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                                             Combinations 

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

这是一道基础数论题，组合数取模。题意就是输入一个n表示有n组测试数据，然后输入a,b(1<=b<=a<=1e6)，求a,b的组合数C(a,b)模1000003后的值，并按格式输出。这道题要用到Lucas定理。

首先给出这个Lucas定理：
A、B是非负整数，p是质数。AB写成p进制：A=a[n]a[n-1]...a[0]，B=b[n]b[n-1]...b[0]。
则组合数C(A,B)与C(a[n],b[n])*C(a[n-1],b[n-1])*...*C(a[0],b[0])  modp同余
即：Lucas(n,m,p)=c(n%p,m%p)*Lucas(n/p,m/p,p) 

一般对于大组合数取模，a,b不大于10^5的话，用逆元的方法，可以解决。对于a,b大于10^5的话，要求p<10^5，这样就是Lucas定理了，将a,b转化到10^5以内解。而这题a,b,p虽然都大于10^5，但也没有大太多，依然可以用Lucas定理来解决，以下就是AC代码。

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#include <set>using namespace std;typedef long long LL;typedef unsigned long long ULL;const int INF = 1e9 + 5;const int MAXN = 1e6 + 5;const int MOD = 1e9 + 7;const double eps = 1e-7;const double PI = acos(-1);using namespace std;LL quick_mod(LL a, LL b, LL c)//费马小定理+快速幂求逆元{LL ans = 1;while (b){if (b%2==1)ans = (ans*a) % c;b /=2;a = (a*a) % c;}return ans;}LL fac[MAXN];void Get_Fac(LL m)//阶乘打表{fac[0] = 1;for (int i = 1; i <= m; i++)fac[i] = (fac[i - 1] * i) % m;}LL Lucas(LL n, LL m, LL p)//卢卡斯定理{LL ans = 1;while (n && m){LL a = n % p;LL b = m % p;if (a < b)return 0;ans = ((ans*fac[a] % p) * (quick_mod(fac[b] * fac[a - b] % p, p - 2, p))) % p;n /= p;m /= p;}return ans;}int main(){int a, b;int T;scanf("%d", &T);Get_Fac(1000003);for (int i = 1; i <= T; i++){scanf("%d%d", &a, &b);printf("Case %d: %d\n", i, Lucas(a,b, 1000003));}}