LighttOJ-1067-Combinations （费马小定理）

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原题链接：
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
题意：
计算组合数C（n，m），最后求余1000003的结果
题解：
如果利用公式直接算阶乘的话，结果会很大。这里要用到同余定理，但是除法不能使用同余定理，所以要把分母转换为其逆元，再利用乘法同余定理计算即可。
中间计算逆元时可以利用快速幂。
附上AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define LL long long
const int N=1000005;
const int mod=1000003;
LL a[N],b[N];
LL quickpow(LL a,LL b)//快速幂
{
    LL ans=1;
    while(b)//一层一层的算
    {
        if(b&1) ans=(ans*a)%mod;//如果b的二进制最右端是1
        a=(a*a)%mod;
        b>>=1;
    }
    return ans;
}
int main()
{
    a[0]=1,b[0]=1;
    for(int i=1;i<=N;++i)
    {
        a[i]=(a[i-1]*i)%mod;
        b[i]=quickpow(a[i],mod-2);//费马小定理
    }
    int t,n,k;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        scanf("%d%d",&n,&k);
        LL ans=((a[n]*b[k]%mod)*b[n-k])%mod;//同余
        printf("Case %d: %lld\n",cas,ans);
    }
    return 0;
}

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