【洛谷】P1349广义斐波那契

题目链接：https://www.luogu.org/problemnew/show/P1349

题意：现在定义fib数列为 an = p * an-1 + q * an-2求第n项%m的答案。

题解：

egin{pmatrix} p & 1\ q & 0 end{pmatrix} *

egin{pmatrix} f(n-1)& f(n-2) end{pmatrix} =

egin{pmatrix} p*f(n-1) + q*f(n-2)& f(n-1) end{pmatrix}=

egin{pmatrix} f(n)& f(n-1) end{pmatrix}

这题有个玄学mod。矩阵乘法的时候记得要多mod一次。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define ll long long
const int maxn = 3;

ll n,mod;

//矩阵结构体 
struct Matrix{
    ll a[maxn][maxn];
    void init(){    //初始化为单位矩阵 
        memset(a, 0, sizeof(a));
        for(int i = 1; i <= maxn;i++){
            a[i][i] = 1;
        }
    }
};

//矩阵乘法 
Matrix mul(Matrix a, Matrix b){
    Matrix ans;
    for(int i = 1;i <= 2;++i){
        for(int j = 1;j <= 2;++j){
            ans.a[i][j] = 0;
            for(int k = 1;k <= 2;++k){
                ans.a[i][j] = ans.a[i][j] % mod + a.a[i][k] * b.a[k][j] % mod;
                ans.a[i][j] %= mod;
            }
        }
    } 
    return ans;
}

//矩阵快速幂 
Matrix qpow(Matrix a,ll b){
    Matrix ans;
    ans.init();
    while(b){
        if(b & 1)
            ans = mul(ans,a);
        a = mul(a,a);
        b >>= 1;
    }
    return ans;
}

void print(Matrix a){
    for(int i = 1; i <= n;++i){
        for(int j = 1;j <= n;++j){
            cout << a.a[i][j]%mod<< " ";
        }
        cout << endl;
    }
}

int main(){
    ll p,q,a1,a2;
    cin>>p>>q>>a1>>a2>>n>>mod;
    if(n == 1){
        cout<<a1%mod<<endl;
        return 0;
    }
    if(n == 2){
        cout<<a2%mod<<endl;
        return 0;
    }
    Matrix base;
    Matrix ans;
    ans.a[1][1] = a2;ans.a[1][2] = a1;

    base.a[1][1] = p;base.a[1][2] = 1;
    base.a[2][1] = q;base.a[2][2] = 0;

    ans = mul(ans,qpow(base,n-2));
    //print(ans);
    cout<<ans.a[1][1]%mod<<endl;
    return 0;
}