Description
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length \(x\), and he wishes to perform operations to obtain an equilateral triangle of side length \(y\).
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length \(y\)?
Input
The first and only line contains two integers \(x\) and \(y\) \((3 \le y < x \le 100 000)\) — the starting and ending equilateral triangle side lengths respectively.
Output
Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length \(y\) if he starts with the equilateral triangle of side length \(x\).
Sample Input
22 4
Sample Ouput
6
我还是too old了。怎么都没想出正的怎么贪心，改了1.5h。TM正解居然是逆向思考，从\(y\)到\(x\)，最佳的方法肯定是用边界三角形来每次更新一条边\((a = b+c-1)\)，这样子轮流更新三角形变大最快，以此来贪心。
代码如下：
#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;
int aim,l[3],res = 3,ans;
int main()
{
freopen("C.in","r",stdin);
freopen("C.out","w",stdout);
scanf("%d %d",&aim,&l[0]); l[1] = l[2] = l[0];
while (res)
{
for (int i = 0;i < 3&&res;++i)
{
l[i] = l[(i+1)%3]+l[(i+2)%3]-1; ++ans;
if (l[i] >= aim) --res,l[i] = aim;
}
}
printf("%d",ans);
fclose(stdin); fclose(stdout);
return 0;
}