NP[階乘取餘]

這題還是挺親切的
暴力+特判
輕鬆90

然後
剩下了10分
根據題目的特殊性——大數據時模數是確定的
進行分塊打表

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<vector>
#include<climits>
#include<string>
#include<cstdlib>
#include<ctime>
#define MOD 1000000007
#define LL long long
using namespace std;

LL n,p,ans=1,i,d;
LL pre[100]={682498929,491101308,76479948,723816384,67347853,27368307,625544428,199888908,888050723,927880474,281863274,661224977,623534362,970055531,261384175,195888993,66404266,547665832,109838563,933245637,724691727,368925948,268838846,136026497,112390913,135498044,217544623,419363534,500780548,668123525,128487469,30977140,522049725,309058615,386027524,189239124,148528617,940567523,917084264,429277690,996164327,358655417,568392357,780072518,462639908,275105629,909210595,99199382,703397904,733333339,97830135,608823837,256141983,141827977,696628828,637939935,811575797,848924691,131772368,724464507,272814771,326159309,456152084,903466878,92255682,769795511,373745190,606241871,825871994,957939114,435887178,852304035,663307737,375297772,217598709,624148346,671734977,624500515,748510389,203191898,423951674,629786193,672850561,814362881,823845496,116667533,256473217,627655552,245795606,586445753,172114298,193781724,778983779,83868974,315103615,965785236,492741665,377329025,847549272,698611116};

int main()
{
/*
    for(i=1;i<=100;i++)
    {
        for(j=((i-1)*10000000)+1;j<=10000000*i;j++) d=(d*j)%MOD;

        printf("%lld,",d);
    }   
*/  

    scanf("%lld%lld",&n,&p);

    if(n>=p)
    {
        printf("0");
        return 0;
    }

    if(p==1000000007)
    {
        if(n<10000000) ans=1;
        else ans=pre[(n/10000000)-1];

        for(i=(n/10000000)*10000000+1;i<=n;i++)
        {
            d=i%p;  
            ans=(ans*d)%p;

            if(ans==0)
            {
                printf("0");
                return 0;
            }               
        }       
    }

    else
    {
        for(i=2;i<=n;i++)
            {
                d=i%p;  
                ans=(ans*d)%p;

                if(ans==0)
                {
                    printf("0");
                    return 0;
                }   
            }       
    }

    printf("%lld",ans);

    return 0;
}