hdu 5446 Unknown Treasure 卢卡斯+中国剩余定理

Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)


[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
pick m 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

[align=left]Source[/align]
2015 ACM/ICPC Asia Regional Changchun Online
题意：求c（n,m)%(p1*p2*...*pk);
思路：求出每个c（n，m）%p1=a1......求出a数组；
　　　然后根据a求中国剩余即是答案；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=1e5+10,M=1e6+10,inf=1e9+10;
const ll INF=1e18+10,mod=2147493647;
ll p[20],a[20];
ll n,m;
ll mulmod(ll x,ll y,ll m)
{
ll ans=0;
while(y)
{
if(y%2)
{
ans+=x;
ans%=m;
}
x+=x;
x%=m;
y/=2;
}
ans=(ans+m)%m;
return ans;
}
ll ff(ll x,ll p)
{
ll ans=1;
for(int i=1;i<=x;i++)
ans*=i,ans%=p;
return ans;
}
ll pow_mod(ll a, ll x, ll p)   {
ll ret = 1;
while (x)   {
if (x & 1)  ret = ret * a % p;
a = a * a % p;
x >>= 1;
}
return ret;
}

ll Lucas(ll n, ll k, ll p) {       //C (n, k) % p
ll ret = 1;
while (n && k) {
ll nn = n % p, kk = k % p;
if (nn < kk) return 0;                   //inv (f[kk]) = f[kk] ^ (p - 2) % p
ret = ret * ff(nn,p) * pow_mod (ff(kk,p) * ff(nn-kk,p) % p, p - 2, p) % p;
n /= p, k /= p;
}
return ret;
}
void exgcd(ll a, ll b, ll &x, ll &y)
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
exgcd(b, a % b, x, y);
ll tmp = x;
x = y;
y = tmp - (a / b) * y;
}
ll CRT(ll a[],ll m[],ll n)
{
ll M = 1;
ll ans = 0;
for(ll i=1; i<=n; i++)
M *= m[i];
for(ll i=1; i<=n; i++)
{
ll x, y;
ll Mi = M / m[i];
exgcd(Mi, m[i], x, y);
//ans = (ans + Mi * x * a[i]) % M;
ans = (ans +mulmod( mulmod( x , Mi ,M ), a[i] , M ) ) % M;
}
ans=(ans + M )% M;
return ans;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int k;
scanf("%lld%lld%d",&n,&m,&k);
for(int i=1;i<=k;i++)
scanf("%lld",&p[i]);
for(int i=1;i<=k;i++)
a[i]=Lucas(n,m,p[i]);
printf("%lld\n",CRT(a,p,k));
}
return 0;
}