ZOJ 3702 Gibonacci number 2017-04-06 23:28 28人阅读 评论(0) 收藏

时间：2021-12-31 13:52:08

Gibonacci number

Time Limit: 2 Seconds      Memory Limit: 65536 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

2

8

-1

516847

题目的意思是给出Gibonacci某一项的值，求出给定项

设G(1)=x;

则G(0)=1,G(1)=x,G(2)=1+x,G(3)=2x+1,G(4)=3x+2;

我们发现x的系数和常数符合斐波那契数列，推导公式即可

注意G(1)=0的时候输出-1；

#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <vector>

#include <set>

#include <stack>

#include <map>

#include <climits>

using namespace std;

#define LL long long

const int INF=0x3f3f3f3f;

int a[30],b[30];

void init()

{

    a[0]=0,a[1]=1;

    b[0]=1,b[1]=0;

    for(int i=2;i<=22;i++)

        a[i]=a[i-1]+a[i-2],b[i]=b[i-1]+b[i-2];

}

int main()

{

    init();

    int t;

    scanf("%d",&t);

    while(t--)

    {

        int x,y,z;

        scanf("%d%d%d",&x,&y,&z);

        y-=b[x];

        if(y%a[x]!=0||y==0) {printf("-1\n");continue;}

        int xx=y/a[x];

        printf("%lld\n",1LL*xx*a[z]+b[z]);

    }

    return 0;

}

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