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最新推荐文章于 2019-07-11 16:56:37 发布

原创最新推荐文章于 2019-07-11 16:56:37 发布 · 187 阅读

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本文介绍了一种类似于斐波那契数列的Gibo数列，并提供了一个算法解决方案来求解特定条件下的数列值。通过对给定的Gibo数列种子值进行分析，文章详细解释了如何快速有效地计算出任意位置的数列值。

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Gibonacci number
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

2
8
-1
516847

斐波那契的变形，推个公式，G（n)=F(n-1)*G(1)+F(n-2)
注意n取1的时候和G(n)为负值的时候，注意数组边界

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long i,n,t,tmp1,tmp2,tmp3,b,c,d;
    long long a[100];
    a[0]=1;
    a[1]=1;
    a[2]=2;
    for(i=3; i<=40; i++)
        a[i]=a[i-1]+a[i-2];
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&tmp1,&tmp2,&tmp3);
        b=1;
        c=0;
        d=0;
        if(tmp1==1)
        {
            c=tmp2;
            if(tmp3==1)
                cout<<c<<'\n';
            else
            {
                d=a[tmp3-1]*c+a[tmp3-2];
                cout<<d<<'\n';
            }
        }
        else
        {
            c=(tmp2-a[tmp1-2])/a[tmp1-1];
            if(c*a[tmp1-1]!=tmp2-a[tmp1-2]||c<=0)
            {
                cout<<"-1"<<'\n';
            }
            else
            {
                if(tmp3==1)
                    cout<<c<<'\n';
                else
                {
                    d=a[tmp3-1]*c+a[tmp3-2];
                    cout<<d<<'\n';
                }
            }
        }
    }
    return 0;
}