BZOJ3129: [Sdoi2013]方程

拓展Lucas+容斥原理

 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 #include<cmath>
 #include<queue>
 #define MAXN 10000+10
 #define INF 0x7f7f7f7f
 #define LINF 0x7f7f7f7f7f7f7f7f
 #define ll long long
 #define pb push_back
 #define ft first
 #define sc second
 #define mp make_pair
 #define pil pair<int,ll>
 #define pll pair<ll,ll>
 using namespace std;
 struct Lucas{
     void extgcd(ll a,ll b,ll &x,ll &y){
         if(!b){x=,y=;}
         else{
             ll xx,yy;
             extgcd(b,a%b,xx,yy);
             x=yy;
             y=xx-a/b*yy;
         }
     }
     ll Inv(ll a,ll b){
         ll x,y;
         extgcd(a,b,x,y);
         x=(x%b+b)%b;
         if(!x)x+=b;
         return x;
     }
     ll Pow(ll a,ll b,ll p){
         ll ret=1LL;
         while(b){
             if(b&){(ret*=a)%=p;}
             (a*=a)%=p;
             b>>=;
         }
         return ret;
     }
     ll fac(ll n,ll pi,ll pk){
         if(!n)return 1LL;
         ll ret=1LL;
         for(ll i=;i<pk;i++){
             if(i%pi)(ret*=i)%=pk;
         }
         ret=Pow(ret,n/pk,pk);
         for(ll i=;i<=(n%pk);i++){
             if(i%pi)(ret*=i)%=pk;
         }
         return ret*fac(n/pi,pi,pk)%pk;
     }
     ll C(ll n,ll m,ll pi,ll pk){
         ll a=fac(n,pi,pk),b=fac(m,pi,pk),c=fac(n-m,pi,pk);
         ll t=0LL;
         for(ll i=n/pi;i;i/=pi)t+=i;
         for(ll i=m/pi;i;i/=pi)t-=i;
         for(ll i=(n-m)/pi;i;i/=pi)t-=i;
         ll ret=a*Inv(b,pk)*Inv(c,pk)%pk;
         (ret*=Pow(pi,t,pk))%=pk;
         return ret;
     }
     ll n,m,p;
     vector<pll> pn;
     ll init(ll pp){
         p=pp;
         ll x=sqrt(pp*1.0);
         for(ll i=;i<=x;i++){
             if(pp%i==){
                 ll pk=1LL;
                 while(pp%i==){
                     pp/=i;
                     pk*=i;
                 }
                 pn.pb(mp(i,pk));
             }
         }
         if(pp^){
             pn.pb(mp(pp,pp));
         }
     }
     ll solve(ll n,ll m){
         ll ans=0LL,pi,pk;
         for(int i=;i<pn.size();i++){
             pi=pn[i].ft,pk=pn[i].sc;
             ll t=C(n,m,pi,pk);
             (t*=(p/pk))%=p;
             (t*=Inv(p/pk,pk))%=p;
             (ans+=t)%=p;
         }
         return ans;
     }
 }L;
 int T,n,n1,n2,m;
 int a[];
 ll ans,p;
 ll calc(ll n,ll m){
     return L.solve(m+n-,min(m,n-));
 }
 void rc(int k,int m,int f){
     if(m<)return;
     ans+=f*calc(n,m);
     ans=(ans%p+p)%p;
     for(int i=k+;i<=n1;i++){
         rc(i,m-a[i],-f);
     }
 }
 void solve(){
     scanf("%d%d%d%d",&n,&n1,&n2,&m);
     m-=n;
     for(int i=;i<=n1;i++){
         scanf("%d",&a[i]);
     }
     int t;
     for(int i=;i<=n2;i++){
         scanf("%d",&t);
         m-=(t-);
     }
     if(m<){
         printf("0\n");
         return;
     }
     ans=0LL;
     rc(,m,);
     printf("%lld\n",ans);
 }
 int main()
 {
     //freopen("data.in","r",stdin);
     scanf("%d%lld",&T,&p);
     L.init(p);
     while(T--){
         solve();
     }
     return ;
 }