cmp(const BN& o) const {
    if (d.size() != o.d.size()) return d.size() - o.d.size();
    for (int i = d.size() - 1; i >= 0; --i) {
        if (d[i] != o.d[i]) return d[i] - o.d[i];
    }
    return 0;
}

string ts() const {
    if (d.empty()) return "0";
    string r;
    if (neg) r += '-';
    for (int i = d.size() - 1; i >= 0; --i) r += d[i] + '0';
    return r;
}

};

pair<string, string> vi(const string& s) {
size_t p = s.find(' ');
if (p == string::npos) throw invalid_argument("");

string a = s.substr(0, p);
string b = s.substr(p + 1);

auto ivn = [](const string& s) {
    if (s.empty()) return false;
    size_t st = (s[0] == '-' || s[0] == '+') ? 1 : 0;
    if (st >= s.size()) return false;
    for (size_t i = st; i < s.size(); ++i) {
        if (!isdigit(s[i])) return false;
    }
    return true;
};

if (!ivn(a) || !ivn(b)) throw invalid_argument("");

return {a, b};

}

int main() {
try {
string l;
if (!getline(cin, l)) throw runtime_error("");

    auto [a, b] = vi(l);
    
    BN x(a);
    BN y(b);
    BN s = x + y;
    
    cout << s.ts() << endl;
} catch (const exception& e) {
    cerr << "Error" << endl;
    return 1;
}

return 0;

}