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;
}