矩阵快速幂（纯数学递推）

Problem of Precision

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2127    Accepted Submission(s): 1304

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

 

Source

HDOJ 2008 Summer Exercise（4）- Buffet Dinner

 

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#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <queue>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 1024
using namespace std;
typedef long long ll ;
struct node{
    int a[2][2];
    node()
    {
        memset(a , 0 , sizeof(a));
    }
};

node mul(node A , node B)
{
    node C ;
    for(int i = 0 ; i < 2 ; i ++)
    {
        for(int j = 0 ; j < 2 ; j++)
        {
            for(int k = 0 ; k < 2 ; k++)
            {
                C.a[i][j] = (C.a[i][j] + A.a[i][k] * B.a[k][j]) % mod;
            }
        }
    }
    return C;
}

node pow1(node A , int m)
{
    node ans ;
    for(int i = 0 ; i < 2 ; i++)
        ans.a[i][i] = 1 ;
    while(m)
    {
        if(m&1)
        {
            ans = mul(ans , A);
        }
        m >>= 1 ;
        A = mul(A , A);
    }
    return ans ;
}



int main()
{
    int t ;
    scanf("%d" , &t);
    while(t--)
    {
        int n ;
        scanf("%d" , &n);
        node A ,B ,C ;
        A.a[0][0] = 5 , A.a[0][1] = 12 ;
        A.a[1][0] = 2 , A.a[1][1] = 5 ;

        B.a[0][0] = 5 , B.a[1][0] = 2 ;

       
        C = mul(pow1(A , n-1) , B);
        printf("%d
" , (C.a[0][0]*2 - 1)%mod);

    }


    return 0 ;
}