HDU 5446 Unknown Treasure(Lucas定理+CRT)

题目链接：传送门 

题意：

求C(n,m)%(p1*p2*p3*...*pk).

分析：

用Lucas求 A1 = C(n,m)%p1,A2 = C(n,m)%p2,...,Ak = C(n,m)%pk

则用同余方程x = A1 (mod p1) ,x = A2 (mod p2) ,...,x = Ak(mod pk) ,然后用CRT解一下就好了。

代码如下：

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <cmath>
using namespace std;
typedef long long LL;

const int maxn = 20;

LL a[maxn];
LL M[maxn] ;

LL mul(LL a,LL b,LL 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)
{
    LL ans = 1;
    a %= m;
    while(b)
    {
        if(b & 1)
        {
            ans = ans * a % m;
            b--;
        }
        b >>= 1;
        a = a * a % m;
    }
    return ans;
}

LL C(LL n, LL m,int cur)
{
    LL p = M[cur];
    if(m > n) return 0;
    if(m>n-m)
        m= n-m;
    LL ans = 1;
    for(LL 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的逆元可以不用扩展欧几里得进行求解  re=i^(p-2)
    }
    return ans%p;
}

LL Lucas(LL n,LL k,int cur)
{
    LL p = M[cur];
    if(k == 0)  return 1%p;
    return mul(C(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 RemindChina(LL a[],LL m[],int k)
{
    LL M = 1;
    LL ans = 0;
    for(int i=0; i<k; i++)
        M *= m[i];
    for(int i=0; i<k; i++)
    {
        LL x, y;
        LL Mi = M / m[i];
        extend_Euclid(Mi, m[i], x, y);
        LL tmp;
        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()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        LL n,m;
        int k;
        scanf("%I64d%I64d%d",&n,&m,&k);
        for(int i=0;i<k;i++){
            scanf("%I64d",&M[i]);
        }
        for(int i=0;i<k;i++){
            a[i] = Lucas(n,m,i)%M[i];
        }
        printf("%I64d\n",RemindChina(a,M,k));
    }
    return 0;
}