杭电1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 62216    Accepted Submission(s): 14226

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

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

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JGShining

#include <iostream>
#include <vector>
using namespace std;

void fn(int A, int B, int n);
void fn(int A, int B, int n)
{
    
    int cirlce[50];
    cirlce[1] = 1;
    cirlce[2] = 1;
    if (n == 1 || n == 2)  
    {
        cout << 1 << endl;
        return;
    }
    int i;
    for ( i = 3;i < 50 ; ++ i) //找到循环节
    {
        cirlce[i] = (A*cirlce[i -1] + B*cirlce[i -2])%7;
        if (cirlce[i -1] == 1 && cirlce[ i ] ==1)
            break;
    }
    if (n%(i -2) == 0)
    {
        cout << cirlce[i -2]<< endl;
    }
    else
    {
        cout << cirlce[n%(i -2)] << endl;
    }
}


int main(int argc, char **argv)
{
    int A, B, n;
    while (cin>>A>>B>>n) {
        if (A == 0 && B == 0 && n == 0)
            break;
        fn(A, B, n);
        
    }
    return 0;
}