2018 江苏省邀请赛 H

题目链接 https://nanti.jisuanke.com/t/28872

解析 递推 直接套杜教板子

AC代码

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <map>
 #include <set>
 #include <iostream>
 #include <cassert>
 using namespace std;
 #define rep(i,a,n) for (int i=a;i<n;i++)
 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 #define pb push_back
 #define mp make_pair
 #define all(x) (x).begin(),(x).end()
 #define fi first
 #define se second
 #define SZ(x) ((int)(x).size())
 typedef vector<int> VI;
 typedef long long ll;
 typedef pair<int,int> PII;
 const int maxn=1e6+;
 const ll mod=;
 ll powmod(ll a,ll b) {ll res=;a%=mod; assert(b>=); for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
 // head
 
 int _,n,k;
 int a[maxn],c[maxn];
 namespace linear_seq {
     const int N=;
     ll res[N],base[N],_c[N],_md[N];
 
     vector<int> Md;
     void mul(ll *a,ll *b,int k) {
         rep(i,,k+k) _c[i]=;
         rep(i,,k) if (a[i]) rep(j,,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
         for (int i=k+k-;i>=k;i--) if (_c[i])
             rep(j,,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
         rep(i,,k) a[i]=_c[i];
     }
     int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
 //        printf("%d\n",SZ(b));
         ll ans=,pnt=;
         int k=SZ(a);
         assert(SZ(a)==SZ(b));
         rep(i,,k) _md[k--i]=-a[i];_md[k]=;
         Md.clear();
         rep(i,,k) if (_md[i]!=) Md.push_back(i);
         rep(i,,k) res[i]=base[i]=;
         res[]=;
         while ((1ll<<pnt)<=n) pnt++;
         for (int p=pnt;p>=;p--) {
             mul(res,res,k);
             if ((n>>p)&) {
                 for (int i=k-;i>=;i--) res[i+]=res[i];res[]=;
                 rep(j,,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
             }
         }
         rep(i,,k) ans=(ans+res[i]*b[i])%mod;
         if (ans<) ans+=mod;
         return ans;
     }
     VI BM(VI s) {
         VI C(,),B(,);
         int L=,m=,b=;
         rep(n,,SZ(s)) {
             ll d=;
             rep(i,,L+) d=(d+(ll)C[i]*s[n-i])%mod;
             if (d==) ++m;
             else if (*L<=n) {
                 VI T=C;
                 ll c=mod-d*powmod(b,mod-)%mod;
                 while (SZ(C)<SZ(B)+m) C.pb();
                 rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                 L=n+-L; B=T; b=d; m=;
             } else {
                 ll c=mod-d*powmod(b,mod-)%mod;
                 while (SZ(C)<SZ(B)+m) C.pb();
                 rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                 ++m;
             }
         }
         return C;
     }
     int gao(VI a,ll n) {
         VI c=BM(a);
         c.erase(c.begin());
         rep(i,,SZ(c)) c[i]=(mod-c[i])%mod;
         return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
     }
 };
 
 int main() {
     while(scanf("%d%d",&n,&k)==)
     {
         for(int i=;i<=n;i++)scanf("%d",&a[i]);
         for(int i=;i<=n;i++)scanf("%d",&c[i]);
         for(int i=n+;i<=n*;i++)
         {
             a[i]=;
             for(int j=;j<=n;j++)
                 a[i]=(a[i]+1ll*a[i-j]*c[j]%mod)%mod;
         }
         VI g;g.clear();
         for(int i=;i<=*n;i++)g.pb(a[i]);
         cout<<linear_seq::gao(g,k-)<<endl;
     }
 }

................................................................................................................................................................

 #include <bits/stdc++.h>
 #define fir first
 #define se second
 #define pb push_back
 #define ll long long
 #define mp make_pair
 #define rep(i,a,n) for (int i=a;i<n;i++)
 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 #define all(x) (x).begin(),(x).end()
 #define SZ(x) ((int)(x).size())
 using namespace std;
 typedef vector<ll> VI;
 typedef pair<int,int> PII;
 const int maxn=1e5+;
 const int maxm=1e6+;
 const int inf=0x3f3f3f3f;
 const ll mod=1e9+;
 const double eps=1e-;
 ll powmod(ll a,ll b) {ll res=;a%=mod; assert(b>=); for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
 namespace linear_seq {
     const int N=;
     ll res[N],base[N],_c[N],_md[N];
 
     vector<int> Md;
     void mul(ll *a,ll *b,int k) {
         rep(i,,k+k) _c[i]=;
         rep(i,,k) if (a[i]) rep(j,,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
         for (int i=k+k-;i>=k;i--) if (_c[i])
             rep(j,,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
         rep(i,,k) a[i]=_c[i];
     }
     ll solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
 //        printf("%d\n",SZ(b));
         ll ans=,pnt=;
         int k=SZ(a);
         assert(SZ(a)==SZ(b));
         rep(i,,k) _md[k--i]=-a[i];_md[k]=;
         Md.clear();
         rep(i,,k) if (_md[i]!=) Md.push_back(i);
         rep(i,,k) res[i]=base[i]=;
         res[]=;
         while ((1ll<<pnt)<=n) pnt++;
         for (int p=pnt;p>=;p--) {
             mul(res,res,k);
             if ((n>>p)&) {
                 for (int i=k-;i>=;i--) res[i+]=res[i];res[]=;
                 rep(j,,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
             }
         }
         rep(i,,k) ans=(ans+res[i]*b[i])%mod;
         if (ans<) ans+=mod;
         return ans;
     }
     VI BM(VI s) {
         VI C(,),B(,);
         int L=,m=,b=;
         rep(n,,SZ(s)) {
             ll d=;
             rep(i,,L+) d=(d+(ll)C[i]*s[n-i])%mod;
             if (d==) ++m;
             else if (*L<=n) {
                 VI T=C;
                 ll c=mod-d*powmod(b,mod-)%mod;
                 while (SZ(C)<SZ(B)+m) C.pb();
                 rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                 L=n+-L; B=T; b=d; m=;
             } else {
                 ll c=mod-d*powmod(b,mod-)%mod;
                 while (SZ(C)<SZ(B)+m) C.pb();
                 rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                 ++m;
             }
         }
         return C;
     }
     int gao(VI a,ll n) {
         VI c=BM(a);
         c.erase(c.begin());
         rep(i,,SZ(c)) c[i]=(mod-c[i])%mod;
         return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
     }
 };
 
 int n,k;
 vector<ll> a,b;
 
 int main(){
     while (~scanf("%d %d",&n,&k)){
         a.clear();
         b.clear();
         for (int i=;i<n;i++){
             int num;
             scanf("%d",&num);
             b.pb(num);
         }
         for (int i=;i<n;i++){
             int num;
             scanf("%d",&num);
             a.pb(num);
         }
         printf("%lld\n",linear_seq::solve(k-,a,b));
     }
     return ;
 }