Recursive sequence 矩阵快速幂 + 组合数 非线性变线性，利用到了组合数（杨辉三角求解快）

Recursive sequence

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status

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  . 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 <   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.

这个做了好久，真是醉了，就是查不出结果，自己动手算也没错，弄了大半天结果是因为自己的在求组和数的时候出错了，我。。。。

求组合数的时候可以用杨辉三角进行求解，代码如下：

void yanghui() {
	int i, j;
	C[0][0] = 1;
	for(i=1; i<=50; ++i) {
		C[i][0] = C[i][i] = 1;
		for(j=1; j

原理图：



AC代码：

#include
#include
#include
#include
#define maxn 8
using namespace std;
typedef long long ll;
ll  k;
const ll mod = 2147493647;
int n = 7;
int rol, col;
struct Matrix {
	ll ma[maxn][maxn];
};

Matrix multi(Matrix a, Matrix b) {
	Matrix ans;
	memset(ans.ma, 0, sizeof(ans.ma));
	int i, j, k;
	for(i=1; i<=n; ++i)
		for(j=1; j<=n; ++j)
			for(k=1; k<=n; ++k)
				ans.ma[i][j] = (ans.ma[i][j] + a.ma[i][k]*b.ma[k][j]) % mod;
    return ans;
}

Matrix pow(Matrix a, ll x) {
	Matrix ans;
	memset(ans.ma, 0, sizeof(ans.ma));
	int i, j;
	for(i=1; i<=n; ++i)
		ans.ma[i][i] = 1;
	while(x) {
		if(x & 1)
			ans = multi(ans, a);
		a = multi(a, a);                         
		x >>= 1;
	}
	return ans;
}

int main()
{
	int t;
	ll f[3];
	scanf("%d",&t);
	while(t--) {
		Matrix ans, b;
		memset(b.ma, 0, sizeof(b.ma));
		scanf("%lld%lld%lld",&k, &f[1], &f[2]);
		if(k<=2) {
			printf("%lld\n",f[k]%mod);
			continue;
		}
		int i, j;
		ll num1[7] = {f[1], f[2], 16, 8, 4, 2, 1};
		ll num2[7][7] = {{0,2,0,0,0,0,0}, {1,1,0,0,0,0,0},{0,1,1,0,0,0,0},{0,4,4,1,0,0,0},{0,6,6,3,1,0,0},{0,4,4,3,2,1,0},{0,1,1,1,1,1,1}};
		for(i=1; i<=n; ++i)
			b.ma[1][i] = num1[i-1];
		for(i=1; i<=n; ++i)
			for(j=1; j<=n; ++j)
				ans.ma[i][j] = num2[i-1][j-1];
		ans = pow(ans, k-2);
		ans =multi(b, ans);
		printf("%lld\n",ans.ma[1][2]);
	}
	return 0;
}