求前n个Catalan数的和mod m的值——可作为求a/b mod m的模版

#include <iostream>
#include <string.h>
#include <stdio.h>
usingnamespace std;

#define N 110

typedef longlong lld;

int n,m,p;
int sm[N],sa[N];
lld sum,res;

void extgcd(int a, int b, int& x, int& y)
{
    if (b ==0) { x=1; y=0; return; }
    extgcd(b, a % b, x, y);
    int t = x; x = y; y = t - a / b * y;
    return ;
}

void work()
{
    int k=m,i;
    n-=2;sum=res=1LL;p=0;
    for(i=2;i*i<=k;i++)
          if(k%i==0)
          {
            sm[p++]=i;
            do{
                k/=i;
            }while(k%i==0);
              }
          if(k>1) sm[p++]=k;
        memset(sa,0,sizeof(sa));
         for(int i=2;i<=n;i++)
         {
            int t=4*i-2;
            for(int j=0;j<p;j++)
            {
                  while(t%sm[j]==0)
                  {
                    sa[j]++;
                    t/=sm[j];
                   }
            }
        res=res*t%m;
        t=i+1;
        for(int j=0;j<p;j++)
        {
             while(t%sm[j]==0)
             {
                sa[j]--;
                t/=sm[j];
               }
        }
        if(t>1)
        {
             int x,y;
               extgcd(t,m,x,y);
               x=(x%m+m)%m;
               res=(res*x)%m;
        }
        lld fin=res;
        for(int j=0;j<p;j++)
            for(int k=0;k<sa[j];k++)
                fin=(fin*sm[j])%m;
        sum=(sum+fin)%m;
      }
      printf("%lld\n",sum);
}

int main()
{
    while(scanf("%d%d",&n,&m)&&(n||m))work();
    return0;
}