问题：用高精度计算出 S = 1! + 2! + 3! + \cdots + n!S=1!+2!+3!+⋯+n!（n \le 50n≤50）。 其中 ! 表示阶乘，定义为 n!=n\times (n-1)\times (n-2)\times \cdots \times 1n!=n×(n−1)×(n−2)×⋯×1。例如，5! = 5 \times 4 \times 3 \times 2 \times 1=1205!=5×4×3×2×1=120。

下面是chatgpt的答案：

以下是使用 C++ 语言实现高精度计算求解S的代码：

#include <iostream>
#include <cstring>
using namespace std;

const int MAXN = 10000; // 最大位数
const int BASE = 10; // 进制

struct BigInt {
    int len, s[MAXN];

    BigInt() {
        memset(s, 0, sizeof(s));
        len = 1;
    }

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

    BigInt(const char* num) {
        *this = num;
    }

    BigInt operator = (int num) {
        char s[MAXN];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }

    BigInt operator = (const char* num) {
        memset(s, 0, sizeof(s));
        len = strlen(num);
        for (int i = 0; i < len; i++) {
            s[i] = num[len - i - 1] - '0';
        }
        return *this;
    }

    BigInt operator + (const BigInt& b) const{
        BigInt c;
        c.len = 0;
        for (int i = 0, g = 0; g || i < max(len, b.len); i++) {
            int x = g;
            if (i < len) x += s[i];
            if (i < b.len) x += b.s[i];
            c.s[c.len++] = x % BASE;
            g = x / BASE;
        }
        return c;
    }

    void clean() {
        while (len > 1 && !s[len-1]) len--;
    }

    BigInt operator * (const BigInt& b) {
        BigInt c; c.len = len + b.len; 
        for (int i = 0; i < len; i++)
            for (int j = 0; j < b.len; j++)
                c.s[i+j] += s[i] * b.s[j];
        for (int i = 0; i < c.len-1; i++)
            c.s[i+1] += c.s[i]/BASE, c.s[i] %= BASE;
        c.clean();
        return c;
    }

    BigInt operator - (const BigInt& b) {
        BigInt c; c.len = 0;
        for (int i = 0, g = 0; i < len; i++) {
            int x = s[i] - g;
            if (i < b.len) x -= b.s[i];
            if (x >= 0) g = 0;
            else {
                g = 1;
                x += BASE;
            }
            c.s[c.len++] = x;
        }
        c.clean();
        return c;
    }

    bool operator < (const BigInt& b) const{
        if (len != b.len) return len < b.len;
        for (int i = len-1; i >= 0; i--)
            if (s[i] != b.s[i])
                return s[i] < b.s[i];
        return false;
    }

    bool operator > (const BigInt& b) const{return b < *this;}
    bool operator <= (const BigInt& b) {return !(b < *this);}
    bool operator >= (const BigInt& b) {return !(*this < b);}
    bool operator != (const BigInt& b) {return b < *this || *this < b;}
    bool operator == (const BigInt& b) {return !(b < *this) && !(b > *this);}

    void print() {
        for (int i = len-1; i >= 0; i--) printf("%d",s[i]);
    }
};

BigInt fac[55];

int main() {
    int n;
    cin >> n;

    fac[0] = fac[1] = 1;
    for (int i = 2; i <= n; i++) {
        fac[i] = fac[i-1] * i;
    }

    BigInt ans;
    for (int i = 1; i <= n; i++) {
        ans = ans + fac[i];
    }

    ans.print();
    cout << endl;

    return 0;
}

其中，BigInt结构体表示高精度整数，支持加、减、乘等运算。在主函数中，我们先预处理出$1!$到$n!$的值，然后依次累加即可得到答案。最后输出结果即可。