组合数取模（逆元+快速幂）

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Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

LightOJ 1067

Description
Given n different objects, you want to take k of them. How many ways to can do it?
For example, say there are 4 items; you want to take 2 of them. So, you can do it 6 ways.
Take 1, 2
Take 1, 3
Take 1, 4
Take 2, 3
Take 2, 4
Take 3, 4

Input
Input starts with an integer T (≤ 2000), denoting the number of test cases.

Each test case contains two integers n (1 ≤ n ≤ 106), k (0 ≤ k ≤ n).

Output
For each case, output the case number and the desired value. Since the result can be very large, you have to print the result modulo 1000003.

Sample Input
3
4 2
5 0
6 4
Sample Output
Case 1: 6
Case 2: 1

Case 3: 15

除法求模不能类似乘法，对于（A/B）mod C，直接（A mod C）/ （B mod C）是错误的；找到B的逆元b(b=B^-1);求出(A*b)modC即可；

由费马小定理：B 关于 P 的逆元为 B^(p-2);

费马小定理(Fermat Theory)是数论中的一个重要定理，其内容为：假如p是质数，且gcd(a,p)=1，那么 a^(p-1)≡1（mod p）。即：假如a是整数，p是质数，且a,p互质(即两者只有一个公约数1)，那么a的(p-1)次方除以p的余数恒等于1。所以，a^-1*a=1=a^(p-1),所以：a^-1=a^(p-2);

数学排列组合公式：C（n，m）= n！/（m！*（n-m）！）

代码：

#include<cstdio>  #include<cstring>  #include<algorithm>  using namespace std;  #define LL long long  #define G  1100000  #define mod 1000003  LL pri[G];  LL ni[G],ans;  LL pow(LL a,int b)  {      LL ans=1,base=a;      while (b>0)      {          if (b%2==1)              ans=(base*ans)%mod;          base=(base*base)%mod;          b/=2;      }      return ans;  }  void s()   //打表 {      pri[0]=1;      ni[0]=1;      for (int i=1;i<G ;i++)      {          pri[i]=pri[i-1]*i%mod;  //N！         ni[i]=pow(pri[i],mod-2);      }  }  int main()  {      s();      int t,n,b,k=1;scanf("%d",&t);      while (t--)      {          scanf("%d%d",&n,&b);          ans=((pri[n]*ni[b]%mod)*ni[n-b])%mod; // C（n，m）= n！/（m！*（n-m）！）        printf("Case %d: %lld\n",k++,ans);      }      return 0;  }

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