HDU 1005 Number Sequence(矩阵快速幂，快速幂模板)

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

 

以下是快速幂的模板， 求a^n， 为防止溢出，中间最好mod一哈

int main(){
	int n;
	int a;
	cin >> a>>n;
	int ans = 1;
	while(n){//a^n
		if(n&1){
			ans *= a; 
		}
		a*=a;
		n>>=1;
		ans %= 10000007;
	}
	cout << ans<

本题的递推式：

代码（含矩阵快速幂）：

#include
#include
#include
#include
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#include
#include