NP[阶乘取余]

这题还是挺亲切的
暴力+特判
轻松90

然后
剩下了10分
根据题目的特殊性——大数据时模数是确定的
进行分块打表

#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 longusing 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;}