題目傳送門

題目描述：給出一個數列的第一項和第二項，計算第n項。

遞推式是 f(n)=f(n-1)+2*f(n-2)+n^4.

由於n很大，所以肯定是矩陣快速冪的題目，但是矩陣快速冪只能解決線性的問題，n^4在這個式子中是非線性的，後一項和前一項沒有什麼直接關係，所以模擬賽的時候想破頭也不會做。

這裡要做一個轉換，把n^4變成一個線性的，也就是和（n-1）^4有關係的東西，而這個辦法就是：

n^4=(n-1+1)^4=(n-1)^4+4*(n-1)^3+6*(n-1)^2+4*(n-1)^1+(n-1)^0;

這個轉換就建立了某一項和前一項的關係，矩陣的F陣列就是  f[7]={ b , a , 81 , 27 , 9 , 3 , 1 };，整體的矩陣也很好構造，程式碼裡有。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string.h>
#include<sstream>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<bitset>
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
ll mod=2147493647;
inline int rd(void) {
	int x=0; int f=1;char s=getchar();
	while(s<'0'||s>'9') {	if(s=='-')f=-1;	s=getchar();}
	while(s>='0'&&s<='9') {	x=x*10+s-'0';s=getchar();}
	x*=f;return x;
}
ll n,a,b;
void mul(ll f[7],ll a[7][7]){
	ll c[7];
	CLR(c,0);
	for(int j=0;j<7;j++){
		for(int k=0;k<7;k++){
			c[j]=(c[j]+f[k]*a[k][j]%mod)%mod;
		}
	}
	memcpy(f,c,sizeof(c));
}
void mulself(ll a[7][7]){
	ll c[7][7];
	CLR(c,0);
	for(int i=0;i<7;i++){
		for(int j=0;j<7;j++){
			for(int k=0;k<7;k++){
				c[i][j]=(c[i][j]+a[i][k]*a[k][j]%mod)%mod;
			}
		}
	}
	memcpy(a,c,sizeof(c));
}
int main(){
	int T;
	cin>>T;
	while(T--){
		scanf("%lld%lld%lld",&n,&a,&b);
		if(n==1){
			printf("%lld\n",a);
		}else if(n==2){
			printf("%lld\n",b);
		}else{
			ll f[7]={b,a,81,27,9,3,1};
			ll a[7][7]=
			{{1,1,0,0,0,0,0},
			 {2,0,0,0,0,0,0},
			 {1,0,1,0,0,0,0},
			 {0,0,4,1,0,0,0},
			 {0,0,6,3,1,0,0},
			 {0,0,4,3,2,1,0},
			 {0,0,1,1,1,1,1}};
			n-=2;
			for(;n;n>>=1){
				if(n&1)mul(f,a);
				mulself(a);
			}
			printf("%lld\n",f[0]);
		}
	}
}
 

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3374    Accepted Submission(s): 1485

 

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.