HDU1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 172545    Accepted Submission(s): 42607

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 31 2 100 0 0

 

Sample Output

25
//按照数学公式进行计算 找到循环 即从循环周期为48
#include <iostream>using namespace std;int main(){    int A,B,n,i;    int a[48];    cin >> A >> B >> n;    while( A!=0 || B!=0 || n!=0)    {        a[0]=a[1]=1;        for(i=2;i<48;i++)            a[i] = ( A * a[i-1] + B * a[i-2]) % 7;        cout << a[(n-1)%48] <<endl;        cin >> A >> B >> n;    }    return 0;}