ACM-ICPC 2018 焦作赛区网络预赛 L题 BM杜教线性递推
Poor God Water
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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 33continuous 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 33continuous 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 NNthat 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.
样例输入复制
3 3 4 15
样例输出复制
20 46 435170
题目来源
ACM-ICPC 2018 焦作赛区网络预赛
思路：
听别人说这题直接BM杜教线性递推板子就能过了，于是网上搜了下板子，哇，tql，对于线性式自动递推，只需要给出至少8
项（越多越准确）就可自动递推（具体原理未知）
代码：
#include
using namespace std;
#define rep(i,a,n) for (long long i=a;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 VI;
typedef long long ll;
typedef pair PII;
const ll mod=(int)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;}
// head
long long _,n;
namespace linear_seq
{
const long long N=10010;
ll res[N],base[N],_c[N],_md[N];
vector 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<=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)