1067

1067 - Combinations

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

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Problem Setter: Jane Alam Jan

思路：费马小定理。

这个是组合数取模，有卢卡斯定理可以解决，但还没学，所以用费马小定理和快速幂水了一发。

当然先打表求阶乘取模，然后根据组合数公式Cⁿ_{m=（m!）/((n!)*(m-n)!);}

由于所给的数是1000003，素数，（n!）*(m-n)！，不能整除，根据（p/q）%N=k%N；其中k为所要求的数，

那么可以得到（p）%N=k*q%N；所以用费马小定理求下q的逆元就可以了，复杂度(N*log(N));

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<string.h>
using namespace std;
typedef long long LL;
const long long N=1e6+3;
long long MM[1000005];
long long quick(long long n,long m);
int main(void)
{
    long long p,q;MM[0]=1;
    MM[1]=1;int i,j;
    for(i=2;i<=1000000;i++)
    {
        MM[i]=(MM[i-1]%N*(i))%N;
    }int v;
    scanf("%d",&v);
    for(j=1;j<=v;j++)
    {scanf("%lld %lld",&p,&q);
        long long x=MM[q]*MM[p-q]%N;
        long long cc=quick(x,N-2);
        long long ans=MM[p]*cc%N;
        printf("Case %d: ",j);
        printf("%lld
",ans);
    }
    return 0;
}

long long quick(long long n,long m)
{
    long long k=1;
    while(m)
    {
        if(m&1)
        {
            k=(k%N*n%N)%N;
        }
        n=n*n%N;
        m/=2;
    }
    return k;
}