题目链接：
传送门

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N = 1e6 + 10;
const ll p = 1000003;
ll t, n, x, y, ca;
ll fact[N], infact[N];
//快速幂模板
ll qmul(ll a, ll k) {
	ll res = 1;
	while (k) {
		if (k & 1) res = (ll)res * a % p;
		a = (ll) a * a % p;
		k >>= 1;
	}
	return res;
}

int main() {
	fact[0] = infact[0] = 1;
	//打表求出所有数的阶乘及其逆元
	for (ll i = 1; i < N; i ++ ) {
		fact[i] = (ll)fact[i - 1] * i % p;
		infact[i] = (ll)infact[i - 1] * qmul(i, p - 2) % p;
	}
	scanf("%lld", &t);
	while(t--) {
		scanf("%lld %lld", &x, &y);
		//套公式计算
		ll ans = fact[x] * infact[y] % p * infact[x - y] % p;
		printf("Case %lld: %lld\n", ++ca, ans);
	}
}

这道题算是很明显的组合数问题，套组合数的公式，打个表即可。注意数据范围。数组不要开太大或太小。