Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30    Accepted Submission(s): 20


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亚洲区沈阳站-重现赛（感谢东北大学）
题意：f[i]=2*f[i-2]+f[i-1]+i^4
题解：[f[i],f[i-1],i^4,i^3,i^2,i,1]
  1 1 0 0 0 0 0
  2 0 0 0 0 0 0
  1 0 1 0 0 0 0
  4 0 4 1 0 0 0
  6 0 6 3 1 0 0
  4 0 4 3 2 1 0
  1 0 1 1 1 1 1
构造矩阵  矩阵快速幂
  1 /******************************
  2 code by drizzle
  3 blog: www.cnblogs.com/hsd-/
  4 ^ ^    ^ ^
  5  O      O
  6 ******************************/
  7 #include<bits/stdc++.h>
  8 #include<map>
  9 #include<set>
 10 #include<cmath>
 11 #include<queue>
 12 #include<bitset>
 13 #include<math.h>
 14 #include<vector>
 15 #include<string>
 16 #include<stdio.h>
 17 #include<cstring>
 18 #include<iostream>
 19 #include<algorithm>
 20 #pragma comment(linker, "/STACK:102400000,102400000")
 21 using namespace std;
 22 #define  A first
 23 #define B second
 24 const int mod=1000000007;
 25 const int MOD1=1000000007;
 26 const int MOD2=1000000009;
 27 const double EPS=0.00000001;
 28 typedef __int64 ll;
 29 const ll MOD=1000000007;
 30 const int INF=1000000010;
 31 const ll MAX=1ll<<55;
 32 const double eps=1e-8;
 33 const double inf=~0u>>1;
 34 const double pi=acos(-1.0);
 35 typedef double db;
 36 typedef unsigned int uint;
 37 typedef unsigned long long ull;
 38 const ll k=2147493647;
 39 struct matrix
 40 {
 41    ll m[10][10];
 42 } ans,exm;
 43 struct matrix matrix_mulit1(struct matrix aa,struct matrix bb)
 44 {
 45     struct matrix there;
 46     for(int i=0;i<7;i++)
 47     {
 48         for(int j=0;j<7;j++)
 49         {
 50             there.m[i][j]=0;
 51             for(int u=0;u<7;u++)
 52             there.m[i][j]=(there.m[i][j]+aa.m[i][u]*bb.m[u][j]%k)%k;
 53         }
 54     }
 55     return there;
 56 }
 57 struct matrix matrix_mulit2(struct matrix aa,struct matrix bb)
 58 {
 59     struct matrix there;
 60         for(int j=0;j<7;j++)
 61         {
 62             there.m[0][j]=0;
 63             for(int u=0;u<7;u++)
 64             there.m[0][j]=(there.m[0][j]+aa.m[0][u]*bb.m[u][j]%k)%k;
 65         }
 66     return there;
 67 }
 68 ll matrix_quick(ll aa,ll bb,ll gg)
 69 {
 70 exm.m[0][0]=1;exm.m[0][1]=1;exm.m[0][2]=0;exm.m[0][3]=0;exm.m[0][4]=0;exm.m[0][5]=0;exm.m[0][6]=0;
 71 exm.m[1][0]=2;exm.m[1][1]=0;exm.m[1][2]=0;exm.m[1][3]=0;exm.m[1][4]=0;exm.m[1][5]=0;exm.m[1][6]=0;
 72 exm.m[2][0]=1;exm.m[2][1]=0;exm.m[2][2]=1;exm.m[2][3]=0;exm.m[2][4]=0;exm.m[2][5]=0;exm.m[2][6]=0;
 73 exm.m[3][0]=4;exm.m[3][1]=0;exm.m[3][2]=4;exm.m[3][3]=1;exm.m[3][4]=0;exm.m[3][5]=0;exm.m[3][6]=0;
 74 exm.m[4][0]=6;exm.m[4][1]=0;exm.m[4][2]=6;exm.m[4][3]=3;exm.m[4][4]=1;exm.m[4][5]=0;exm.m[4][6]=0;
 75 exm.m[5][0]=4;exm.m[5][1]=0;exm.m[5][2]=4;exm.m[5][3]=3;exm.m[5][4]=2;exm.m[5][5]=1;exm.m[5][6]=0;
 76 exm.m[6][0]=1;exm.m[6][1]=0;exm.m[6][2]=1;exm.m[6][3]=1;exm.m[6][4]=1;exm.m[6][5]=1;exm.m[6][6]=1;
 77 ans.m[0][0]=bb;ans.m[0][1]=aa;ans.m[0][2]=16;ans.m[0][3]=8;ans.m[0][4]=4;ans.m[0][5]=2;ans.m[0][6]=1;
 78      gg-=2;
 79      while(gg)
 80      {
 81         if(gg&1)
 82         ans=matrix_mulit2(ans,exm);
 83         exm=matrix_mulit1(exm,exm);
 84         gg >>= 1;
 85      }
 86      return ans.m[0][0];
 87 }
 88 int t;
 89 ll n,a,b;
 90 int main()
 91 {
 92     scanf("%d",&t);
 93     for(int i=1;i<=t;i++)
 94     {
 95         scanf("%I64d %I64d %I64d",&n,&a,&b);
 96         if(n==1)
 97             printf("%I64d\n",a);
 98         else if(n==2)
 99             printf("%I64d\n",b);
100         else
101         printf("%I64d\n",(matrix_quick(a, b, n)+k)%k);
102     }
103     return 0;
104 }