hdu 6172（递推式

传送门

Array Challenge

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 143    Accepted Submission(s): 64

Problem Description

There’s an array that is generated by following rule.
h0=2,h1=3,h2=6,hn=4hn−1+17hn−2−12hn−3−16
And let us define two arrays bnandan as below.
bn=3hn+1hn+9hn+1hn−1+9h2n+27hnhn−1−18hn+1−126hn−81hn−1+192(n>0)
an=bn+4n
Now, you have to print ⌊√(an)⌋ , n>1.
Your answer could be very large so print the answer modular 1000000007.

 

Input

The first line of input contains T (1 <= T <= 1000) , the number of test cases.
Each test case contains one integer n (1 < n <= 1015) in one line.

 

Output

For each test case print &#8970;√(a_n )&#8971; modular 1000000007.

 

Sample Input

3479

 

Sample Output

125532472513185773

 

Source

2017 Multi-University Training Contest - Team 10

 

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草爆一切递推式。

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <map>#include <set>#include <cassert>using namespace std;#define rep(i,a,n) for (long long i=a;i<n;i++)#define per(i,a,n) for (long long 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) ((long long)(x).size())typedef vector<long long> VI;typedef long long ll;typedef pair<long long,long long> PII;const ll mod=1e9+7;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;}// headlong long _,n;namespace linear_seq {    const long long N=10010;    ll res[N],base[N],_c[N],_md[N];    vector<long long> Md;    void mul(ll *a,ll *b,long long 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 (long long 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];    }    long long 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=0,pnt=0;        long long 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 (long long p=pnt;p>=0;p--)         {            mul(res,res,k);            if ((n>>p)&1)             {                for (long long 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);        long long 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;    }    long long 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() {    for (scanf("%I64d",&_);_;_--)     {        scanf("%I64d",&n);        printf("%I64d\n",linear_seq::gao(VI{31,197,1255,7997},n-2));    }}