思路就是高精度x低精度 + 高精度加法

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

vector<int> B,C;

vector<int> mul(vector<int> A, int b)
{
    vector<int> C;
    if(b == 0)
        C.push_back(0);
    else
    {
        int t = 0;
        for(int i = 0; i < A.size(); i ++)
        {
            t += A[i]*b;
            C.push_back(t%10);
            t /= 10;
        }
        while(t)
        {
            C.push_back(t%10);
            t /= 10;
        }
    }
    return C;
}
vector<int> add(vector<int> A, vector<int> B)
{
    int t = 0;
    vector<int> C;
    for(int i = 0; i < A.size() || i < B.size(); i ++)
    {
        if(i < A.size())
            t += A[i];
        if(i < B.size())
            t += B[i];
        C.push_back(t %10);
        t /= 10;
    }
    if(t)
        C.push_back(t);
    return C;
}
int main()
{
    int n;
    cin >> n;
    if(n==1)
        cout << "1";
    else
    {
        C.push_back(1);
        B = C;
        for(int i = 2; i <= n; i ++)
        {
            C = mul(C,i);//下一个数的阶乘
            B = add(B,C);//累加
        }
        for(int i = B.size()-1; i>=0;i--)
            cout << B[i];
    }
    return 0;
}