0分，TLE

大致看了一下题解，和我想的一样就是一道高精乘低精的模板题啊？阶乘也用递推优化过了，怎么会超时呢？

代码中的add和mul完全是照模板写的，求助

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

string add(string, string);
string mul(string, int);

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int t;
    int n, maxn = 1;
    char a;
    string fac[1001] = {"0", "1"};
    cin >> t;
    for (int i = 1; i <= t; i++)
    {
        cin >> n >> a;
        if (n <= maxn)
            cout << count(fac[n].begin(), fac[n].end(), a) << "\n";
        else
        {
            for (int j = maxn + 1; j <= n; j++)
                fac[j] = mul(fac[j - 1], j);
            maxn = n;
            cout << count(fac[n].begin(), fac[n].end(), a) << "\n";
        }
    }
    return 0;
}

string add(string str1, string str2)
{
    string str = "";
    int str1_len = str1.length(), str2_len = str2.length(), temp = 0;
    bool carry = 0;
    if (str1_len < str2_len)
        for (int i = 1; i <= str2_len - str1_len; i++)
            str1 = '0' + str1;
    else
        for (int i = 1; i <= str1_len - str2_len; i++)
            str2 = '0' + str2;
    str1_len = str1.length(), str2_len = str.length();
    for (int i = str1_len - 1; i >= 0; i--)
    {
        temp = str1[i] - '0' + str2[i] - '0' + carry;
        carry = temp / 10;
        temp %= 10;
        str = char(temp + '0') + str;
    }
    if (carry == 1)
        str = '1' + str;
    return str;
}

string mul(string a, int b)
{
    string str = "", temp_num = "";
    int temp_digit = 0;
    int a_len = a.length(), carry = 0;
    for (int i = a_len - 1; i >= 0; i--)
    {
        temp_num = "";
        temp_num.insert(0, a_len - 1 - i, '0');
        temp_digit = (a[i] - '0') * b + carry;
        carry = temp_digit / 10;
        temp_digit %= 10;
        temp_num = char(temp_digit + '0') + temp_num;
        str = add(str, temp_num);
    }
    if (carry != 0)
        str = to_string(carry) + str;
    str.erase(0, str.find_first_not_of('0'));
    if (str.empty())
        str = "0";
    return str;
}