HDU 4990 Reading comprehension 简单矩阵快速幂

Problem Description

Read the program below carefully then answer the question.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>

const int MAX=100000*2;
const int INF=1e9;

int main()
{
  int n,m,ans,i;
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    ans=0;
    for(i=1;i<=n;i++)
    {
      if(i&1)ans=(ans*2+1)%m;
      else ans=ans*2%m;
    }
    printf("%d\n",ans);
  }
  return 0;
}

Input

Multi test cases，each line will contain two integers n and m. Process to end of file.
[Technical Specification]
1<=n, m <= 1000000000

Output

For each case，output an integer，represents the output of above program.

Sample Input

1 10
3 100

Sample Output

1
5

题意：给出了原程序，显然，看到内涵的递推式明显可以使用矩阵快速幂了。

思路：原递推式是当n为偶数时fn=2*f(n-1)+1 奇数时fn=2*f(n-1) 找规律得到递推式为 f(n) = *f(n-1)+*f(n-2) +*1

（下面是没学过线代的鶸的一点理解）

其实可以把矩阵看做一个储存多数据的容器，例如这题

运算为等式右红框分别乘以左边三个红框得到的三个值

对应的就是等式左侧的f(n),f(n-1),1。

化简

 #include <stdio.h>
 #include <algorithm>
 #include <iostream>
 #include <string.h>
 #define ll __int64
 using namespace std;

 ll mod;
 struct matrix
 {
     ll x[][];
     void init()
     {
         for(int i = ; i < ; i++)
             for(int j = ; j < ; j++)
                     x[i][j] = ;
     }
 };

 matrix mul(matrix a, matrix b)
 {
     matrix c;
     c.init();
     for( int i = ; i < ; i++)
         for(int j = ; j < ; j++)
         {
             for(int k = ; k < ; k++)
             {
                 c.x[i][j] += a.x[i][k] * b.x[k][j];
             }
             c.x[i][j] %= mod;
         }
     return c;
 }
 matrix powe(matrix x, ll n)
 {
     matrix r;
     r.init();
     r.x[][] = r.x[][] = r.x[][] = ; //初始化

     while(n)
     {
         if(n & )
             r = mul(r , x);
         x = mul(x , x);
         n >>= ;
     }
     return r;
 }
 int main()
 {

     ll x, y, n, ans;
     //while(~scanf("%I64d%I64d", &n, &mod))
     while(cin >> n >> mod)
     {
         if(n == )
             printf("%I64d\n", %mod);
         else if(n == )
             printf("%I64d\n", %mod);
         else
         {
             matrix d;
             d.init();
             d.x[][] = ;
             d.x[][] = ;
             d.x[][] = ;
             d.x[][] = ;
             d.x[][] = ;

             d = powe(d, n - );

             ans = d.x[][] *  + d.x[][] *  + ; //如果使用手动乘，不知为何还要再判断

             matrix e;
             e.init();
             e.x[][] = ;
             e.x[][] = ;
             e.x[][] = ;
             d = mul(e , d);
             /*if( n % 2 ) //再判断
                 ans-=2;
             else
                 ans-=1;*/
             cout << d.x[][] % mod << endl;
         }
     }
 }