Description
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
一开始只知道很大 然后就一直优化 最后还是wa 后来才知道是一个数学公式
a/b%mod=a*(b^(mod-2))%mod 表示a乘b的（mod-2）次方 mod是素数
即s[n]/(s[n-m]*s[m])%mod==s[n]*pow(s[n-m],mod-2)*pow(s[m],mod-2)%mod 因为幂比较大 所以要用大指数幂的公式
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<math.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000006
#define mod 1000003
#define LL long long
LL s[N];
int pow_mod(LL a,int b)
{
LL c=1;
while(b)
{
if(b&1) c=c*a%mod;
b=b/2;
a=a*a%mod;
}
return (int)c;
}
void q()
{
s[0]=1;
for(int i=1;i<N;i++)
s[i]=s[i-1]*i%mod;
}
int qq(int n,int m)
{
LL ans=1;
ans=s[n]*pow_mod(s[n-m],mod-2)%mod*pow_mod(s[m],mod-2)%mod;
return (int)ans;
}
int main()
{
int T,n,m,t=1;
q();
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
printf("Case %d: %d\n",t++,qq(n,m));
}
return 0;
}