hdu 5950

http://acm.hdu.edu.cn/showproblem.php?pid=5950

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 329    Accepted Submission(s): 183

Problem Description

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and i4. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right. 

 

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < 231 as described above.

 

Output

For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.

 

Sample Input

2 3 1 2 4 1 10

 

Sample Output

85 369

Hint

In the first case, the third number is 85 = 2*1十2十3^4. In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.

 

Source

2016ACM/ICPC亞洲區瀋陽站-重現賽（感謝東北大學）

已知f(n) = f(n-1) + 2*f(n-2) + n^4，給你f(1) = a , f(2) = b，以及n；求f(n)%mod;

以前不會這種有次方的矩陣快速冪，還以爲可以化簡找到什麼公式，看了別人的解法才懂，原來也不難。

把f（n）等於啥寫出來，再把f（n+1）寫出來，再構造矩陣去套就行了。

|1   2   1   4   6   4   1|     

|1   0   0   0   0   0   0|     

|0   0   1   4   6   4   1|     

|0   0   0   1   3   3   1|  

|0   0   0   0   1   2   1|     

|0   0   0   0   0   1   1|     

|0   0   0   0   0   0   1|    

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL mod=2147493647;
struct Mat{
    LL mat[10][10];
    Mat()
    {
        memset(mat,0,sizeof(mat));
    }
};
Mat ec(Mat a,Mat b)
{
    Mat c;
    for(int i=0;i<7;i++)
    {
        for(int k=0;k<7;k++)
        {
            for(int j=0;j<7;j++)
            {
                c.mat[i][j]=(c.mat[i][j]+(a.mat[i][k]*b.mat[k][j])%mod)%mod;
            }
        }
    }
    return c;
}
Mat ex(Mat a,LL b)
{
    Mat c;
    for(int i=0;i<7;i++)
    {
        c.mat[i][i]=1;
    }
    while(b)
    {
        if(b%2==1)
        {
            c=ec(c,a);
        }
        b=b/2;
        a=ec(a,a);
    }
    return c;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        LL n,a,b;
        scanf("%lld%lld%lld",&n,&a,&b);
        if(n==1)
        {
            printf("%lld\n",a);
            continue;
        }
        if(n==2)
        {
            printf("%lld\n",b);
            continue;
        }
        Mat c;
        c.mat[0][0]=1;c.mat[0][1]=2;c.mat[0][2]=1;c.mat[0][3]=4;c.mat[0][4]=6;c.mat[0][5]=4;c.mat[0][6]=1;
        c.mat[1][0]=1;
        c.mat[2][2]=1;c.mat[2][3]=4;c.mat[2][4]=6;c.mat[2][5]=4;c.mat[2][6]=1;
        c.mat[3][3]=1;c.mat[3][4]=3;c.mat[3][5]=3;c.mat[3][6]=1;
        c.mat[4][4]=1;c.mat[4][5]=2;c.mat[4][6]=1;
        c.mat[5][5]=1;c.mat[5][6]=1;
        c.mat[6][6]=1;

        Mat d;
        d.mat[0][0]=b;
        d.mat[1][0]=a;
        d.mat[2][0]=16LL;
        d.mat[3][0]=8LL;
        d.mat[4][0]=4LL;
        d.mat[5][0]=2LL;
        d.mat[6][0]=1LL;

        Mat e=ex(c,n-2);
        e=ec(e,d);
        printf("%lld\n",e.mat[0][0]%mod);
    }
    return 0;
}