1067 - Combinations
1067 - Combinations
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 |
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Cnm=(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));
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<math.h> 5 #include<stdlib.h> 6 #include<string.h> 7 using namespace std; 8 typedef long long LL; 9 const long long N=1e6+3; 10 long long MM[1000005]; 11 long long quick(long long n,long m); 12 int main(void) 13 { 14 long long p,q;MM[0]=1; 15 MM[1]=1;int i,j; 16 for(i=2;i<=1000000;i++) 17 { 18 MM[i]=(MM[i-1]%N*(i))%N; 19 }int v; 20 scanf("%d",&v); 21 for(j=1;j<=v;j++) 22 {scanf("%lld %lld",&p,&q); 23 long long x=MM[q]*MM[p-q]%N; 24 long long cc=quick(x,N-2); 25 long long ans=MM[p]*cc%N; 26 printf("Case %d: ",j); 27 printf("%lld\n",ans); 28 } 29 return 0; 30 } 31 32 long long quick(long long n,long m) 33 { 34 long long k=1; 35 while(m) 36 { 37 if(m&1) 38 { 39 k=(k%N*n%N)%N; 40 } 41 n=n*n%N; 42 m/=2; 43 } 44 return k; 45 }
油!油!you@