time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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;
}
