1005 hdoj Number Sequence (java函数格式)

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 145944    Accepted Submission(s): 35462

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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题目的意思很明白，就是给你表达式让你推出给的数，运用函数的递归，我把java函数温习了一遍，在敲了遍，希望有所提高。为了防止忘记留下函数语法知识java 函数
奉上新鲜出炉的代码

<pre name="code" class="java">import java.util.*;
import java.math.*;
import java.io.*;
public class Main {
public static int fun(int x,int y,int z)
{
	if(z==1||z==2)
	{
		return 1;
	}
	else
	{
	 z = (x * fun(x, y, z - 1) + y * fun(x, y, z - 2)) %7;//递归
	return z;
	}
	
}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
    Scanner in = new Scanner(System.in);
    while(in.hasNext())
    {
    	int a,b,c;
    	a=in.nextInt();
    	b=in.nextInt();
    	c=in.nextInt();
    	if(a==0&&b==0&&c==0)
    	break;
    	else
    	{
    		System.out.println(fun(a,b,c%49));//防止超时%49和%7效果一样·，但%49不会超时
    	}
    }
    

	}

}