HDU 1005 Number Sequence（找规律）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 179056    Accepted Submission(s): 44500

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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题意：给出A， B，n， 求出f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7的值。其中1 <= A, B <= 1000, 1 <= n <= 100,000,000。显然是不能直接计算的，我们可以找规律。f(n)是要mod7的，所以f(n)的值最多只会有7*7 = 49种。找出了规律，写起来就会简短很多。

# include<iostream>
# include<cstdio>
# include<cstring>
using namespace std;
int f[50];

int main() {
	int a, b, n;
	while(~scanf("%d %d %d", &a, &b, &n)) {
		if(a == 0 && b == 0 && n == 0)
			break;
		
		f[1] = 1;
		f[2] = 1;
		for(int i = 3; i <= 48; i ++)
			f[i] = (a*f[i - 1] + b*f[i - 2])%7;
			
		printf("%d\n", f[n%49]);
	}
	return 0;
}