​​原题链接​​

考察:矩阵快速幂

fw本f,想到了把f[i-1]拆开,但没想到只拆一个啊!!!

思路:


\[f[i] = 2 \times f[i-1]+1\qquad(i为奇数)\]


\[f[i] = f[i-1] + 2 \times f[i-2] +1\]


\[f[i] = 2 \times f[i-1] \qquad (i为偶数)\]


\[f[i] = f[i-1] + 2 \times f[i-2]+1 \]


Code

#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 3; 
int n,m;
void mul(int f[],int a[][N])
{
  int res[N] = {0};
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++)
      res[i] = (res[i]+(LL)f[j]*a[j][i])%m;
  memcpy(f,res,sizeof res);
}
void mul(int a[][N])
{
  int res[N][N] = {0};
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++)
     for(int k=0;k<N;k++)
       res[i][j] = (res[i][j]+(LL)a[i][k]*a[k][j])%m;
  memcpy(a,res,sizeof res);
}
int main()
{
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    int f[N] = {1,0,1};
    int a[N][N] = {
        {1,1,0},
        {2,0,0},
        {1,0,1}
    };
    while(n)
    {
      if(n&1) mul(f,a);
      mul(a);
      n>>=1;
    }
    printf("%d\n",f[1]);
  }
  return 0;
}