ZOJ 3702 Gibonacci number 找规律

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 forG(1) is a random integert, (t>=1). Given thei-th Gibonacci number valueG(i), and the numberj, your task is to output the value forG(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 contains3 integersi, 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 fort, output -1.

Sample Input

41 1 23 5 43 4 612 17801 19

Sample Output

28-1516847

    题目是说定义一个类似于Fibonacci数列的数列Gibonacci，这个数列的计算方式和Fibonacci一样，不同的是G(0)=1，而G(1)是一个未知数t，题目则是告诉你G(i)的值，然后计算G(j)的值，其实这个题并不复杂，列几个数就会发现，G(0)=1,G(1)=t,G(2)=1+t,G(3)=1+2t,G(4)=2+3t,G(5)=3+5t......前面的系数仍然符合Fibonacci数列的规律，而这个题目的范围只到20就结束了，所以简单打个表把Fibonacci数列前面的值都算出来，再通过规律计算t的值，就能很容易得出结果了。

    下面AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int f[105];int fibonacci(){    int i;    memset(f,0,sizeof(f));    f[0]=0;    f[1]=1;    for(i=2;i<30;i++)    {        f[i]=f[i-1]+f[i-2];    }    return 0;}int main(){    int T;    int i,j;    long long t;    long long ans;    int g;    fibonacci();    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&i,&g,&j);        t=(g-f[i-1])%f[i];        if(t==0)            t=(g-f[i-1])/f[i];        else        {            cout<<-1<<endl;            continue;        }        if(t<1)        {            cout<<-1<<endl;            continue;        }        ans=f[j-1]+f[j]*t;        cout<<ans<<endl;    }    return 0;}