hdu-4990 Reading comprehension(快速幂+矩阵)

• 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
  

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  题意：

  按照题目的代码运行肯定会超时，而且数据会爆int，所以我用了快速幂矩阵。 通项公式为f[n] = f[n -1] + 2 * f[n -2] + 1;

  ps：写矩阵相乘函数的时候自己脑抽竟然写反了两个参数。

  #include<iostream>
  #include<cstdio>
  #define ll long long
  using namespace std;
  
  struct node{
  	ll m[3][3];
  };
  
  int n,m;
  node mult(node a, node b){
  	node te;
  	for(int i = 0; i < 3; i++)
  		for(int j = 0; j < 3; j++){
  			te.m[i][j] = 0;
  			for(int k = 0; k < 3; k++){
  				te.m[i][j] += a.m[i][k] * b.m[k][j];
  				te.m[i][j] %= m; 
  			} 
  		}
  	return te;
  }
  
  ll powermod(){
  	node base={0, 2, 0, 1, 1, 0, 0, 1, 1}, ans={0, 1, 1, 0, 0, 0, 0, 0, 0};
  	while(n){
  		if(n&1) ans = mult(ans, base);
  		n = n / 2;
  		base =	mult(base, base);		
  	}	
  	return ans.m[0][0];
  }
  int main(){
  	while(scanf("%d%d", &n,&m) != EOF){
  		
  		printf("%I64d\n",powermod());
  	}
  	return 0;
  }
  

  
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