杭电1005

小鱼儿 2022-06-05 11:52 157阅读 0赞

# 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][]

Recommend

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;
    }

[ZJCPC2004]: http://acm.hdu.edu.cn/search.php?field=problem&key=ZJCPC2004&source=1&searchmode=source