Codeforces Round #370 (Div. 2) C. Memory and De-Evolution【逆向思维＋贪心】

传送门：C. Memory and De-Evolution

描述：

C. Memory and De-Evolution

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ≤ y < x ≤ 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.

Examples

input

6 3

output

4

input

8 5

output

3

input

22 4

output

6

Note

In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and c as (a, b, c). Then, Memory can do .

In the second sample test, Memory can do .

In the third sample test, Memory can do: 

.

题意：

给你一个以x为边长的等边三角形，你每一次操作都可以减少一条边的长度，当然减少之后需要保证还是一个三角形才行，问你最少使用多少步操作就能够使得以x为边长的等边三角形变成一个以y为边长的等边三角形。

思路：

1、直接想如何减少边长显得很麻烦也不能马上就找到最优解。

2、采用逆向思维，如果一个以y为边长的等边三角形，将这个三角形增加边长，问最少需要操作多少次能够使得边长为y的等边三角形，变成一个以x为边长的等边三角形，那么其实我们直接将：

a，b，c三边从小到大排序好了之后，使得a=b+c-1即可，当a>=x&&b>=x&&c>=x的时候，就是可以跳出循环的时候，我们就能够得到一个那样的解。

PS：刚开始还想正推，倍数关系什么的_(:зゝ∠)_ ，换个思路瞬间AC

代码：

#include <bits/stdc++.h>using  namespace  std;int  main(){  std::ios::sync_with_stdio(false);  std::cin.tie(0);  int x,y;  while(cin>>x>>y){    int a,b,c,cnt=0;    a=b=c=y;    while(1){      if(a>b)swap(a, b);      if(b>c)swap(b, c);      a=b+c-1;      cnt++;      if(a>=x && b>=x && c>=x)break;    }    printf("%d\n",cnt);  }  return 0;}