hdu-5446(中国剩余定理+lucas）

[align=left]Problem Description[/align]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 pickm different apples among n of them and modulo it with M.M is the product of several different primes.[align=left]Input[/align]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 k different primes p1,...,pk. It is guaranteed that M=p1⋅p2⋅⋅⋅pk≤1018 and pi≤105 for every i∈{1,...,k}.
[align=left]Output[/align]For each test case output the correct combination on a line.[align=left]Sample Input[/align]

1
9 5 2
3 5[align=left]Sample Output[/align]

6
题目题意:题目意思很简单，就是让我们求组合数C(n,m)%M的数值。
题目分析:我们知道，对于组合数C(n,m)%mod，很容易想到的就是lucas，但是lucas有一个限定，就是后面的mod必须是素数，而这个题目中的M肯定不会是素数，因为它是几个素数的乘积，那么应该怎么办了，其实M的这种形式（几个素数的乘积形式），给了我们一点提示，就是不能直接用M去模。应该分开去模，而且单个的pi也是素数，正好满足Lucas的条件。b[1]=C(n,m)%p[1];.....b[i]=C(n,m)%p[i];解出了所有的b[i],现在我们就反解回去，得到X=b[1]%p[i]，，，，X=b[i]%p[i]几组同于方程，求解这几组同于方程，很明显得用中国剩余定理了。代码写得不好，里面用了很多的函数，注释会表明每个函数的用途的！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;

const int maxn=1e5+10;
ll p[15],b[15],n,m;
ll fac[maxn],inv[maxn];

ll fast_pow(ll base,ll k,ll mod)//快速幂运算
{
ll ans=1;
while (k) {
if (k&1)
ans=ans*base%mod;
base=base*base%mod;
k>>=1;
}
return ans;
}

ll fast_mul(ll a,ll b,ll mod)//快速乘法a*b%mod，因为在代码里面运算时，乘法会炸long long
{
ll ans = 0;
while(b){
if(b&1)
ans = (ans+a)%mod;
a = (a+a)%mod;
b >>= 1;
}
return ans;
}

void init(ll n,ll m,ll mod)//预处理，求n!,和m!(n-m)!的逆元,逆元是费马小定理
{
fac[0]=fac[1]=1;
for (ll i=2;i<=n;i++)
fac[i]=fac[i-1]*i%mod;
inv[m]=fast_pow(fac[m],mod-2,mod);
inv[n-m]=fast_pow(fac[n-m],mod-2,mod);
}

ll com(ll n,ll m,ll mod)//求C(n,m)%mod
{
if (m>n) return 0;
if (m==n) return 1;
init(n,m,mod);
return fac
*inv[n-m]%mod*inv[m]%mod;
}

ll lucas(ll n,ll m,ll mod)//lucas
{
if (m==0)
return 1;
return fast_mul(lucas(n/mod,m/mod,mod),com(n%mod,m%mod,mod),mod);//乘法会爆ll,

}

ll exgcd(ll a,ll b,ll &x,ll &y)
{
if (b==0) {
x=1;
y=0;
return a;
}
ll r=exgcd(b,a%b,x,y);
ll t=x;
x=y;
y=t-a/b*y;
return r;
}

void CRT(ll a[],ll m[],int n)
{
ll w=1;
ll ans=0;
for (int i=0;i<n;i++)
w=w*m[i];
for (int i=0;i<n;i++) {
ll x,y;
ll mi=w/m[i];
exgcd(mi,m[i],x,y);
ans=(ans+fast_mul(fast_mul(x,mi,w),a[i],w))%w;//这里的乘法会爆ll
}
if (ans<0) ans+=w;
printf("%lld\n",ans);
}

int main()
{
int t,k;
scanf("%d",&t);
while (t--) {
scanf("%lld%lld%d",&n,&m,&k);
for (int i=0;i<k;i++)
scanf("%lld",&p[i]);
for (int i=0;i<k;i++)
b[i]=lucas(n,m,p[i]);//求出每一组bi
CRT(b,p,k);
}
return 0;
}