HDU-6172: array Challenge (BM linear recursion)

Question:Given, there are three functions, H, B, A, and then t queries. Each time n is given, SQRT (an) is obtained );

Ideas:I don't know how to push it, but I feel that a should be linear. At this time, we can use BM linear recursion to find the first few items and put them in the template. Then we can find them.

The data range is 1e15, and 1000 groups can all be in seconds.

The main problem is to ensure that the data is linear and the first few items must be obtained.

#include<bits/stdc++.h>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 ll mod=1000000007;ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}ll n;namespace linear_seq {    const int N=10010;    ll res[N],base[N],_c[N],_md[N];    vector<int> Md;    void mul(ll *a,ll *b,int k) {        rep(i,0,k+k) _c[i]=0;        rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;        for (int i=k+k-1;i>=k;i--) if (_c[i])            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;        rep(i,0,k) a[i]=_c[i];    }    int solve(ll n,VI a,VI b) {        ll ans=0,pnt=0;        int k=SZ(a);        assert(SZ(a)==SZ(b));        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;        Md.clear();        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);        rep(i,0,k) res[i]=base[i]=0;        res[0]=1;        while ((1ll<<pnt)<=n) pnt++;        for (int p=pnt;p>=0;p--) {            mul(res,res,k);            if ((n>>p)&1) {                for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;            }        }        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;        if (ans<0) ans+=mod;        return ans;    }    VI BM(VI s) {        VI C(1,1),B(1,1);        int L=0,m=1,b=1;        rep(n,0,SZ(s)) {            ll d=0;            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;            if (d==0) ++m;            else if (2*L<=n) {                VI T=C;                ll c=mod-d*powmod(b,mod-2)%mod;                while (SZ(C)<SZ(B)+m) C.pb(0);                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;                L=n+1-L; B=T; b=d; m=1;            } else {                ll c=mod-d*powmod(b,mod-2)%mod;                while (SZ(C)<SZ(B)+m) C.pb(0);                rep(i,0,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,0,SZ(c)) c[i]=(mod-c[i])%mod;        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));    }};int main() {    int T; scanf("%d",&T);    while(T--){        scanf("%lld",&n);        printf("%lld\n", linear_seq::gao(VI{31,197,1255,7997,50959,324725,2069239,13185773,84023455,535421093,411853810},n-2))
;    }}//

 

 

For example, hdu6185 is obviously linear. We use a 16*16 matrix to perform T. I can use the pressure DP to obtain the first few items and then set this one.

#include<bits/stdc++.h>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 ll mod=1000000007;ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}ll _,n;namespace linear_seq {    const int N=10010;    ll res[N],base[N],_c[N],_md[N];    vector<int> Md;    void mul(ll *a,ll *b,int k) {        rep(i,0,k+k) _c[i]=0;        rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;        for (int i=k+k-1;i>=k;i--) if (_c[i])            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;        rep(i,0,k) a[i]=_c[i];    }    int solve(ll n,VI a,VI b) {        ll ans=0,pnt=0;        int k=SZ(a);        assert(SZ(a)==SZ(b));        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;        Md.clear();        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);        rep(i,0,k) res[i]=base[i]=0;        res[0]=1;        while ((1ll<<pnt)<=n) pnt++;        for (int p=pnt;p>=0;p--) {            mul(res,res,k);            if ((n>>p)&1) {                for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;            }        }        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;        if (ans<0) ans+=mod;        return ans;    }    VI BM(VI s) {        VI C(1,1),B(1,1);        int L=0,m=1,b=1;        rep(n,0,SZ(s)) {            ll d=0;            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;            if (d==0) ++m;            else if (2*L<=n) {                VI T=C;                ll c=mod-d*powmod(b,mod-2)%mod;                while (SZ(C)<SZ(B)+m) C.pb(0);                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;                L=n+1-L; B=T; b=d; m=1;            } else {                ll c=mod-d*powmod(b,mod-2)%mod;                while (SZ(C)<SZ(B)+m) C.pb(0);                rep(i,0,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,0,SZ(c)) c[i]=(mod-c[i])%mod;        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));    }};int main() {    while(~scanf("%lld",&n)){        printf("%d\n",linear_seq::gao(VI{1,5,11,36,95,281,781,2245},n-1));    }}

 

HDU-6172: array Challenge (BM linear recursion)