HDOJ--1005Number Sequence

原题链接

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

 

思路：因为对7求余，每种变量只有7种可能性，题目有两个变量，就有7*7=49种可能性

代码

#include<stdio.h>
int main()
{
	int fn[51]={1,1,1};
	int A,B,N;
	int i,n;
	while(scanf("%d%d%d",&A,&B,&N)&&(A|B|N))
	{
		for(i=3;i<=N%49;i++)
		{
			fn[i]=((A*fn[i-1])+(B*fn[i-2]))%7;
		}
		printf("%d\n",fn[N%49]);
	}
	return 0;
}

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