HDU 2256 Problem of Precision 构造矩阵

Problem of Precision

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 948    Accepted Submission(s): 554

Problem Description

 

Input

The first line of input gives the number of cases, T. T test cases follow, each on a separate line. Each test case contains one positive integer n. (1 <= n <= 10^9)

Output

For each input case, you should output the answer in one line.

Sample Input

3 1 2 5

 

Sample Output

9 97 841

/*
HDU 2256 构造矩阵 
题解 见上图  
*/
#include<iostream>
#include<stdio.h>
using namespace std;
#define M 1024

struct matrix{
  int map[2][2];
};
matrix mat,ans;

void init()
{
  int i,j;
  mat.map[0][0]=5;
  mat.map[0][1]=12;
  mat.map[1][0]=2;
  mat.map[1][1]=5;
  
  for(i=0;i<2;i++)
    for(j=0;j<2;j++)
      ans.map[i][j]=(i==j);
}

matrix mul(matrix a,matrix b)
{
  int i,j,k;
  matrix c;
  for(i=0;i<2;i++)
    for(j=0;j<2;j++)
    {
      c.map[i][j]=0;
      for(k=0;k<2;k++)
      c.map[i][j]+=a.map[i][k]*b.map[k][j];
      c.map[i][j]%=M;
    }
  return c;
}

void pow(int n)
{
  for(;n;n>>=1)
  {
    if(n&1)
      ans=mul(ans,mat);
    mat=mul(mat,mat);
  }
}

int main()
{
  int t,i,j,n,sum;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d",&n);
    init();
    pow(n);
    sum=(ans.map[0][0]*2-1)%M;
    printf("%d\n",sum);
  }
  return 0;
}