Kirai[HDOJ6172] Array Challenge（线性递推，黑科技）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6172
题意：给一堆东西，就是求个线性递推式，求第n项%1e9+7
杜教板真牛逼啊，线性递推式用某特征值相关的论文板，打表前几项丢进去就出结果了。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef long long ll; 5 #define rep(i,a,n) for (ll i=a;i<n;i++) 6 #define per(i,a,n) for (ll i=n-1;i>=a;i--) 7 #define pb push_back 8 #define mp make_pair 9 #define all(x) (x).begin(),(x).end() 10 #define fi first 11 #define se second 12 #define SZ(x) ((ll)(x).size()) 13 typedef vector<ll> VI; 14 typedef pair<ll,ll> PII; 15 const ll mod=1000000007; 16 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;} 17 // head 18 19 ll _,n; 20 namespace linear_seq { 21 const ll N=10010; 22 ll res[N],base[N],_c[N],_md[N]; 23 24 vector<ll> Md; 25 void mul(ll *a,ll *b,ll k) { 26 rep(i,0,k+k) _c[i]=0; 27 rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod; 28 for (ll i=k+k-1;i>=k;i--) if (_c[i]) 29 rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod; 30 rep(i,0,k) a[i]=_c[i]; 31 } 32 ll solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+... 33 // printf("%d\n",SZ(b)); 34 ll ans=0,pnt=0; 35 ll k=SZ(a); 36 assert(SZ(a)==SZ(b)); 37 rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1; 38 Md.clear(); 39 rep(i,0,k) if (_md[i]!=0) Md.push_back(i); 40 rep(i,0,k) res[i]=base[i]=0; 41 res[0]=1; 42 while ((1ll<<pnt)<=n) pnt++; 43 for (ll p=pnt;p>=0;p--) { 44 mul(res,res,k); 45 if ((n>>p)&1) { 46 for (ll i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0; 47 rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod; 48 } 49 } 50 rep(i,0,k) ans=(ans+res[i]*b[i])%mod; 51 if (ans<0) ans+=mod; 52 return ans; 53 } 54 VI BM(VI s) { 55 VI C(1,1),B(1,1); 56 ll L=0,m=1,b=1; 57 rep(n,0,SZ(s)) { 58 ll d=0; 59 rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod; 60 if (d==0) ++m; 61 else if (2*L<=n) { 62 VI T=C; 63 ll c=mod-d*powmod(b,mod-2)%mod; 64 while (SZ(C)<SZ(B)+m) C.pb(0); 65 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod; 66 L=n+1-L; B=T; b=d; m=1; 67 } else { 68 ll c=mod-d*powmod(b,mod-2)%mod; 69 while (SZ(C)<SZ(B)+m) C.pb(0); 70 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod; 71 ++m; 72 } 73 } 74 return C; 75 } 76 ll gao(VI a,ll n) { 77 VI c=BM(a); 78 c.erase(c.begin()); 79 rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod; 80 return solve(n,c,VI(a.begin(),a.begin()+SZ(c))); 81 } 82 }; 83 84 signed main() { 85 // freopen("in", "r", stdin); 86 for (scanf("%lld",&_);_;_--) { 87 scanf("%lld",&n); 88 printf("%lld\n",linear_seq::gao(VI{31,197,1255,7997,50959,324725,2069239,13185773,84023455},n-2)); 89 } 90 }
