HDU 4565 So Easy! 矩阵快速幂 + 共轭数

  1 // File Name: temp.cpp
  2 // Author: darkdream
  3 // Created Time: 2014年09月17日 星期三 11时35分45秒
  4 
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25 using namespace std;
 26 LL n ,m; 
 27 LL ta,tb;
 28 struct Matrix
 29 {
 30    LL  mat[2][2];
 31    void clear()
 32    {
 33       memset(mat,0,sizeof(mat));
 34    }
 35    void output()
 36    {
 37      for(int i =0  ;i < n ;i ++)
 38      {
 39        for(int j = 0 ;j < n ;j ++)
 40            printf("%I64d ",mat[i][j]);
 41      printf("\n");
 42      }
 43    }
 44    void init()
 45    {
 46       clear();
 47       n = 2 ;
 48       mat[0][0] = ta % m ; 
 49       mat[0][1] = 1;
 50       mat[1][0] = tb % m ;
 51       mat[1][1] = ta % m ;
 52    }
 53    Matrix operator *(const Matrix &b) const
 54    {
 55        Matrix ret;
 56        ret.clear();
 57        for(int i = 0 ;i < n ;i ++)
 58            for(int j = 0;j < n;j ++)
 59            {
 60                for(int k = 0 ;k < n ;k ++)
 61                {
 62                  ret.mat[i][j] =(ret.mat[i][j] + mat[i][k] * b.mat[k][j]) % m ; // 第I 行  第J  列        
 63                }
 64            }
 65        return ret;
 66    }
 67 };
 68 Matrix Pow( Matrix a ,LL t )
 69 {
 70   Matrix ret;
 71   ret.clear();
 72   for(int i = 0 ;i < n ;i ++)
 73        ret.mat[i][i] = 1;
 74   Matrix tmp = a; 
 75   while(t)
 76   {
 77       if(t&1) ret = ret * tmp;
 78       tmp = tmp * tmp;
 79       t >>=1;
 80   }
 81   return ret;
 82 }
 83 int main(){
 84     LL l ;
 85     while(scanf("%I64d %I64d %I64d %I64d",&ta,&tb,&l,&m) != EOF)
 86     {
 87         Matrix  a;
 88         a.init();
 89         if(l == 1)
 90         {
 91           printf("%I64d\n",(2* ta)%m);
 92           continue;
 93         }
 94         a = Pow(a,l-1);
 95         //a.output();
 96         LL ans = (((a.mat[1][0]  + (a.mat[1][1] * ta)%m)%m)*2)%m; 
 97         printf("%I64d\n",ans);
 98     }
 99     return 0;
100 }