[LOJ3086] [GXOI2019] 逼死强迫症

题目链接

LOJ:https://loj.ac/problem/3086

洛谷:https://www.luogu.org/problemnew/show/P5303

Solution

显然不考虑\(1\times 1\)的矩形就是斐波那契数列,设为\(g\),则\(g_n=g_{n-1}+g_{n-2}\)

设考虑的方案数为\(f\),那么可以枚举放哪里得到一个暴力式子:

\[\begin{align}f_n&=2\sum_{i=1}^{n}\sum_{j=i+2}^{n}g_{i-1}\cdot g_{n-j}\\ f_n&=2\sum_{i=1}^{n}g_{i-1}\sum_{j=i+2}^{n}g_{n-j}\\ f_n&=2\sum_{j=3}^{n}g_{n-j}\sum_{i=1}^{j-2}g_{i-1}\\ f_n&=2\sum_{j=3}^{n}g_{n-j}s_{j-3} \end{align} \]

\(s\)\(g\)的前缀和。

然后我们考虑\(f_{n+1}\)用已知量凑出来:

\[\begin{align} \frac{1}{2}f_{n+1}&=\sum_{i=3}^{n+1}g_{n+1-i}s_{i-3}\\ \frac{1}{2}f_{n+1}&=s_{n-2}+s_{n-3}+\sum_{i=3}^{n-1}g_{n+1-i}s_{i-3}\\ \frac{1}{2}f_{n+1}&=s_{n-2}+s_{n-3}+\sum_{i=3}^{n-1}(g_{n-i}+g_{n-i-1})s_{i-3}\\ \frac{1}{2}f_{n+1}&=s_{n-2}+s_{n-3}+\sum_{i=3}^{n}g_{n-i}s_{i-3}-s_{n-3}+\sum_{i=3}^{n-1}g_{n-i-1}s_{i-3}\\ f_{n+1}&=2s_{n-2}+f_n+f_{n-1} \end{align} \]

其实到这里已经可以矩阵维护了,但是那样矩阵会设的很大,对于前缀和我们考虑如何递推:

\[s_n=2+\sum_{i=2}^{n}g_i=2+\sum_{i=2}^{n}g_{i-1}+g_{i-2}=s_{n-1}+s_{n-2}+1 \]

那么变一下就是:

\[s_n=s_{n+2}-s_{n-1}-1=g_{n+2}-1 \]

那么我们只需要开\(5\times 5\)的矩阵维护就好了,复杂度\(O(125T\log n)\)

#include<bits/stdc++.h>
using namespace std;

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}

void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++) 

const int maxn = 2e5+10;
const int inf = 1e9;
const lf eps = 1e-8;
const int mod = 1e9+7;

int n,T;

const int tmp[5][5] =
    {{1,1,0,0,0},
     {1,0,0,0,0},
     {2,0,1,1,0},
     {0,0,1,0,0},
     {mod-2,0,0,0,1}};

int add(int x,int y) {return x+y>=mod?x+y-mod:x+y;}
int del(int x,int y) {return x-y<0?x-y+mod:x-y;}
int mul(int x,int y) {return 1ll*x*y-1ll*x*y/mod*mod;}

struct Matrix {
    int a[6][6];
    Matrix () {memset(a,0,sizeof a);}
    Matrix operator * (const Matrix &r) const {
        Matrix res;
        for(int i=0;i<5;i++)
            for(int j=0;j<5;j++)
                for(int k=0;k<5;k++)
                    res.a[i][j]=add(res.a[i][j],mul(a[i][k],r.a[k][j]));
        return res;
    }
    void print() {
        FOR(i,0,4){FOR(j,0,4)printf("%d ",a[i][j]);puts("");}puts("");
    }
};

Matrix qpow(Matrix a,int x) {
    Matrix res;for(int i=0;i<5;i++) res.a[i][i]=1;
    for(;x;x>>=1,a=a*a) if(x&1) res=res*a;
    return res;
}

void solve() {
    read(n);if(n<=2) return puts("0"),void();
    Matrix ans,res;
    FOR(i,0,4) FOR(j,0,4) res.a[i][j]=tmp[i][j];
    ans.a[0][2]=2,ans.a[0][3]=1,ans.a[0][4]=1;
    write((ans*qpow(res,n-2)).a[0][0]);
}

int main() {
    read(T);while(T--) solve();
    return 0;
}
posted @ 2019-05-10 16:32  Hyscere  阅读(112)  评论(0编辑  收藏  举报