Light OJ 1058

题意： 简单的就组合数 C（m，n）；

数据多，大， 要预处理；

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e6 + 2;
const LL MOD = 1000003;

LL Pow_Mod(LL a, LL n)
{
    LL ret = 1;
    while(n)
    {
        if(n & 1) ret = ret * a % MOD;
        n >>=1;
        a = a * a % MOD;
    }
    return ret;
}

LL Num[maxn], Inv[maxn];
LL Init()
{
    Num[0] = 1;
    for(LL i = 1; i < maxn; ++i) Num[i] = Num[i-1] * i % MOD;
    //cout << Num[maxn-1] << endl;
    Inv[maxn-1] = Pow_Mod(Num[maxn-1], MOD-2);
    for(LL i = maxn-2; i >= 0; --i)
        Inv[i] = Inv[i+1] * (i+1) % MOD;
}

LL C(LL m, LL n)
{
    LL ret = 1;
    if(n == 0 || n == m) return ret;
    else
    {
        LL s = m-n;
        //cout << s << endl;
        ret = Num[m] * Inv[s] % MOD;
        //cout << Num[m] << "   " << Inv[s] << endl;
        ret = ret * Inv[n] % MOD;
        //cout << Inv[n] << endl;
        return ret;
    }
}

int main()
{
    Init();
    int t;
    LL m, n;
    scanf("%d",&t);
    for(int kase = 1; kase <= t; ++kase)
    {
        scanf("%lld %lld",&m,&n)    ;
        printf("Case %d: %lld
",kase, C(m,n));
    }
}