BZOJ3129: [Sdoi2013]方程

1 #include<cstdio>
2 #include<cstdlib>
3 #include<algorithm>
4 #include<cstring>
5 #include<vector>
6 #include<cmath>
7 #include<queue>
8 #define MAXN 10000+10
9 #define INF 0x7f7f7f7f
10 #define LINF 0x7f7f7f7f7f7f7f7f
11 #define ll long long
12 #define pb push_back
13 #define ft first
14 #define sc second
15 #define mp make_pair
16 #define pil pair<int,ll>
17 #define pll pair<ll,ll>
18 using namespace std;
19 struct Lucas{
20     void extgcd(ll a,ll b,ll &x,ll &y){
21         if(!b){x=1,y=0;}
22         else{
23             ll xx,yy;
24             extgcd(b,a%b,xx,yy);
25             x=yy;
26             y=xx-a/b*yy;
27         }
28     }
29     ll Inv(ll a,ll b){
30         ll x,y;
31         extgcd(a,b,x,y);
32         x=(x%b+b)%b;
33         if(!x)x+=b;
34         return x;
35     }
36     ll Pow(ll a,ll b,ll p){
37         ll ret=1LL;
38         while(b){
39             if(b&1){(ret*=a)%=p;}
40             (a*=a)%=p;
41             b>>=1;
42         }
43         return ret;
44     }
45     ll fac(ll n,ll pi,ll pk){
46         if(!n)return 1LL;
47         ll ret=1LL;
48         for(ll i=2;i<pk;i++){
49             if(i%pi)(ret*=i)%=pk;
50         }
51         ret=Pow(ret,n/pk,pk);
52         for(ll i=2;i<=(n%pk);i++){
53             if(i%pi)(ret*=i)%=pk;
54         }
55         return ret*fac(n/pi,pi,pk)%pk;
56     }
57     ll C(ll n,ll m,ll pi,ll pk){
58         ll a=fac(n,pi,pk),b=fac(m,pi,pk),c=fac(n-m,pi,pk);
59         ll t=0LL;
60         for(ll i=n/pi;i;i/=pi)t+=i;
61         for(ll i=m/pi;i;i/=pi)t-=i;
62         for(ll i=(n-m)/pi;i;i/=pi)t-=i;
63         ll ret=a*Inv(b,pk)*Inv(c,pk)%pk;
64         (ret*=Pow(pi,t,pk))%=pk;
65         return ret;
66     }
67     ll n,m,p;
68     vector<pll> pn;
69     ll init(ll pp){
70         p=pp;
71         ll x=sqrt(pp*1.0);
72         for(ll i=2;i<=x;i++){
73             if(pp%i==0){
74                 ll pk=1LL;
75                 while(pp%i==0){
76                     pp/=i;
77                     pk*=i;
78                 }
79                 pn.pb(mp(i,pk));
80             }
81         }
82         if(pp^1){
83             pn.pb(mp(pp,pp));
84         }
85     }
86     ll solve(ll n,ll m){
87         ll ans=0LL,pi,pk;
88         for(int i=0;i<pn.size();i++){
89             pi=pn[i].ft,pk=pn[i].sc;
90             ll t=C(n,m,pi,pk);
91             (t*=(p/pk))%=p;
92             (t*=Inv(p/pk,pk))%=p;
93             (ans+=t)%=p;
94         }
95         return ans;
96     }
97 }L;
98 int T,n,n1,n2,m;
99 int a[10];
100 ll ans,p;
101 ll calc(ll n,ll m){
102     return L.solve(m+n-1,min(m,n-1));
103 }
104 void rc(int k,int m,int f){
105     if(m<0)return;
106     ans+=f*calc(n,m);
107     ans=(ans%p+p)%p;
108     for(int i=k+1;i<=n1;i++){
109         rc(i,m-a[i],-f);
110     }
111 }
112 void solve(){
113     scanf("%d%d%d%d",&n,&n1,&n2,&m);
114     m-=n;
115     for(int i=1;i<=n1;i++){
116         scanf("%d",&a[i]);
117     }
118     int t;
119     for(int i=1;i<=n2;i++){
120         scanf("%d",&t);
121         m-=(t-1);
122     }
123     if(m<0){
124         printf("0\n");
125         return;
126     }
127     ans=0LL;
128     rc(0,m,1);
129     printf("%lld\n",ans);
130 }
131 int main()
132 {
133     //freopen("data.in","r",stdin);
134     scanf("%d%lld",&T,&p);
135     L.init(p);
136     while(T--){
137         solve();
138     }
139     return 0;
140 }