A. Perfect Pair
time limit per test
memory limit per test
input
output
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.
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).
m-perfect?
Input
x, y and m ( - 1018 ≤ x, y, m ≤ 1018).
%lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64dspecifier.
Output
-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: