hdu3123actorial in GCC求阶乘

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

题意：给出n，m，求 (0! + 1! + …. + n!) % m的值
0!是0的阶乘
分析：一开始让n给吓住了，太大了。
当n >= m的时候，只需要求0-m的阶乘和就可以，因为当n>=m时候
阶乘为1 * 2 … *m…n 中间会乘到m，然而最后结果会取余m，结果是0
所以没必要算。（原因 (x + y) % m =( x %m + y %m ) % m
这里当y >= m时候 y %m = 0所以没必要算
因为输入的n特别大，所以先用数组存，当长度大于7的时候，超过了m的最大值，将n赋值为m

#include <stdio.h>#include <stdlib.h>#include <string.h>char str[200];int main(void){    int T;    int m, cnt, i;    long long sum, n;    scanf("%d", &T);    while (T--)    {        scanf(" %s %d", str, &m);        if (strlen(str) > 7) cnt = m;        else sscanf(str, "%lld", &cnt);        sum = n = 1;        if (m == 1) {printf("0\n"); continue;}        for (i = 1; i <= cnt; i++)        {            n = (n * i) % m;            sum = (sum + n) % m;        }        printf("%lld\n", sum);    }    return 0;}