A. Perfect Pair
http://codeforces.com/problemset/problem/317/A
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
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;
else
y += x ;
++count;
}
printf("%I64d", count);
}
}
return 0;
}
