Combinations (乘法逆元)

题目：https://vjudge.net/contest/242287#problem/A

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

利用扩展欧几里得和费马小定理求乘法逆元。需要给分母求乘法逆元，将出发变成乘逆元得乘法。

扩展欧几里得

#include <iostream>
#include <stdio.h>
#include <string.h>
#define LL long long
#define MAXN 1000005
#define mod 1000003
using namespace std;
LL com[MAXN];
void C()
{
    com[1]=com[0]=1;
    for(LL i=2;i<MAXN;i++)
        com[i]=(i*com[i-1])%mod;
}

LL exgcd(LL a, LL b, LL &x, LL &y) {
    if(b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    else {
        LL d = exgcd(b, a % b, y, x);
        y -= a / b * x;
        return d;
    }
}

LL inv(LL a, LL p) {
    LL x, k;
    LL d = exgcd(a, p, x, k);
    if(d == 1) return (x % p + p) % p;
    else return -1;
}


int main()
{
    C();
    int t;
    int Case=0;
    scanf("%d",&t);
    while(t--)
    {
            int n,k;
            scanf("%d%d",&n,&k);
            //cout<<com[k]<<endl;
            LL a=com[n];
            if(k*2>n)
                k=n-k;
            LL b=inv(com[k], mod);
            LL c=inv(com[n-k],mod);
           // cout<<c<<endl;
            LL ans=(a*b*c)%mod;
            printf("Case %d: %lld\n",++Case,ans);
    }
    return 0;
}

费马小定理：

#include <iostream>
#include <stdio.h>
#include <string.h>
#define LL long long
#define MAXN 1000005
#define mod 1000003
using namespace std;
LL com[MAXN];
void C()
{
    com[1]=com[0]=1;
    for(LL i=2;i<MAXN;i++)
        com[i]=(i*com[i-1])%mod;
}
LL pow_mod(LL a, LL n, LL p){
    LL ret = 1;
    LL tmp = a;
    while(n) {
        if(n & 1) ret = (ret * tmp) % p;
        tmp = tmp * tmp % p;
        n >>= 1;
    }
    return ret;
}


LL inv(LL a, LL p) {
    return pow_mod(a, p-2, p);
}


int main()
{
    C()
    int t;
    int Case=0;
    scanf("%d",&t);
    while(t--)
    {
            int n,k;
            scanf("%d%d",&n,&k);
            //cout<<com[k]<<endl;
            LL a=com[n];
            if(k*2>n)//组合数对称性
                k=n-k;
            LL b=inv(com[k], mod);
            LL c=inv(com[n-k],mod);
            //cout<<c<<endl;
            LL ans=(a*b*c)%mod;
            printf("Case %d: %lld\n",++Case,ans);
    }
    return 0;
}