codeforces 317 A Perfect Pair

来源：互联网发布：漫步者煲音箱软件编辑：程序博客网时间：2024/05/22 01:42

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;}