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Combinations LightOJ - 1067

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题目

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:

求组合数C(n,k).

思路:

此题的关键在于除法取模的问题,由于模是一个素数,所以我们可以直接由费马小定理得出 
        a / b % mod = a * pow(b, mod -2) % mod;
ps:组合数取模应该也可以用lucas来解决.

代码:

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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;

typedef long long LL;
const int maxn = 1e6 +10;
const int mod = 1e6 + 3;
LL a[maxn];
LL b[maxn];

LL Pow(LL x, int n){
    LL base = x;
    LL ans = 1;
    while(n){
        if(n&1) ans =(ans * base ) % mod;
        base = base * base % mod;
        n = n >> 1;
    }
    return ans;
}

void init(){
     a[0] = 1;
     b[0] = 1;
    for(int i = 1; i < maxn; ++i){
        a[i] =( a[i - 1] * i) % mod;
        b[i] = Pow(a[i], mod - 2);
    }
}

int main(){
    int t;
    init();
    scanf("%d", &t);
    int cas = 0;
    while(t--){
        int n, k;
        scanf("%d %d", &n, &k);
        printf("Case %d: %lld",++cas,(a[n] * b[k] % mod )* b[n - k] % mod);
        printf("\n");
    }
    return 0;
}