codeforces 317 A Perfect Pair

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概述

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 %I64dspecifier.

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 test(s) 
 
     input 
   
1 2 5

     output 
   
2

     input 
   
-1 4 15

     output 
   
4

     input 
   
0 -1 5

     output 
   
-1

题意：

给你x 和 y 还有M，让你通过变换x和y，使得其中一个大于或者等于m，变换的方法就是自身加上另外一个，如果可以通过若干步使满足条件，输出最小的步数，否则输出-1。

思路：
如果xy小于0且m大于0，那么肯定不可能变换到m，如果 x < m && y < m && x+y<0也肯定没办法达到m。

我先排除了输出-1的，然后再考虑如何计算最小的步数。我们主要在每一步中最小一个加上另一个就可以了，这是朴素的求法，但可能出现这样的情况比如 -100000000 1 10000000 这样的话会循环100000000多次，肯定超时，所以我们要加快速度。

代码：

//cf 317 A
//2013-06-22-16.43
#include <iostream>

using namespace std;

int main()
{
    __int64 x, y, m;
    while (cin >> x >> y >> m)
    {
        if (x <= 0 && y <= 0)
        {
            if (x < m && y < m && x+y <= 0)
            {
                cout << "-1" << endl;
                continue;
            }
        }
        __int64 ans = 0;
        int cnt = 0;
        while (x < m && y < m)
        {
            if (x > y)
            {
                __int64 t = x; x = y; y = t;
            }
            if (x < 0 && y > 0 && -x > y)
            {
                ans += (-x)/y;
                x += (-x)/y * y;

            }
            else
            {
                x = x+y;
                ans++;
            }
        }
        cout << ans << endl;
    }
    return 0;
}