http://www.lydsy.com/JudgeOnline/problem.php?id=1485
卡特兰数。
把第1,3,...,2N-1个位置看做左括号,第2,4,...,2N个位置看成右括号。
考虑从1到2N把数放进去,其实就变成了括号序列。
所以是卡特兰数。
求$\frac{C_{2n}^{n}}{n+1}\%P$可以看我下一篇博客。
#include<cstdio> #include<cstdlib> #include<iostream> #include<fstream> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<queue> #include<stack> #include<map> #include<utility> #include<set> #include<bitset> #include<vector> #include<functional> #include<deque> #include<cctype> #include<climits> #include<complex> //#include<bits/stdc++.h>适用于CF,UOJ,但不适用于poj using namespace std; typedef long long LL; typedef double DB; typedef pair<int,int> PII; typedef complex<DB> CP; #define mmst(a,v) memset(a,v,sizeof(a)) #define mmcy(a,b) memcpy(a,b,sizeof(a)) #define fill(a,l,r,v) fill(a+l,a+r+1,v) #define re(i,a,b) for(i=(a);i<=(b);i++) #define red(i,a,b) for(i=(a);i>=(b);i--) #define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++) #define fi first #define se second #define m_p(a,b) make_pair(a,b) #define SF scanf #define PF printf #define two(k) (1<<(k)) template<class T>inline T sqr(T x){return x*x;} template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;} template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;} const DB EPS=1e-9; inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;} const DB Pi=acos(-1.0); inline int gint() { int res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z==‘-‘){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar()); return (neg)?-res:res; } inline LL gll() { LL res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z==‘-‘){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar()); return (neg)?-res:res; } const int maxN=2000000; int N;LL P; LL ans; inline LL power(LL a,LL k){LL x=1,y=a;while(k){if(k&1)x=x*y%P;k>>=1;y=y*y%P;}return x;} int flag[maxN+100],cnt,prime[maxN+100]; int a[maxN+100]; LL b[maxN+100]; int main() { freopen("bzoj1485.in","r",stdin); freopen("bzoj1485.out","w",stdout); int i,j; cin>>N>>P; re(i,2,2*N) { if(!flag[i])prime[++cnt]=i; for(j=1;j<=cnt && i*prime[j]<=2*N;j++) { flag[i*prime[j]]=1; a[i*prime[j]]=prime[j]; if(i%prime[j]==0)break; } } re(i,2,N)b[i]=-1; re(i,N+2,2*N)b[i]=1; ans=1; red(i,2*N,2) if(!flag[i]) ans=ans*power(LL(i),b[i])%P; else b[a[i]]+=b[i],b[i/a[i]]+=b[i]; cout<<ans<<endl; return 0; }
时间: 2024-11-03 03:45:08