http://codeforces.com/problemset/problem/712/C

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我是靠谱客的博主眼睛大小笼包，最近开发中收集的这篇文章主要介绍http://codeforces.com/problemset/problem/712/C，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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 $\text{[math]}$ .

In the second sample test, Memory can do $\text{[math]}$ .

In the third sample test, Memory can do: $\text{[math]}$

$\text{[math]}$ .

题目大意：给你一个x和y，让你把边长为x的等边三角形通过最短的步数变为长度为y 的等边三角形

我的方法是逆过来构造的，详细的看代码，我也不好解释，这题感觉比较水的说

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
int main(){
    int x,y;
    int a[4];
    int i,j;
    int coun;
    int tmp;
    while(~scanf("%d %d",&x,&y)){
        for(i=0;i<3;i++){
            a[i]=y;
        }
        coun=0;
        while(a[2]<x){
            a[2]=a[0]+a[1]-1;
            tmp=a[2];
            a[2]=a[0];
            a[0]=a[1];
            a[1]=tmp;
            coun++;
            /*for(i=0;i<3;i++){
                printf("%d ",a[i]);
            }
            printf("n");*/
        }
        printf("%dn",coun);
    }
    return 0;
}