soj 3252 大组合数取模

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

解题思路：首先，应该注意到10007是一个素数。而且，m，n范围很大，那么直接分解素因子的方法不可行。

                  由此，我们联想到lacus定理,即：对于C(m,n)%p,p为素数，令m,n在p进制下的表示形式分别为：

                                                                       (x0x1x2...xz),

                                                                       (y0y1y2...yz)                                                                   

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

                 所以，我们可以应用上述定理将m,n的范围限制在10007以内，再应用拓展欧几里得求解。

                此题时间卡的紧，可先将10007以内数的阶乘对10007取模的结果保存在数组里，此过程只需遍历一遍数组即可。

         代码如下：

#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
*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;
}