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[1000005];
long long inv[1000005];
long long x,y,gcd;
void extend_gcd(long long  a, long long b)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        gcd = a;
    }
    else {
        extend_gcd(b, a%b);
        long long temp = x;
        x = y;
        y = temp - a/b*y;
    }
}
void factorial()
{
    f[0]=1;inv[0]=1;
    for(int i=1;i<=1000000;i++)
    {
       f[i]=f[i-1]*i%m;
       extend_gcd(f[i],m);
       inv[i]=(x%m+m)%m;
    }
}
int main()
{
    factorial();
    int t;
    int cnt=0;
    scanf("%d
",&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
",++cnt,ans);

    }

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

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