HDU 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 102285    Accepted Submission(s): 24710

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


这是一道给公式，求答案类型的题目。

看到这道题第一个反应就是递推，不断的求直到答案的位置。

但是题目数据量太大，最多一亿的数据，用递推的话至少一亿次循环，限定时间一秒肯定超时。

很显然需要找规律。

对于每一个fn来说，有7种结果（0 1 2 3 4 5 6），所以对于公式  f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.则最多有7 * 7 = 49种可能

也就是说至多每49个结果就循环一次，所以对于给出的A，B，N；每次求到f[49]然后用Nmod49，输出f[N%49]即可。

代码如下：

#include <stdio.h>
#include <iostream>
#include <queue>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
int a, b, n;
int ans[50];
int main()
{
	int i;
	while(~scanf("%d%d%d", &a, &b, &n))
	{
		if(a == 0 && b == 0 && n == 0)
			break;
		ans[1] = 1, ans[2] = 1;
		for(i = 3; i < 50; i++)
		{
			ans[i] = (a * ans[i - 1] + b * ans[i - 2]) % 7;
		}
		printf("%d\n", ans[n % 49]);
	}
	return 0;
}