Codeforces Round #188 (Div. 1) / 317A Perfect Pair(数学&优化)

A. Perfect Pair

http://codeforces.com/problemset/problem/317/A

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

注意x，y异号的情况可以优化。

优化后复杂度：O(log m)

完整代码：

/*30ms,0KB*/#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int main(void){__int64 x, y, m, count = 0;scanf("%I64d%I64d%I64d", &x, &y, &m);if (m <= max(x, y))printf("0");else{if (x <= 0 && y <= 0)printf("-1");else{    ///x,y异号的情况可以优化一下if (x < 0 && y > 0){    count = ceil((double) - x / y);                x+=count*y;}else if (x > 0 && y < 0){count = ceil((double) - y / x);y+=count*x;}while (max(x, y) < m){if (x < y)x += y;elsey += x ;++count;}printf("%I64d", count);}}return 0;}