(0! + 1! + 2! + 3! + 4! + ... + n!)%m

Description
The GNU Compiler Collection (usually shortened to GCC) is a compiler system produced by the GNU Project supporting various
programming languages. But it doesn’t contains the math operator “!”. In mathematics the symbol represents the factorial
operation. The expression n! means "the product of the integers from 1 to n". For example, 4! (read four factorial) is
 4 × 3 × 2 × 1 = 24. (0! is defined as 1, which is a neutral element in multiplication, not
multiplied by anything.) We want you to help us with this formation: (0! + 1! + 2! + 3! + 4! + ... + n!)%m

Input

The first line consists of an integer T, indicating the number of test cases. Each test on a single consists of two integer n and m.
Output

Output the answer of (0! + 1! + 2! + 3! + 4! + ... + n!)%m. Constrains 0 < T <= 20 0 <= n < 10^100 (without leading zero) 0 < m < 1000000
Sample Input
1 10 861017
Sample Output
593846
解题思路：我采用的是化简法：
0! + 1! + 2! + 3! + 4! + ... + n!
=1 + 1 * (1+2*~2+2*~3+....+2*~n)
=1 + 1 * (1+2 * (1 + 3*~3+....+3*~n))
=1 + 1 * (1+2 * (1+ 3 * (......(1 + n * (1 + 0)))))
这样化简后，可以从后往前递推记s初始为1;然后
for (int i = n; i >= 1; i--)
 { s = 1 % m + ((i %m) * (s%m))%m ; }
s %=m;
方法一：
我的代码如下：
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

#define MAX_LEN 110
int main()
{
    long m;
    long n = 0;
    char ch[MAX_LEN];
    int testNum, i;

    cin >> testNum;
    for (int test = 1; test <= testNum; test++)
    {
        scanf("%s %ld", ch, &m);
        //***************确定n的值*********************************
        for (i = 0; ch[i] != '\0'; i++)
            ;
        int len = i;

        //cout << "ch =" << ch << endl;
       // cout << "len = " << len << endl;
        if (len >= 7)
        {
            n = m-1;
        }
        else
        {
            int k = 1;
            n = ch[len-1]-'0';
            for (int j = len-2; j >= 0; --j)
            {
                n += long((ch[j] - '0') * pow(10.0, double(k)));
                k ++;
                if (n >= m)
                {
                    n = m-1;
                    break;
                }
            }
        }
        //***********************************************************
        //cout << "n = " << n << endl;
        long long s = 1;
        for (i = n; i >= 1; i--)
        {
           //s = (1 + i * s) % m;
           s = 1 % m + ((i % m) * (s % m)) % m;
        }
        s %= m;
        cout << s << endl;
    }

    return 0;
}

方法二：
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

#define MAX_LEN 110
int main()
{
    long m;
    long n = 0;
    char ch[MAX_LEN];
    int testNum, i;

    cin >> testNum;
    for (int test = 1; test <= testNum; test++)
    {
        scanf("%s %ld", ch, &m);
        //***************确定n的值*********************************
        for (i = 0; ch[i] != '\0'; i++)
            ;
        int len = i;

        //cout << "ch =" << ch << endl;
       // cout << "len = " << len << endl;
        if (len >= 7)
        {
            n = m-1;
        }
        else
        {
            int k = 1;
            n = ch[len-1]-'0';
            for (int j = len-2; j >= 0; --j)
            {
                n += long((ch[j] - '0') * pow(10.0, double(k)));
                k ++;
                if (n >= m)
                {
                    n = m-1;
                    break;
                }
            }
        }
        //***********************************************************
        //cout << "n = " << n << endl;
        long long s = 1;//0!=1
        long long t = 1;//0!=1
        long sum = 1;
        for (i = 1; i <= n; i++)
        {
            t = (t % m) * (i % m);
            s = t % m;
            sum = (s  + sum % m) % m;
        }
        cout << sum % m << endl;
    }

    return 0;
}