hdu 4990 Reading comprehension（矩阵乘法）

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

根据上面的代码可以发现就是让求一个数组的第n项

奇数项等于前一项*2+1，偶数项等于前一项*2：

a0 = 0；

a1 = a0*2+1;

a2 = a1*2；

a3 = a2*2+1;

a4 = a3*2;

如果直接构造矩阵求第n项好像不太好求，但是我们发现偶数项或者奇数项之间的公式是固定的，我们就是偶数项为例吧。

可以推出：

a2 = 4*a0 + 2;

a4 = 4*a2 + 2;

这样就可以很轻松的构造出矩阵，如果n是偶数的话，直接就可以求出答案，如果是奇数项，那么就*2+1.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;

int mod;

struct Matrix
{
long long m[2][2];
int n;
Matrix(int x)
{
n = x;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
m[i][j] = 0;
}
Matrix(int _n,int a[2][2])
{
n = _n;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
m[i][j] = a[i][j];
}
}
};
Matrix operator *(Matrix a,Matrix b)
{
int n = a.n;
Matrix ans = Matrix(n);
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
for(int k=0;k<n;k++)
{
ans.m[i][j] += (a.m[i][k]%mod)*(b.m[k][j]%mod)%mod;
ans.m[i][j] %= mod;
}
return ans;
}
Matrix operator ^(Matrix a,int k)
{
int n = a.n;
Matrix c(n);
int i,j;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
c.m[i][j] = (i==j);
for(;k;k>>=1)
{
if(k&1)
c=c*a;
a = a*a;
}
return c;
}

int main(void)
{
int n,m,i,j;
while(scanf("%d%d",&n,&m)==2)
{
mod = m;
int a[2][2] = { 4,0,
2,1};
Matrix A(2,a);
A = A^(n/2);
LL ans = A.m[1][0];
if(n % 2 == 1)
ans = (ans*2+1)%mod;
cout << ans << endl;
}

return 0;
}