杭电OJ 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 212774    Accepted Submission(s): 53788

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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).

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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.

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Output

For each test case, print the value of f(n) on a single line.

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Sample Input

1 1 3

1 2 10

0 0 0

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Sample Output

2

5

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错误思路：递归

一开始看这道题，想都没想直接函数递归，VS上跑的不亦乐乎，样本测试用例都通过了，欸心想肯定过，拿到OJ上提交，显示内存溢出，眉头一皱发现事情并不简单，仔细分析，n可能是一个超大的数，当数很大时，函数递归的深度不可想象，把n换成大数果然，VS上也出现了栈溢出，先把错误代码po出来

#include<math.h>
#include<iostream>
#include<string.h>
using namespace std;

long long f(long long n,int A,int B)
{
	long long result;
	if (n == 1 || n == 2)
		return 1;
	else
	{
		result = (A*f(n - 1,A,B) + B * f(n - 2,A,B)) % 7;
		return result;
	}
}
int main() {
	int A, B;
	long long n;
	while (cin >> A >> B >> n)
	{
		if (A == 0 && B == 0 && n == 0)
			return 0;
		if (A >= 1 && A <= 1000 && B >= 1 && B <= 1000 && n >= 1 && n <= 100000000)
		{
			cout << f(n,A,B)<<endl;
		}
	}
	return 0;
}

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正确思路：如题所述，发现数的顺序规律

仔细观察所给函数可以发现，无论f(n)怎么变，最终取模7后的答案，结果一定是[0,6]，这个地方理解到很重要；也就是说，f(n-1)和f(n-2)的范围同样是在[0,6]，那么当n>2时，两两组合得到的f(n)会有7*7=49种不同的答案，即结果取值会是49种一周期，之后所有的结果都会从者49种结果中产生，那么新代码的思路也就出来了，只需要将前49种结果都算出来，最后匹配n%49传参即可得到正确答案。

#include<math.h>
#include<iostream>
#include<string.h>
using namespace std;

int f(int n,int A,int B)
{
	int result;
	if (n == 1 || n == 2)
	{
		return 1;
	}
	else
	{
		result= (A*f(n - 1, A, B) + B * f(n - 2, A, B)) % 7;
		return result;
	}
}
int main() {
	int A, B;
	long n;
	while (cin >> A >> B >> n)
	{
		if (A == 0 && B == 0 && n == 0)
			return 0;
		if (A >= 1 && A <= 1000 && B >= 1 && B <= 1000 && n >= 1 && n <= 100000000)
		{
			cout << f((n%49),A,B)<<endl;
		}
	}
	return 0;
}