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
",B.t[1][0]);
else if(n==2)
printf("%lld
",B.t[0][0]);
else{
mat C=pow(n-2,B,A);
printf("%lld
",C.t[0][0]%inf);
}
}
return 0;
}