Fibonacci number’s ratio (the hard version)-优快云博客

Problem C. Fibonacci number’s ratio (the hard version)
Input file:
Output file:
Time limit:
Memory limit:
stdin
stdout
2s
512 Mb
Fibonacci numbers have many remarkable properties. For example, the ratio of two successive Fibonacci numbers
tends to the golden ratio:
√
F n+1
1+ 5
lim
=
.
n→∞ F n
2
You are going to test this property for generalized Fibonacci numbers
G n = aG n−1 + bG n−2 ,
n ≥ 2.
For given a, b, G 0 and G 1 calculate the limit
G n+1
,
n→∞ G n
lim
if it exists.
Input
The first line contains four integers a, b, G 0 , G 1 from −1000 to 1000.
Output
Print “YES” in the first line, if the limit exists, and “NO” otherwise. In the case of the positive result, print the
limit itself in the second line. Absolute or relative error must not exceed 10 −6 .
Examples
stdin
1 1 1 1
0 0 1 1
stdout
YES
1.6180339887
NO
Note
If the sequence G n contains an infinite number of zeros, the limit does not exist.

求极限。

细节比较坑，主要是可能为0

为不乱，可以把a,b,g0,g1为0 的情况分别列出来。

g[n]=a*g[n-1]+b*g[n-2];

g[n]/g[n-1] = a + b *(g[n-2]/g[n-1])

假设 g[n]/g[n-1]的极限存在，社为k

那么有 k = a + b * (1/k)

k*k-a*k-b=0;

delta = sqrt(a*a+4(b)

lim = (1+delta)/2.0 or (1-delta)/2.0;

/*************************************************************************
    > File Name: code/whust/##1/A.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月12日 星期三 13时38分02秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const long double C = 1.6180339887;
const double eqs = 1E-6;
int a,b,g0,g1;
int main()
{
    cin>>a>>b>>g0>>g1;
    double delta = a*a*1.0+4*b*1.0;
    if (delta<0)
    {

    cout<<"NO"<<endl;
    return 0;
    }

    if (a*g1==0&&b*g0==0)  //前两项为0,就会有无数个0,也是仅有的有无数个0的情况，根据提议，极限不存在。
    {
    cout<<"NO"<<endl;
    return 0;
    }
    if (b==0)   //根据公式，此时极限为常数a
    {
    cout<<"YES"<<endl;
    cout<<fixed<<setprecision(10)<<a*1.0<<endl;
    return 0;
    }
    if (a==0)
    {
        if (b*g0*g0==g1*g1)  //(sqrt(b)==(g1/g0))
        {
        if (g0*g1<0)   //初始值囧丁了极限的符号
        {
            cout<<"YES"<<endl;
            cout<<fixed<<setprecision(10)<<-1.0*sqrt(b)<<endl;
            return 0;
        }
        else
        {
            cout<<"YES"<<endl;
            cout<<fixed<<setprecision(10)<<1.0*sqrt(b)<<endl;
            return 0;
        }

        }
    }
    double lim1 = (a + sqrt(delta*1.0))/2.0;//两个可能的极限值
    double lim2 = (a - sqrt(delta*1.0))/2.0;
    double c1 = (g1-lim2*g0)/(lim1-lim2);
    double c2 = (lim1*g0-g1)/(lim1-lim2);
    if (fabs(c1)<eqs)
    {
    cout<<"YES"<<endl;
    cout<<fixed<<setprecision(10)<<lim2<<endl;
    return 0;
    }
    if (fabs(c2)<eqs)
    {
    cout<<"YES"<<endl;
    cout<<fixed<<setprecision(10)<<lim1<<endl;
    return 0;
    }
    double r = lim2/lim1;
    if (fabs(r)<1)
    {
    cout<<"YES"<<endl;
    cout<<fixed<<setprecision(10)<<lim1<<endl;
    }
    else
    {
    cout<<"YES"<<endl;
    cout<<fixed<<setprecision(10)<<lim2<<endl;
    }


    return 0;
}