hdu

题意：

然后求fn取p的余之后的答案，n 为10^18次方；

思路： 设xn为a的幂： 则可以得出xn = c * x(n -1) + x(n - 2) + b;

由费马小定理可以得出，fn %= p, 那么xn %= (mod - 1);

矩阵为: c b 1

              1 0 0 

              0 0  1

#include <iostream>#include <set>#include<bits/stdc++.h>using namespace std;typedef long long ll;ll n,p;struct Matrix{Matrix(){memset(maze,0,sizeof(maze));for(int i  = 0; i < 3; i ++)maze[i][i] = 1;}Matrix(ll c,ll b){maze[0][1] = maze[1][0] = maze[2][2] = 1;maze[0][0] = c;maze[0][2] = b;maze[1][1] = maze[1][2] = maze[2][0] = maze[2][1] = 0;}void show(){for(int i = 0; i < 3; i ++){for(int j = 0; j < 3; j ++){cout << maze[i][j] << " ";}cout << endl;}}ll maze[3][3];inline Matrix operator *(const Matrix& b){Matrix c;for(int i = 0; i < 3; i ++){for(int j = 0; j < 3; j ++){c.maze[i][j] = 0;for(int k = 0; k < 3; k ++){c.maze[i][j] += (maze[i][k] * b.maze[k][j]) % (p - 1);c.maze[i][j] %= (p - 1);}}}return c;}};ll a,b,c;Matrix pow(ll len){//cout << "Pass _ by" << endl;//cout << len << endl;Matrix ans;//ans.show();Matrix t(c,b);while(len){if(len & 1)ans = ans * t;len >>= 1;t = t * t;}return ans;}ll pows(ll x,ll n){ll ans = 1;ll t = x;while(n){if(n & 1)ans = ans * t % p;n >>= 1;t = (t * t) % p;}return ans;}ll solve(){if(n == 1)return 1;if(n == 2)return pow(a,b);Matrix temp = pow(n - 2);ll ans = temp.maze[0][0] * b + temp.maze[0][2];ans = pows(a,ans);return ans;}int main(){int Tcase;scanf("%d",&Tcase);while(Tcase --){scanf("%I64d%I64d%I64d%I64d%I64d",&n,&a,&b,&c,&p);cout << solve() << endl; }return 0;}