soj 3252 大組合數取模

題意：給定 m，n，求C(m,n)%10007的值，其中0<=n<=m<=200000000.

解題思路：首先，應該注意到10007是一個素數。而且，m，n範圍很大，那麼直接分解素因子的方法不可行。

                  由此，我們聯想到lacus定理,即：對於C(m,n)%p,p爲素數，令m,n在p進制下的表示形式分別爲：

                                                                       (x0x1x2...xz),

                                                                       (y0y1y2...yz)                                                                   

                                                              則：C(m,n)%p=[C(x0,y0)*C(x1,y1)*C(x2,y2)...C(xz,yz)]%p

                 所以，我們可以應用上述定理將m,n的範圍限制在10007以內，再應用拓展歐幾里得求解。

                此題時間卡的緊，可先將10007以內數的階乘對10007取模的結果保存在數組裏，此過程只需遍歷一遍數組即可。

         代碼如下：

#include <stdio.h>
#include <string.h>
#define MOD 10007

long long keep[10008];

long long change(long long n,long long a[])
{
    long long count = 0;
    while(n)
    {
        a[count++] = n%MOD;
        n/=MOD;
    }
    return count;
}

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

long long C_mod(long long m,long long n)
{
  long long x,y,a,b;
   
   if(n>m)
   return 0;
   n = n<m-n?n:m-n;
      a = keep[m];
      b = (keep[n]*keep[m-n])%MOD;
      extgcd(b,MOD,x,y);
      x*=a;
      x%=MOD;
      if(x<0)
          x+=MOD;
      return x;
}

void solve(long long m,long long n)
{
    long long a[100],b[100];
    long long counta,countb,i;
    long long res = 1;
    n = n<m-n?n:m-n;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    counta = change(n,a);
    countb = change(m,b);
    counta = counta>countb?counta:countb;
    for(i = 0;i<counta;i++)
    {
        res = (res*C_mod(b[i],a[i]))%MOD;
    }
    printf("%lld\n",res);
}

int main()
{
  long long T,m,n,i;
  
  keep[0] = 1;
  for(i = 1;i<=10007;i++)
      keep[i] = (keep[i-1]*i)%MOD;
  scanf("%lld",&T);
  while(T--)
  {
      scanf("%lld%lld",&m,&n);
      solve(m,n);
  }
  return 0;
}