http://blog.youkuaiyun.com/helloworld10086/article/details/48446457
这题解讲的超级好,我这数学白痴都明白了。。
#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<stack>
#include<set>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<sstream>
#include<queue>
#define ll __int64
#define lll unsigned long long
#define MAX 1000009
#define MAXN 2009
#define eps 1e-8
#define INF 0x7fffffff
#define clr(a) memset(a,0,sizeof(a))
#define clr1(a) memset(a,-1,sizeof(a))
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
using namespace std;
inline ll Max(ll a,ll b)
{
return a>b?a:b;
}
inline ll Min(ll a,ll b)
{
return a<b?a:b;
}
ll fac[MAX];//因子
ll p[109];
ll lucas[109];
ll n,m,k;
ll modpow(ll a,ll b,ll mod)//a^b%mod
{
ll ret = 1;
while(b)
{
if(b&1) ret = (ret*a)%mod;
a = (a*a)% mod;
b>>=1;
}
return ret;
}
ll modmul(ll a,ll b,ll mod)//a*b%mod;
{
ll ret = 0;
while(b)
{
if(b&1) ret = (ret + a)%mod;
a = (a + a)% mod;
b>>=1;
}
return ret;
}
ll getFactor(ll p)//求因子
{
fac[0] = 1;
for(int i = 1;i<=p;i++)
{
fac[i] = (fac[i - 1]*i)%p;
}
}
ll Lucas(ll n,ll m,ll p)//大组合数取余
{
ll ret = 1;
while(n&&m)
{
ll a = n%p, b = m%p;
if(a<b) return 0;
ret = (ret*fac[a]*modpow(fac[b]*fac[a - b]%p , p - 2, p)) %p;
n/=p;
m/=p;
}
return ret;
}
ll exgcd(ll a,ll b,ll &x,ll &y)//拓展欧几里得
{
if(!b)
{
x = 1 , y = 0;
return a;
}
int ans = exgcd(b , a%b, y , x);
y-=a/b*x;
return ans;
}
ll CRT(ll *a,ll *m,int len)//中国剩余定理
{
ll d,x,y,ret = 0;
ll M = 1;
for(int i = 0;i < len;i++) M*=m[i];
for(int i = 0;i < len;i++)
{
ll w = M/m[i];
d = exgcd(m[i],w,x,y);
ret = (ret + modmul(modmul(y, w, M), a[i], M) ) % M;
}
return (ret + M) % M;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("an.txt","r", stdin);
#endif
int T;
scanf("%d",&T);
while(T--)
{
scanf("%lld%lld%d",&n,&m,&k);
for(int i = 0; i<k; i++)
{
scanf("%I64d",&p[i]);
getFactor(p[i]);
lucas[i] = Lucas(n,m,p[i]);
}
ll ans = CRT(lucas,p,k);
printf("%I64d\n",ans);
}
return 0;
}