1005 Number Sequence

最新推荐文章于 2021-03-20 16:05:55 发布

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本文详细介绍了如何通过给定的初始值和递推公式，计算特定数列的第n项，同时提供了求解数列周期性的方法。

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

Sample Output

2
5

Author

CHEN, Shunbao

Source

ZJCPC2004

找规律题，因为对7取余数，因此49必定是一个循环。也可以像如下代码一样，求出具体的循环的最短周期。

#include<iostream>
using namespace std;

int main()
{
	int a[100],A,B,I,n;
	a[1]=1;
	a[2]=1;
	while(cin>>A>>B>>n)
	{
		if(A==0&&B==0&&n==0)	break;
		int cnt=0;
		for(i=3;i<=99;i++)
		{
			a[i]=(A*a[i-1]+B*a[i-2])%7;
			if(a[i]==1&&a[i-1]==1)	break;
			if(a[i]==0&&a[i-1]==0) {cnt=1;break;}
		}
		if(cnt&&n>i-2){cout<<"0"<<endl;continue;}
		i-=2;
		n%=i;
		if(n==0)n=i;
		cout<<a[n]<<endl;
	}
}