求助大佬，C++，0分，2WA 2RE

用的自写的字符串加法和乘法

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

namespace str {
[[nodiscard]] bool isAllZeroString(const string &str) {
    bool flag = true;
    for (auto &ch : str) {
        flag &= (ch == '0');
        if (!flag) {
            break;
        }
    }
    return flag;
}

void operator+=(string &str0, string str1) {
    ll sz0 = str0.size(), sz1 = str1.size();
    for (ll i = sz0; i < sz1; ++i) {
        str0.insert(str0.begin(), str1[0]);
        str1.erase(str1.begin());
    }
    if (sz0 < sz1) {
        reverse(str0.begin(), str0.begin() + sz1 - sz0);
    }
    if (str::isAllZeroString(str1)) {
        return;
    }
    for (auto str0Iter = str0.rbegin(), str1Iter = str1.rbegin();
         str0Iter != str0.rend() && str1Iter != str1.rend();
         ++str0Iter, ++str1Iter) {
        ll sum = (*str0Iter ^ 48) + (*str1Iter ^ 48);
        if (sum < 10) {
            *str0Iter = char(sum ^ 48);
        } else {
            *str0Iter = char((sum % 10) ^ 48);
            ++str0Iter;
            *str0Iter = char((sum / 10) ^ 48);
        }
    }
}
} // namespace str

void operator*=(string &str, const ll &n) {
    if (n == 0) {
        str.clear();
        str.push_back('0');
    }
    if (n > 9) {
        string temp{str};
        temp *= (n % 10);
        str  *= (n / 10);
        str.push_back('0');
        str::operator+=(str, temp);
        return;
    }
    for (ll i = str.size() - 1; i >= 0; --i) {
        ll ans = (str.begin()[i] ^ 48) * n;
        if (ans <= 9) {
            str.begin()[i] = char(ans ^ 48);
        } else {
            str.begin()[i] = char(ans / 10 ^ 48);
            str.push_back(char(ans % 10 ^ 48));
        }
    }
}

[[nodiscard]] string f(ll n) {
    string ans{'1'};
    for (; n; --n) {
        ans *= n;
    }
    return ans;
}

[[nodiscard]] string x(ll n) {
    string ans{'0'};
    for (; n; --n) {
        str::operator+=(ans, f(n));
    }
    return ans;
}

int main(void) {
    ll n;
    cin >> n;
    cout << x(n) << '\n';
}