LightOJ - 1067 lucas定理 模板

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：求C(n,m)%p;

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
#define ll long long
const int Max = 1e6;
#define MOD 1000003
ll fac[Max];

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

ll _pow(ll x, ll y, ll p)
{
ll res = 1,tmp = x % p;
while(y)
{
if(y & 1)
res = res * tmp % p;
tmp = tmp * tmp % p;
y >>= 1;
}
return res;
}

ll C(ll n, ll m, ll p)
{
if(m > n)
return 0;
return fac
* _pow(fac[m] * fac[n-m], p-2, p) % p;
}

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;
cin >> t;
int k = 0;
init(MOD);
while(t--)
{
k++;
int n,m;
cin >> n >> m;
cout << "Case " << k << ": " << lucas(n, m, MOD) << endl;
}
return 0;
}