HDU5950 Recursive sequence (矩阵快速幂）

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 3832    Accepted Submission(s): 1662

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

HintIn 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亚洲区沈阳站-重现赛（感谢东北大学）

f(n)=f(n-1)+2f(n-2)+n^4

f(n)	1	2	1	0	0	0	0	f(n-1)
f(n-1)	1	0	0	0	0	0	0	f(n-2)
(n+1)^4	0	0	1	4	6	4	1	n^4
(n+1)^3	0	0	0	1	3	3	1	n^3
(n+1)^2	0	0	0	0	1	2	1	n^2
(n+1)	0	0	0	0	0	1	1	n
1	0	0	0	0	0	0	1	1

#include<iostream>
#include<string.h>
#include<algorithm>
#define inf 2147493647
#define ll long long
using namespace std;
struct mat{
         ll t[7][7];
         mat(){
                  memset(t,0,sizeof(t));
         }
         mat operator*(mat b){
                  mat c;
                  for(int i=0;i<7;i++)
                           for(int j=0;j<7;j++)
                                    for(int k=0;k<7;k++)
                                    c.t[i][j]=(c.t[i][j]%inf+t[i][k]*b.t[k][j])%inf;
                  return c;
         }
};
mat pow(int nn,mat B,mat A)
{
         while(nn){
                  if(nn%2==1)
                           B=A*B;
                  A=A*A;
                  nn/=2;
         }
         return B;
}
int main()
{
         int T,n;
         ll a[7][7]=
         {1,2,1,0,0,0,0,
          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};
         mat A;
         for(int i=0;i<7;i++)
                  for(int j=0;j<7;j++)
                           A.t[i][j]=a[i][j];
         mat B;
         B.t[2][0]=81;
         B.t[3][0]=27;
         B.t[4][0]=9;
         B.t[5][0]=3;
         B.t[6][0]=1;
         scanf("%d",&T);
         while(T--)
         {
                  scanf("%d%lld%lld",&n,&B.t[1][0],&B.t[0][0]);
                  if(n==1)
                           printf("%lld\n",B.t[1][0]);
                  else if(n==2)
                           printf("%lld\n",B.t[0][0]);
                  else{
                           mat C=pow(n-2,B,A);
                           printf("%lld\n",C.t[0][0]%inf);
                  }
         }
         return 0;
}