HDOJ 3123 GCC

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本文探讨了GNU Compiler Collection (GCC) 缺少数学运算符“!”的问题，并提供了一个编程解决方案来计算从0!到n!的累加和对m取模的结果。输入包括测试案例数量、n和m的值，输出为计算结果。

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Problem 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

已AC：
#include <iostream>
#include <string>

int main()
{
    using namespace std;
    int T;
    cin >> T;
    while(T--)
    {
        string nstr;
        int m;
        cin >> nstr >> m;
        int nl;
        nl = nstr.size();
        int k, n = 0;
        if(nl>7)
            k = m-1;
        else
        {
            for(int i=0; i<nl; ++i)
            {
                n = n*10+nstr[i]-'0';
            }
            if(n>=m)
                k = m-1;
            else
                k = n;
        }
        long long temp = 1, sum = 1;
        for(int i=1; i<=k; ++i)
        {
            temp = (temp*i)%m;
            sum = (sum+temp)%m;
            if(!temp)
                break;
        }
        cout << sum%m << endl;
    }
    return 0;
}