bjfu1238 Cattleya number of surplus

The topic is to specify N, Cattleya CA (n)%m. The Cattleya number has recursive formula, general formula and approximate Formula Three, because to take the remainder, so the approximate formula can not be used directly, recursive formula I simply tried, TLE. So we can only start with the general formula.

Ca (n) = (2*n)! /n! /(N+1)!

The idea is to decompose the CA (n) mass factor and then calculate the final answer with a quick power. However, when the calculation of n! from 1 to N in turn, the quality factor decomposition, it must be time-out, fortunately, the factorial to take the rest of the law, and constantly remove the element factor can be.

Finally, or grazing, perhaps the screen is written in general, but also the title of the request too carved.

/** Author:ben*/#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<ctime>#include<iostream>#include<algorithm>#include<queue>#include<Set>#include<map>#include<stack>#include<string>#include<vector>#include<deque>#include<list>#include<functional>#include<numeric>#include<cctype>using namespacestd; #ifdef on_local_debug#else#endiftypedefLong LongLL;Const intMAXN =80000;Const intN =1000001;BOOLIsprime[n +3];//use two more elements to avoid judging boundariesintPN, PT[MAXN], NUM[MAXN];voidinit_prime_table () {memset (IsPrime,true,sizeof(IsPrime)); intp =2, Q, del; Doubletemp;  while(P <=N) { while(!isprime[p]) {p++; }        if(P > N) {//is over             Break; } temp= (Double) p; Temp*=p; if(Temp >N) Break;  while(Temp <=N) {del= (int) temp; Isprime[del] =false; Temp*=p; } q= p +1;  while(Q <N) { while(!isprime[q]) {q++; }            if(q >= N) { Break;} Temp= (Double) p; Temp*=Q; if(Temp > N) Break;  while(Temp <=N) {del= (int) temp; Isprime[del]=false; Temp*=p; } q++; } P++; } PN=0;  for(inti =2; I <= N; i++) {        if(Isprime[i]) {PT[PN++] =i; }    }}intModular_exp (intAintBintc) {LL res, temp; Res=1% c, temp = a%C;  while(b) {if(B &1) {res= Res * Temp%C; } temp= Temp * Temp%C; b>>=1; }    return(int) Res;voidGetnums (intN) {intx =2*N; intLen =PN;  for(inti =0; I < len && pt[i] <= x; i++) {        intR =0, a =x;  while(a) {R+ = A/Pt[i]; A/=Pt[i]; } Num[i]=R; } x=N;  for(inti =0; I < len && pt[i] <= x; i++) {        intR =0, a =x;  while(a) {R+ = A/Pt[i]; A/=Pt[i]; } Num[i]-=R; } x= n +1;  for(inti =0; I < len && pt[i] <= x; i++) {        intR =0, a =x;  while(a) {R+ = A/Pt[i]; A/=Pt[i]; } Num[i]-=R; }}intMain () {#ifdef on_local_debug freopen ("data.in","R", stdin);//freopen ("test.in", "R", stdin);//freopen ("Data.out", "w", stdout);#endif    intN, M;    Init_prime_table ();  while(SCANF ("%d%d", &n, &m) = =2) {getnums (n); LL ans=1; intN2 =2*N;  for(inti =0; Ans && (i < PN) && Pt[i] <= n2; i++) {            if(Num[i] >0) {ans= ans *(LL) modular_exp (Pt[i], num[i], M); Ans%=m; }} printf ("%d\n", (int) ans); }    return 0;}

bjfu1238 Cattleya number of surplus