Gibonacci number

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李逝水

最新推荐文章于 2024-02-26 17:19:49 发布

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本文介绍了一种类似于斐波那契数列的Gibonacci数列问题，通过给定特定位置的数值来逆向求解初始随机整数，并进一步计算数列中任意位置的值。文章提供了一个具体的C++实现示例。

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Gibonacci number

Time Limit: 2 Seconds Memory Limit: 32768 KB

In mathematical terms, the normal sequence F(n) of Fibonacci numbers is defined by the recurrence relation

F(n)=F(n-1)+F(n-2)

with seed values

F(0)=1, F(1)=1

In this Gibonacci numbers problem, the sequence G(n) is defined similar

G(n)=G(n-1)+G(n-2)

with the seed value for G(0) is 1 for any case, and the seed value for G(1) is a random integer t, (t>=1). Given the i-th Gibonacci number value G(i), and the number j, your task is to output the value for G(j)

Input

There are multiple test cases. The first line of input is an integer T < 10000 indicating the number of test cases. Each test case contains 3 integers i, G(i) and j. 1 <= i,j <=20, G(i)<1000000

Output

For each test case, output the value for G(j). If there is no suitable value for t, output -1.

Sample Input

4
1 1 2
3 5 4
3 4 6
12 17801 19

Sample Output

思路：设第二位为t，可以知道第i为得公式，然后算出t,后面就是fibbonacci的计算方法

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--){
		int a,b,c,i,T;
		cin>>a>>b>>c;
		unsigned long long s1[23],s2[23];
		s1[0]=1;	s1[1]=0;
		s2[0]=0;	s2[1]=1;
		for(i=2;i<=20;i++){
			s1[i]=s1[i-1]+s1[i-2];
			s2[i]=s2[i-1]+s2[i-2];
		}
		if(a==1)
			T=b;
		else
			if((b-s1[a])%s2[a]==0)
				T=(b-s1[a])/s2[a];
			else
				T=-1;
		unsigned long long f[23];
		f[0]=1;
		f[1]=T;
		if(T<1)
			cout<<"-1"<<endl;
		else
		{
			for(i=2;i<=c;i++)
				f[i]=f[i-1]+f[i-2];
			//cout<<f[c]<<endl;
			printf("%lld\n",f[c]);
		}
		
	}
	return 0;
}