light oj 1067 费马小定理求逆元

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1067

1067 - 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	Output for Sample Input
3 4 2 5 0 6 4	Case 1: 6 Case 2: 1 Case 3: 15

分析：

时间只有2秒，T组测试数据加上n的10⁶达到了10⁹递推肯定超时，那么考虑组合公式，C(n,k)=n!/(k!*(n-k)!);先打一个阶乘的表（当然要取模，只有10⁶）,然后就是这个除法取模的问题，当然是求逆元，（a/b）%mod=a*(b对mod 的逆元);求逆元可以用扩欧和费马小定理。

费马小定理的使用条件mod必须为素数。

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
#define N 1000009
#define mod 1000003
#define LL long long

LL fact[N];

void init()
{
fact[0] = fact[1] = 1;

for(int i = 2; i < N; i++)
fact[i] = (fact[i-1] * i) % mod;
}
/*LL niyuan(int a, int p)
{
LL ans = 1;

if(p == 0)
return 1;

while(p)
{
if(p & 1)
ans = (ans * a) % mod;
a = (a * a) % mod;
p >>= 1;
}
return ans;

}*/

LL niyuan(int a, int b)///就是一个快速幂。
{
if(b == 0)
return 1;

LL x = niyuan(a, b / 2);

LL ans = x * x % mod;

if(b % 2 == 1)
ans = ans * a % mod;

return ans;
}
LL c(int n, int k)
{
LL fm = (fact[k] * fact[n-k]) % mod;///n! * (n-m)!.
LL ans1 = niyuan(fm, mod - 2);///求n!的逆元。

return (ans1 * fact[n]) % mod;///公式(a / b ) % mod = (a * a ^(mod-2) % mod。
}
int main(void)
{
int T, cas;
int n, k;

init();
scanf("%d", &T);
cas = 0;

while(T--)
{
cas++;
scanf("%d%d", &n, &k);
if(2 * k > n)///组合数性质。
k = n - k;

printf("Case %d: %lld\n", cas, c(n, k));

}
return 0;
}