Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 148003    Accepted Submission(s): 35976

Problem Description A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n). Input The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed. Output For each test case, print the value of f(n) on a single line. Sample Input 1 1 3 1 2 10 0 0 0 Sample Output 2 5

最近學了簡單一點的二維的矩陣快速冪，發現十分好用。於是又找以前做過的遞推式的題目。大部分人做法應該是暴力打表發現迴圈節規律取前49項即可，但是這題也很適合用矩陣來加速求需要的某一項。只要會矩陣或者行列式的乘法就可以。另外為了寫起來自然美觀把函式改成了操作符過載。

（矩陣寫法可以有很多種，F1，F2位置互換、橫著寫或者更離譜也行，只要能得出正確結果即可）

若求出遞推式後只要注意一下指數到底應該是n還是n-1還是n-2，然後稍微特判一下，其他應該沒啥問題。

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
struct mat
{
	int pos[2][2];
	mat(){memset(pos,0,sizeof(pos));}
};
inline mat operator*(const mat &a,const mat &b)
{
	mat c;
	for (int i=0; i<2; i++)
	{
		for (int j=0; j<2; j++)
		{
			for (int k=0; k<2; k++)
			{
				c.pos[i][j]+=(a.pos[i][k]*b.pos[k][j])%7;
			}
		}
	}
	return c;
}
inline mat operator^(mat a,LL b)
{
	mat r;r.pos[0][0]=r.pos[1][1]=1;
	mat bas=a;
	while (b!=0)
	{
		if(b&1)
			r=r*bas;
		bas=bas*bas;
		b>>=1;
	}
	return r;
}
int main(void)
{
	ios::sync_with_stdio(false);
	int n,a,b;
	while (cin>>a>>b>>n&&(a||b||n))
	{
		if(n==1)
		{
			cout<<1<<endl;
			continue;
		}
		mat one,t;
		one.pos[0][0]=one.pos[1][0]=1;
		t.pos[0][0]=a,t.pos[0][1]=b;t.pos[1][0]=1;
		t=t^(n-2);
		one=t*one;
		cout<<one.pos[0][0]%7<<endl;
	}
	return 0;
}