## page was renamed from 程序设计练习20——zju2105——Number Sequence
= Number Sequence =
http://acm.zju.edu.cn/show_problem.php?pid=2105

Time limit: 1 Seconds   

Memory limit: 32768K  

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
}}}
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{{{#!cplusplus
/*written by proby*/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
//336 is the LCM of all cases
vector<int> temList;

int Calc(int a, int b, int n){
	if(temList[n] >= 0){
		return temList[n];
	}
	if(n==1 || n==2){
		temList[n]=1;
		return temList[n];
	}
	temList[n] = (a*Calc(a, b, n-1) + b*Calc(a, b, n-2))%7 ;
	return temList[n];
}

int MyMod(double n){
	while(n > 340){
		n-=336;
	}

	return int(n);
}

void MyClear(int& n){
	n=-1;
}

void MyPrint(int n){
	cout<<n<<" ";
}
int main(){
	int a, b, res;
	double n;
	cin>>a>>b>>n;
	while(a!=0){
		res=MyMod(n);
		if(temList.capacity() < res+1){
			//temList.reserve(res+1);
			temList.resize(res+1);
		}
		for_each(temList.begin(), temList.begin()+res+1, MyClear); //"+1" is very necessary
		cout<<Calc(a, b, res)<<endl;
		//for_each(temList.begin(), temList.begin()+res+1, MyPrint); cout<<endl;
		cin>>a>>b>>n;
	}	
	return 1;
}
}}}


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