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
#include
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<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
#include
#include
#include
#include
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<return 0;
}