Light OJ 1067 Combinations (乘法逆元）

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

知识点：乘法逆元,扩展欧几里得。

题意：求C(n,m)%mod

难点:理解乘法逆元。

扩展：乘法逆元定义：如果a*b=1(mod n),那么b就是a关于模n的乘法逆元,此时，也可称作a与b互为乘法逆元。

思路：1.C(n,m)%mod=n!/m!*(n-m)!%mod除以一个数并取模等价于乘以这个数的逆元然后再去模.可将n!/m!*(n-m)!%mod等价于n!/m!%mod*1/(n-m)!%mod.再等价于n!*inv[m!]%mod(m!的逆元）*inv[(n-m)!]((n-m!)的逆元).

2.也就是说，求出逆元即可求解。c*x=1(mod n)=1-k*n等价于c*x+k*n=1所以可以用扩展欧几里得算法求得x(逆元）的值。为了保证x是正整数，通常需要加上：x=(x%n+n)%n.

 #include<cstdlib>
 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 using namespace std;
 #define m 1000003
 long long  f[];
 long long inv[];
 long long x,y,gcd;
 void extend_gcd(long long  a, long long b)
 {
     if(b == )
     {
         x = ;
         y = ;
         gcd = a;
     }
     else {
         extend_gcd(b, a%b);
         long long temp = x;
         x = y;
         y = temp - a/b*y;
     }
 }
 void factorial()
 {
     f[]=;inv[]=;
     for(int i=;i<=;i++)
     {
        f[i]=f[i-]*i%m;
        extend_gcd(f[i],m);
        inv[i]=(x%m+m)%m;
     }
 }
 int main()
 {
     factorial();
     int t;
     int cnt=;
     scanf("%d\n",&t);
     int a1,b1;
     while(t--)
     {
        scanf("%d%d",&a1,&b1);
        long long ans=f[a1]*inv[b1]%m*inv[a1-b1]%m;
        printf("Case %d: %lld\n",++cnt,ans);

     }

  return ;
 }
 //3
 //
 //4 2
 //
 //5 0
 //
 //6 4

 //3
 //1 1
 //2 3
 //4 3