8.3.5 Problem of Precision

Problem of Precision

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 24 Accepted Submission(s): 20

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

共轭数？？

网上是这么说的

我只是说一下推理过程

(m+n)^k=an+根号6bn

(m-n)^k=an-根号6bn

我们把上面的k次方展开，那么只有奇数次方 （-n）^x(x属于奇数)才会对后面的产生影响，而影响加起来就是根号6bn (谢谢zhx～～)

/*
Author:wuhuajun
*/
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;

typedef long long ll;
typedef double dd;
const int maxn=210,mod=1024;

struct matrix
{
   int m[3][3];
} ans,base,c;
int n,T,sum;

void close()
{

exit(0);
}

matrix mul(matrix a,matrix b)
{
   memset(c.m,0,sizeof(c.m));
   for (int i=1;i<=2;i++)
      for (int j=1;j<=2;j++)
     for (int k=1;k<=2;k++)
     {
        c.m[i][j]+=a.m[i][k]*b.m[k][j];
        c.m[i][j] %= mod;
     }
   return c;
}

void work()
{
   memset(ans.m,0,sizeof(ans.m));
   ans.m[1][1]=ans.m[2][2]=1;
   base.m[1][1]=5;
   base.m[1][2]=12;
   base.m[2][1]=2;
   base.m[2][2]=5; 
   n--;
   while (n!=0)
   {
      if (n & 1)
             ans=mul(base,ans);
      n/=2;
      base=mul(base,base);
   }
   sum=ans.m[1][1]*5+ans.m[1][2]*2;
    sum*=2;
    sum--;
   sum %= mod;
   printf("%d
",sum);
}


void init()
{
   scanf("%d",&T);
   while (T--)
   {
      scanf("%d",&n);
      work();
   }
}

int main ()
{
   init();
   close();
   return 0;
}