【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：

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

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

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

\begin{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 = ;

 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, , sizeof(a));
         for(int i = ; i <= maxn;i++){
             a[i][i] = ;
         }
     }
 };

 //矩阵乘法
 Matrix mul(Matrix a, Matrix b){
     Matrix ans;
     for(int i = ;i <= ;++i){
         for(int j = ;j <= ;++j){
             ans.a[i][j] = ;
             for(int k = ;k <= ;++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 & )
             ans = mul(ans,a);
         a = mul(a,a);
         b >>= ;
     }
     return ans;
 }

 void print(Matrix a){
     for(int i = ; i <= n;++i){
         for(int j = ;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 == ){
         cout<<a1%mod<<endl;
         return ;
     }
     if(n == ){
         cout<<a2%mod<<endl;
         return ;
     }
     Matrix base;
     Matrix ans;
     ans.a[][] = a2;ans.a[][] = a1;

     base.a[][] = p;base.a[][] = ;
     base.a[][] = q;base.a[][] = ;

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