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 ",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) = 1*f(n-1)+2*f(n-2) +1*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[3][3];
    void init()
    {
        for(int i = 0; i < 3; i++)
            for(int j = 0; j < 3; j++)
                    x[i][j] = 0;
    }
};

matrix mul(matrix a, matrix b)
{
    matrix c;
    c.init();
    for( int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++)
        {
            for(int k = 0; k < 3; 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[1][1] = r.x[0][0] = r.x[2][2] = 1; //初始化

    while(n)
    {
        if(n & 1)
            r = mul(r , x);
        x = mul(x , x);
        n >>= 1;
    }
    return r;
}
int main()
{

    ll x, y, n, ans;
    //while(~scanf("%I64d%I64d", &n, &mod))
    while(cin >> n >> mod)
    {
        if(n == 2)
            printf("%I64d
", 2%mod);
        else if(n == 1)
            printf("%I64d
", 1%mod);
        else
        {
            matrix d;
            d.init();
            d.x[0][0] = 1;
            d.x[0][1] = 1;
            d.x[1][0] = 2;
            d.x[2][0] = 1;
            d.x[2][2] = 1;

            d = powe(d, n - 2);

            ans = d.x[0][0] * 2 + d.x[1][0] * 2 + 1; //如果使用手动乘，不知为何还要再判断

            matrix e;
            e.init();
            e.x[0][0] = 2;
            e.x[0][1] = 1;
            e.x[0][2] = 1;
            d = mul(e , d);
            /*if( n % 2 ) //再判断
                ans-=2;
            else
                ans-=1;*/
            cout << d.x[0][0] % mod << endl;
        }
    }
}