Combinations 乘法逆元

G - 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

3

4 2

5 0

6 4

Sample Output

Case 1: 6

Case 2: 1

Case 3: 15

这题 给出a，b求 a,b的组合数mod1000003  我是先用fac打表fac[i]表示i的阶乘%1000003的结果,用inv【i】表示i的阶乘%1000003的乘法逆元；最后用组合公式即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int INF=0x3f3f3f3f;
typedef long long LL;
const int mod=1e6+3;
const int maxn=1e6+100;
LL fac[maxn],inv[maxn];
LL Pow(LL a,LL b)
{
    LL ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ans;
}
int main()
{
    int cas=1;
    int n,a,b;
    fac[0]=inv[0]=1;
    for(int i=1;i<=1000000;i++)
    {
        fac[i]=(fac[i-1]*i)%mod;
        inv[i]=Pow(fac[i],mod-2);
    }
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d%d",&a,&b);
        LL ans=fac[a]*inv[b]%mod*inv[a-b]%mod;
        printf("Case %d: %lld\n",cas++,ans);
    }
    return 0;
}

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