HDU 3123 GCC

GCC

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4204    Accepted Submission(s): 1384

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
   
   
   
   

n大于m的部分不需要求，取模后等于0了嘛

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <cmath>
#define N 1000009
using namespace std;

char s[N];
int m;

int main()
{
    int ca;

    scanf("%d",&ca);
    while(ca--)
    {
        int ff=0;
        scanf("%s%d",s,&m);

        int len=strlen(s);
        int num=0;

        if(len>=7)
        ff=m;
        else
        {
            for(int i=0;i<len;i++)
            {
                int a=s[i]-'0';
                num+=a;
                num*=10;
            }
            num/=10;

            if(num>=m)
            {
                ff=m-1;
            }
            else ff=num;
        }

        __int64 ans=0;
        __int64 t=1;

        for(int i=1;i<=ff;i++)
        {
            t*=i;
            t%=m;
            ans+=t;
            ans%=m;

        }
        ans=(ans+1)%m;//如果开始ans取1后再求mod的话会wa，但是最后加1再取mod就过了

        printf("%I64d\n",ans);

    }
    return 0;
}