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Problem 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 4Sample Output
Case 1: 6
Case 2: 1
Case 3: 15
题意:t 组数据,每组给出 n、k 两个数,求 C(n,k)%1000003
思路:卢卡斯定理模版题
Source Program
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-9
#define INF 0x3f3f3f3f
#define LL long long
const LL MOD=1000003;
const int N=1000000+5;
const int dx[]= {-1,1,0,0};
const int dy[]= {0,0,-1,1};
using namespace std;
LL fac[N];
void getFac(){//构造阶乘
fac[0]=1;
for(int i=1;i<=1000000;i++){
fac[i]=fac[i-1]*i%MOD;
}
}
LL quickPowMod(LL a,LL b,LL mod){//快速幂
LL res=1;
while(b){
if(b&1)
res=res*a%mod;
b>>=1;
a=a*a%mod;
}
return res;
}
LL getC(LL n,LL m,LL mod){//获取C(n,m)%mod
if(m>n)
return 0;
return fac[n]*(quickPowMod(fac[m]*fac[n-m]%mod,mod-2,mod))%mod;
}
LL Lucas(LL n,LL m,LL mod){//卢卡斯定理
if(m==0)
return 1;
return getC(n%mod,m%mod,mod)*Lucas(n/mod,m/mod,mod)%mod;
}
int main(){
getFac();
int t;
scanf("%d",&t);
int Case=1;
while(t--){
LL n,k;
scanf("%lld%lld",&n,&k);
printf("Case %d: %lld\n",Case++,Lucas(n,k,(LL)MOD));
}
return 0;
}