結果

問題 No.1025 Modular Equation
ユーザー tko919tko919
提出日時 2020-04-10 23:59:12
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 5,120 bytes
コンパイル時間 2,811 ms
コンパイル使用メモリ 202,612 KB
実行使用メモリ 71,744 KB
最終ジャッジ日時 2024-09-16 03:48:57
合計ジャッジ時間 10,202 ms
ジャッジサーバーID
(参考情報)
judge2 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 6 TLE * 1 -- * 25
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
//template
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll;
const int inf = 0x3fffffff; const ll INF = 0x1fffffffffffffff; const double eps=1e-12;
void tostr(ll x,string& res){while(x)res+=('0'+(x%10)),x/=10; reverse(ALL(res)); return;}
template<class T> inline bool chmax(T& a,T b){ if(a<b){a=b;return 1;}return 0; }
template<class T> inline bool chmin(T& a,T b){ if(a>b){a=b;return 1;}return 0; }
//end
template<unsigned mod=1000000007>struct mint {
unsigned val;
static unsigned get_mod(){return mod;}
unsigned inv() const{
int tmp,a=val,b=mod,x=1,y=0;
while(b)tmp=a/b,a-=tmp*b,swap(a,b),x-=tmp*y,swap(x,y);
if(x<0)x+=mod; return x;
}
mint():val(0){}
mint(ll x):val(x>=0?x%mod:mod+(x%mod)){}
mint pow(ll t){mint res=1,b=*this; while(t){if(t&1)res*=b;b*=b;t>>=1;}return res;}
mint& operator+=(const mint& x){if((val+=x.val)>=mod)val-=mod;return *this;}
mint& operator-=(const mint& x){if((val+=mod-x.val)>=mod)val-=mod; return *this;}
mint& operator*=(const mint& x){val=ll(val)*x.val%mod; return *this;}
mint& operator/=(const mint& x){val=ll(val)*x.inv()%mod; return *this;}
mint operator+(const mint& x)const{return mint(*this)+=x;}
mint operator-(const mint& x)const{return mint(*this)-=x;}
mint operator*(const mint& x)const{return mint(*this)*=x;}
mint operator/(const mint& x)const{return mint(*this)/=x;}
bool operator==(const mint& x)const{return val==x.val;}
bool operator!=(const mint& x)const{return val!=x.val;}
}; using Mint=mint<>;
struct factorial {
vector<Mint> Fact,Finv,Inv;
factorial(int maxx){
Fact.resize(maxx); Finv.resize(maxx); Inv.resize(maxx);
Fact[0]=Fact[1]=Finv[0]=Finv[1]=Inv[1]=1; unsigned mod=Mint::get_mod();
rep(i,2,maxx){
Fact[i]=Fact[i-1]*i;
Inv[i]=Inv[mod%i]*(mod-mod/i);
Finv[i]=Finv[i-1]*Inv[i];
}
}
Mint fact(int n,bool inv=0){if(inv)return Finv[n];else return Fact[n];}
Mint inv(int n){return Inv[n];}
Mint nPr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[n-r];}
Mint nCr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[r]*Finv[n-r];}
};
template<typename T,unsigned p>struct NTT{
vector<T> rt,irt;
NTT(int lg=21){
unsigned mod=T::get_mod();
T prt=p; rt.resize(1<<lg,1); irt.resize(1<<lg,1);
rep(w,0,lg){
int mask=(1<<w)-1,t=(mod-1)>>w;
T g=prt.pow(t),ig=prt.pow(mod-1-t);
rep(i,0,mask){
rt[mask+i+1]=g*rt[mask+i];
irt[mask+i+1]=ig*irt[mask+i];
}
}
}
void ntt(vector<T>& f,bool inv=0){
int n=f.size();
if(inv){
for(int i=1;i<n;i<<=1)for(int j=0;j<n;j+=i*2)rep(k,0,i){
f[i+j+k]*=irt[i*2-1+k]; const T tmp=f[j+k]-f[i+j+k];
f[j+k]+=f[i+j+k]; f[i+j+k]=tmp;
}
T mul=T(n).inv(); rep(i,0,n)f[i]*=mul;
}else{
for(int i=n>>1;i;i>>=1)for(int j=0;j<n;j+=i*2)rep(k,0,i){
const T tmp=f[j+k]-f[i+j+k];
f[j+k]+=f[i+j+k]; f[i+j+k]=tmp*rt[i*2-1+k];
}
}
}
vector<T> conv(vector<T> a,vector<T> b){
int n=a.size()+b.size()-1,m=(n&(n-1)?1<<(32-__builtin_clz(n)):n);
a.resize(m); b.resize(m); ntt(a); ntt(b);
rep(i,0,m)a[i]*=b[i]; ntt(a,1); a.resize(n); return a;
}
};
using M1=mint<1045430273>; using M2=mint<1051721729>; using M3=mint<1053818881>;
NTT<mint<1045430273>,3> N1; NTT<mint<1051721729>,6> N2; NTT<mint<1053818881>,7> N3;
vector<Mint> conv(vector<Mint> a,vector<Mint> b){
int n=a.size()+b.size()-1; vector<Mint> res(n); vector<int> vals[3];
vector<int> aa(a.size()),bb(b.size());
rep(i,0,a.size())aa[i]=a[i].val; rep(i,0,b.size())bb[i]=b[i].val;
vector<M1> a1(ALL(aa)),b1(ALL(bb)),c1=N1.conv(a1,b1);
vector<M2> a2(ALL(aa)),b2(ALL(bb)),c2=N2.conv(a2,b2);
vector<M3> a3(ALL(aa)),b3(ALL(bb)),c3=N3.conv(a3,b3);
for(M1 x:c1)vals[0].push_back(x.val);
for(M2 x:c2)vals[1].push_back(x.val);
for(M3 x:c3)vals[2].push_back(x.val);
M2 r_12=175287122;
M3 r_13=395182206,r_23=526909943,r_1323=461108887;
Mint w1=1045430273; Mint w2=571989935;
rep(i,0,n){
ll a=vals[0][i];
ll b=(vals[1][i]+M2::get_mod()-a)*r_12.val%M2::get_mod();
ll c=((vals[2][i]+M3::get_mod()-a)*r_1323.val+
(M3::get_mod()-b)*r_23.val)%M3::get_mod();
res[i]=(a+b*w1.val+c*w2.val);
} return res;
}
int mpow(int x,int k,int m){
if(x<=1)return x; int res=1;
while(k){
if(k&1)res=(1LL*res*x)%m;
x=(1LL*x*x)%m; k>>=1;
} return res;
}
int main(){
int p,n,k,b; cin>>p>>n>>k>>b;
vector<int> a(n); rep(i,0,n)scanf("%d",&a[i]);
vector<int> po(p);
rep(i,0,p)po[i]=mpow(i,k,p);
vector<Mint> dp(b+1); dp[0]=1;
rep(i,0,n){
vector<Mint> vs(p);
rep(j,0,p)vs[(1LL*a[i]*po[j])%p]+=1;
dp=conv(dp,vs);
rep(j,p,dp.size())dp[j-p]+=dp[j];
dp.resize(p);
} printf("%d\n",dp[b].val);
return 0;
}
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