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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; }
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原文地址：http://www.cnblogs.com/a719525932/p/5748574.html
