LightOJ - 1067 (卢卡斯定理，打表法)

题意：求C(m,n)%p;
题解：因为n (1 ≤ n ≤ 106), k (0 ≤ k ≤ n).T (≤ 2000)所以用到卢卡斯定理。#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
#define LL __int64
#define N 1000100
#define M 1000000007ll
LL fac
;

void init(LL p)
{
LL i;
fac[0] =1;
for(i =1; i <= p; i++)
fac[i] = fac[i-1]*i % p;
}

LL exp_mod(LL a, LL b, LL p)
{
LL tmp = a % p, ans =1;
while(b)
{
if(b & 1) ans = ans * tmp % p;
tmp = tmp*tmp % p;
b >>=1;
}
return ans;
}

LL C(LL n, LL m, LL p)
{
if(m > n)
return 0;
return fac
*exp_mod(fac[m]*fac[n-m], p-2, p) % p;//逆元
}

LL Lucas(LL n, LL m, LL p)
{
if(m ==0)
return 1;
return (C(n%p, m%p, p)*Lucas(n/p, m/p, p))%p;
}

int main()
{
int T;
LL n,m,p,cas=1;
p=1000003;
init(p);

scanf("%d",&T);
while(T--)
{
scanf("%lld %lld",&n,&m);
printf("Case %lld: %lld\n",cas++,Lucas(n,m,p));
}
return 0;
}