大数阶乘取模

水了90分。。。
如果不会正解的话，直接暴力拿分，无脑暴力可以拿到90分 正解分块打表

暴力

就是直接求阶乘然后取模。。。
加一个比较有用的特判：如果n>=p，那么n的阶乘的因子中一定有p，n的阶乘膜p一定等于0

#include <iostream>#include <cstdio>using namespace std;long long n,p;int js(int n){    long long ans=1;    for(int i=2;i<=n;i++)    {        ans=(1ll*ans*i)%p;    }    return ans;}int main(){    scanf("%lld%lld",&n,&p);    if(n>=p)    {        cout<<'0';        return 0;    }    else    {        cout<<js(n)%p;    }    return 0;}

正解

分块打表。。。
思路很简单 但是不好想，就是每10000000个数打一个表，这样就可以把时间复杂度降到O(10000000)，二打表不计入程序运行时间，完美而又轻松的A掉本题
比如要求29999999的阶乘，就可以从20000000的阶乘的基础上开始计算

#include <cmath>#include <cstdio>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;long long n,p;long long now;const int a[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};//。。。const int MOD=1000000007;int main(){    freopen("np.in","r",stdin);    freopen("np.out","w",stdout);    cin>>n>>p;    if (p==1000000007)    {        if (n>=p)         {            cout<<"0";            return 0;        }        if(n<10000000)             now=1;        else now=a[n/10000000-1];        for(int i=n/10000000*10000000+1;i<=n;i++)            now=now*i%MOD;    } else    {        now=1;        if (n>=p) now=0;         else             for(int i=1;i<=n;i++)                 now=now*i%p;    }    cout<<now<<endl;    return 0;}