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

int num[100];

//1! + 2! + 3! + 4! + 5! = 1 + 2(1 + 3(1 + 4(1 + 5)));

void factorialSum(int n, int num[]) //求n阶乘和
{
    num[0] = 1;
    int len = 1;
    for (int i = n; i >= 2; i--)
    {
        int add = 0;
        for (int j = 0; j < len; j++)
        {
            int k = num[j] * i + add;
            if(j == 0)
                k++;
            num[j] = k % 10;
            add = k / 10;
            if(j == len - 1 && add)
                len++;
        }
    }
    for (int i = len - 1; i >= 0; i--)
        cout << num[i];
    cout << endl;

}

int main()
{
    int n;
    cin >> n;
    factorialSum(n, num);

    return 0;
}

基本思路：使用惰性求值来构建一个无限列表，后面项的值依赖于之前的项（实现递推），然后用库函数sum求和。

main=print.sum.(\n->take (n+1) fac).read=<<getLine
  where fac=0:1:zipWith(*)(tail fac)[2..]

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int Max=10001;
int a[Max],b[Max],len=1,anslen=1,n;

void jiecheng(int t)
{
for(int i=1;i<=len;i++) a[i]*=t;
int i=1;
while((a[i]>9)||(i<len))
{
a[i+1]+=a[i]/10;
a[i]%=10;
i++;
}
len=i;
}

void jia()
{
for(int i=1;i<=len;++i)
{
b[i]+=a[i];
if(b[i]>9)
{
b[i+1]+=b[i]/10;
b[i]%=10;
anslen=max(anslen,i+1);

}
anslen=max(anslen,i);
}
}
int main ()
{
scanf("%d",&n);
a[len]=1;
for(int i=1;i<=n;++i) {
jiecheng(i);jia();
}
for(int i=anslen;i>0;--i) cout<<b[i];
return 0;
}