hdu 3123 GCC 阶乘

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3123

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

 

题意描述：给出n和m，计算（0!+1!+2!+......+n!）%m。

算法分析：由于是模m，m的范围不是很大，即使n很大，想想如果n一旦大于m，那么（n!）%m就等于0了，所以没有必要考虑比m大的时候了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
typedef long long LL;

LL n,m;
char str[111];

int main()
{
    int t ;scanf("%d",&t) ;
    while (t--)
    {
        scanf("%s%I64d",str,&m);
        if (str[0]=='0') {printf("%I64d\n",(LL)1%m);continue; }
        LL sum=0,len=strlen(str);
        for (LL i= len-7>=0 ? len-7 : 0 ;i<=len-1 ;i++) sum=sum*10+str[i]-'0';
        LL ans=1,a=1;
        for (LL i=1 ;i<=sum && i<=m ;i++)
        {
            a=a*i%m;
            ans=(ans+a)%m;
        }
        printf("%I64d\n",ans%m);
    }
    return 0;
}