ACM-ICPC 2018 焦作赛区网络预赛 L 题 Poor God Water
God Water likes to eat meat, fish and chocolate very much, but unfortunately, the doctor tells him that some sequence of eating will make them poisonous.
Every hour, God Water will eat one kind of food among meat, fish and chocolate. If there are 33 continuous hours when he eats only one kind of food, he will be unhappy. Besides, if there are 33 continuous hours when he eats all kinds of those, with chocolate at the middle hour, it will be dangerous. Moreover, if there are 33 continuous hours when he eats meat or fish at the middle hour, with chocolate at other two hours, it will also be dangerous.
Now, you are the doctor. Can you find out how many different kinds of diet that can make God Water happy and safe during NN hours? Two kinds of diet are considered the same if they share the same kind of food at the same hour. The answer may be very large, so you only need to give out the answer module 10000000071000000007.
Input
The fist line puts an integer TT that shows the number of test cases. (T \le 1000T≤1000)
Each of the next TT lines contains an integer NN that shows the number of hours. (1 \le N \le 10^{10}1≤N≤1010)
Output
For each test case, output a single line containing the answer.
样例输入
3 3 4 15
样例输出
20 46 435170
题目来源
题解：递推
// a[i]=2*a[i-1]-a[i-2]+3*a[i-3]+2*a[i-4] (i>2)然后矩阵快速幂
当然也可以用dls的BM求
参考代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define rep(i,a,n) for (int i=a;i<n;i++) 4 #define per(i,a,n) for (int i=n-1;i>=a;i--) 5 #define pb push_back 6 #define mp make_pair 7 #define all(x) (x).begin(),(x).end() 8 #define fi first 9 #define se second 10 #define SZ(x) ((int)(x).size()) 11 typedef vector<int> VI; 12 typedef long long ll; 13 typedef pair<int,int> PII; 14 const ll mod=1000000007; 15 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;} 16 ll _,n; 17 namespace linear_seq{ 18 const int N=10010; 19 ll res[N],base[N],_c[N],_md[N]; 20 21 vector<ll> Md; 22 void mul(ll *a,ll *b,int k) 23 { 24 rep(i,0,k+k) _c[i]=0; 25 rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod; 26 for (int i=k+k-1;i>=k;i--) if (_c[i]) 27 rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod; 28 rep(i,0,k) a[i]=_c[i]; 29 } 30 int solve(ll n,VI a,VI b) 31 { 32 ll ans=0,pnt=0; 33 int k=SZ(a); 34 assert(SZ(a)==SZ(b)); 35 rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1; 36 Md.clear(); 37 rep(i,0,k) if (_md[i]!=0) Md.push_back(i); 38 rep(i,0,k) res[i]=base[i]=0; 39 res[0]=1; 40 while ((1ll<<pnt)<=n) pnt++; 41 for (int p=pnt;p>=0;p--) 42 { 43 mul(res,res,k); 44 if ((n>>p)&1) 45 { 46 for (int 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 int 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 int 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 int T; 84 int main() 85 { 86 scanf("%d",&T); 87 while(T--) 88 { 89 scanf("%lld",&n); 90 vector<int>v; 91 v.push_back(3); 92 v.push_back(9); 93 v.push_back(20); 94 v.push_back(46); 95 v.push_back(106); 96 v.push_back(244); 97 v.push_back(560); 98 v.push_back(1286); 99 v.push_back(2956); 100 v.push_back(6794); 101 printf("%lld\n",linear_seq::gao(v,n-1)%mod); 102 } 103 return 0; 104 }
