Codeforces Round \ # 188 (Div. 2) C. 완벽 한 쌍 수학

B. Strings of Power
Time Limit: 20 Sec
Memory Limit: 256 MB
제목 연결
http://codeforces.com/contest/318/problem/C
Description
Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).
What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?
Input
Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.
Sample Input
1 2 5
Sample Output
2
HINT
 
제목
x, y 를 드 리 겠 습 니 다. 매번 에 하나의 수 를 선택 하여 그 를 x + y 로 만 든 다음 에 최소 몇 걸음 을 물 어보 면 x, y 중의 최대 치 는 k 보다 클 수 있 습 니 다.
문제 풀이:
이 문 제 는 fib 수 와 유사 하기 때문에 우 리 는 (x, y) 를 (y, x + y) 로 바 꾸 면 된다.
주의
코드:
 

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 4000001
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

int main()
{
    ll x=read(),y=read(),k=read();
    if(x>=k||y>=k)
    {
        cout<<"0"<<endl;
        return 0;
    }
    if(x<k&&y<k&&x<=0&&y<=0)
    {
        cout<<"-1"<<endl;
        return 0;
    }
    ll ans=0;
    if(x>y)
        swap(x,y);
    if(x<0)
    {
        ans=(y-x)/y;
        x+=ans*y;
    }
    while(y<k)
    {
        ll tmp=x;
        x=y;
        y=tmp+y;
        ans++;
    }
    cout<<ans<<endl;
}