hdu5446 Unknown Treasure（数论综合题：大组合数取大合数模:Lucas+CRT）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5446

Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1590    Accepted Submission(s): 585

Problem Description

On the way to the next secret treasure hiding place, the mathematician discovered a cave unknown to the map. The mathematician entered the cave because it is there. Somewhere deep in the cave, she found a treasure chest with a combination lock and some numbers on it. After quite a research, the mathematician found out that the correct combination to the lock would be obtained by calculating how many ways are there to pick m different apples among n of them and modulo it with M. Mis the product of several different primes.

 

Input

On the first line there is an integer T(T≤20) representing the number of test cases.

Each test case starts with three integers n,m,k(1≤m≤n≤1018,1≤k≤10) on a line where k is the number of primes. Following on the next line are kdifferent primes p1,...,pk. It is guaranteed that M=p1⋅p2⋅⋅⋅pk≤1018 and pi≤105 for every i∈{1,...,k}.

 

Output

For each test case output the correct combination on a line.

 

Sample Input

19 5 23 5

 

Sample Output

6

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

 

题意：求C(n,m)%M,其中M=(m0*m2*…*m(k-1)),mi为素数。

AC code：

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>#include<queue>#include<map>#include<stack>#include<vector>#define LL long long#define MAXN 1000010using namespace std;const  int N=20;//模方程数 LL a[N],mod[N];LL mul(LL a,LL b,LL mod)//a*b%mod{LL ans=0;while(b){if(b&1)ans=(ans+a)%mod;b>>=1;a=(a+a)%mod;}return ans;}LL quick_mod(LL a,LL b,LL m)//a^b%m {LL ans=1;a%=m;while(b){if(b&1){ans=ans*a%m;}b>>=1;a=a*a%m;}return ans;}LL getC(LL n,LL m,int cur)//C(n,m)%mod[cur]{LL p=mod[cur];if(m>n)return 0;if(m>n-m)m=n-m;LL ans=1;for(int i=1;i<=m;i++){LL a=(n+i-m)%p;LL b=i%p;//ans=mul(ans,mul(a,quick_mod(b,p-2,p),p),p);//p为素数，i对p的逆元可以不用扩张欧几里得进行求解  re=i^(P-2) ans = ans * (a * quick_mod(b, p-2,p) % p) % p;  }return ans; }LL Lucas(LL n,LL k,int cur)//求C(n,k)%mod[cur] {LL p=mod[cur];if(k==0)return 1%p;//return getC(n%p,k%p,cur)*Lucas(n/p,k/p,cur)%p;return getC(n % p, k % p,cur) * Lucas(n / p, k / p,cur) % p;  }void extend_Euclid(LL a,LL b,LL &x,LL &y){if(b==0){x=1;y=0;return;}extend_Euclid(b,a%b,x,y);LL tmp=x;x=y;y=tmp-a/b*y;}LL CRT(LL a[],LL m[],int k)//求C(n,m)%M,其中M=(m0*m2*…*m(k-1)),mi为素数，则先用a[i]存储模方程C(n,m)%mi,{                           //m[]存储所有素数因子mi，k表示总共有k个模方程，返回C(n,m)%M的值 LL M=1;LL ans=0;for(int i=0;i<k;i++)M*=mod[i];for(int i=0;i<k;i++){LL x,y,tmp;LL Mi=M/m[i];extend_Euclid(Mi,m[i],x,y);if(x<0){x=-x;tmp=mul(Mi,x,M);tmp=mul(tmp,a[i],M);tmp=-tmp;}else{tmp=mul(Mi,x,M);tmp=mul(tmp,a[i],M);}ans=(ans+tmp)%M;}while(ans<0)ans+=M;return ans;}int main(){//freopen("D:\\in.txt","r",stdin);int T;scanf("%d",&T);while(T--){LL n,m;int k;scanf("%lld%lld%d",&n,&m,&k);//k=1;for(int i=0;i<k;i++)scanf("%lld",&mod[i]);for(int i=0;i<k;i++)a[i]=Lucas(n,m,i)%mod[i];//printf("%I64d\n",a[0]);printf("%lld\n",CRT(a,mod,k)); }  return 0;}