【hdu 6172】Array Challenge（数列、找规律）

多校10 1002 HDU 6172 Array Challenge

题意

There’s an array that is generated by following rule.
(h_0=2,h_1=3,h_2=6,h_n=4h_{n-1}+17h_{n-2}-12h_{n-3}-16)

And let us define two arrays bnandan as below.
( b_n=3h_{n+1} h_n+9h_{n+1} h_{n-1}+9h_n^2+27h_n h_{n-1}-18h_{n+1}-126h_n-81h_{n-1}+192(n>0) )
( a_n=b_n+4^n )
Now, you have to print (left lfloor sqrt{a_n} ight floor) , n>1.
Your answer could be very large so print the answer modular 1000000007.

题解

首先要打表，然后就可以：
方法1，找规律：
令(f_n=left lfloor sqrt{a_n} ight floor)，打表出来，发现接近7倍关系，再打出 (7f_{n-1}-f_{n})，会发现是(f_{n-2})的4倍，所以(f_n=7f_{n-1}-4f_{n-2})。再用矩阵快速幂。注意取模有负数。
方法2，猜：
前几项和h的前几项差不多，于是模仿h的递推式，(f_n=4f_{n-1}+17f_{n-2}-12f_{n-3})，刚好符合。
方法3，BM算法：
如果递推式是线性的，就把前几项带进去就可以得到递推式。
具体算法介绍建议看这个：Berlekamp–Massey algorithm(From Wikipedia)、
"Shift-register synthesis and BCH decoding"
方法4，数学递推，我不会。

代码

#include <cstdio>
#include <map>
#include <cstdlib>
#include <queue>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,l,r) for (int i=l;i<r;++i)
typedef unsigned long long ull;
int dx[4]={1,1,-1,-1},dy[4]={0,1,0,-1};
map<ull,bool>vis;
struct Sta{
	int a[6][6],step,x,y;
	Sta(){step=x=y=0;}
};
int gujia(Sta s){
	int ans=0;
	rep(i,0,6)rep(j,0,i+1)
	if(s.a[i][j])ans+=abs(s.a[i][j]-i);
	return ans;
}
ull haxi(Sta s){
	ull ans=0;
	rep(i,0,6)rep(j,0,i+1){
		ans<<=3;ans|=s.a[i][j];
	}
	return ans;
}
int bfs(Sta s){
	vis.clear();
	queue<Sta>q;q.push(s);
	while(!q.empty()){
		Sta now=q.front();q.pop();
		if(gujia(now)==0)return now.step;
		rep(i,0,4){
			int x=now.x,y=now.y;
			int nx=x+dx[i],ny=y+dy[i];
			if(nx>=0 && nx<6 && ny>=0 && ny<=nx){
				swap(now.a[x][y],now.a[nx][ny]);
				now.x=nx,now.y=ny,++now.step;
				ull hx=haxi(now);
				if(!vis[hx]&&gujia(now)+now.step<21){
					q.push(now);
					vis[hx]=true;
				}
				swap(now.a[x][y],now.a[nx][ny]);
				now.x-=dx[i],now.y-=dy[i],--now.step;
			}
		}
	}
	return -1;
}
int main() {
	int t;
	scanf("%d",&t);
	while(t--){
		Sta s;
		rep(i,0,6)
		rep(j,0,i+1){
			scanf("%d",&s.a[i][j]);
			if(s.a[i][j]==0)s.x=i,s.y=j;
		}
		int ans=bfs(s);
		if(ans==-1)puts("too difficult");else printf("%d
",ans);
	}
	return 0;
}

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
const ll mod=1000000007;
ll qpow(ll a,ll b) {ll res=1;for(a%=mod;b;b>>=1,a=a*a%mod)if(b&1)res=res*a%mod;return res;}
VI BM(VI s) {//c[0]s[0]+c[1]s[1]+...=0
	VI C(1,1),B(1,1);
	int L=0,m=1,rev=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*rev%mod;
			C.resize(SZ(B)+m);
			rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
			L=n+1-L; B=T; rev=qpow(d,mod-2); m=1;
		} else {
			ll c=mod-d*rev%mod;
			C.resize(SZ(B)+m);
			rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
			++m;
		}
	}
	rep(i,0,SZ(C))printf("%dx[%d]%s",C[i],i,i+1==SZ(C)?"=0
":"+");
	return C;
}
调用：BM(VI{31,197,1255,7997})