zju2105

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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

---

/*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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