HDU 2256 Problem of Precision 解题报告（矩阵快速幂 + 构造）

Problem of Precision

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

Total Submission(s): 844    Accepted Submission(s): 489


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

 

    解题报告： 想了很久，没有思路……看了解题报告

    推理过程如下：



    最终结果表明，中途取模也不会有影响，用矩阵快速幂求的xn即可。代码如下

#include <cstdio>
#include <cstring>
#include <map>
#include <cmath>
#include <algorithm>
using namespace std;

const int mod=1024;

struct Matrix
{
int a[2][2];
Matrix(int t=0)
{
memset(a, 0, sizeof(a));
a[0][0]=a[1][1]=t;
}

Matrix operator*(const Matrix& cmp) const
{
Matrix c;
for(int i=0;i<2;i++) for(int j=0;j<2;j++)
{
long long t=0;
for(int k=0;k<2;k++) t+=a[i][k]*cmp.a[k][j];
c.a[i][j]=t%mod;
}
return c;
}
} p;

Matrix powM(Matrix a, int b)
{
Matrix res(1);
while(b)
{
if(b&1)
res=res*a;
a=a*a;
b>>=1;
}
return res;
}

int main()
{
p.a[0][0]=5;
p.a[0][1]=2;
p.a[1][0]=12;
p.a[1][1]=5;

int T;
scanf("%d", &T);
while(T--)
{
int n;
scanf("%d", &n);

Matrix r = powM(p, n-1);
int res = 2*(5*r.a[0][0]+2*r.a[1][0])-1;
printf("%d\n", res%mod);
}
}