CodeForces 712C Memory and De-Evolution

outputstandard 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
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 (6,6,6)→(6,6,3)→(6,4,3)→(3,4,3)→(3,3,3).

In the second sample test, Memory can do (8,8,8)→(8,8,5)→(8,5,5)→(5,5,5).

In the third sample test, Memory can do：
(22,22,22)→(7,22,22)→(7,22,16)→(7,10,16)→(7,10,4)→(7,4,4)→(4,4,4).

题意

给你一个各边都为x的等边三角形，你有一种魔法，在每一秒钟可以改变当前三角形一边的长度，使之变成一个新的三角形，问你最少需要多少秒？

分析

1. 每一秒钟都可以改变当前三角形一边的长度，那我是不是可以只花一秒就将某一边x减少到y了？这当然是不可能的。题目要求要构成一个新的三角形，三角形三边必须要满足两边之和大于第三边。
2. 既然不能一下减少y，那减少到多少了？不好判断。
3. 那就逆推吧。从x减少到y无法判断，但是从y增加到x可以知道啊？让某一边尽可能大，并且还能和其余两边构成三角形就行了。
4. 求最少时间，贪心，使改变量尽可能多，每次改变最小边，改变到多少? 最小边 = 剩下两边长度之和 - 1 。

Source Code

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{
    int x,y,i;
    int s[4];
    cin>>x>>y;
    s[0]=s[1]=s[2]=y;
    for(i=0; s[0]<x||s[1]<x||s[2]<x; i++)
        s[i%3]=s[(i+1)%3]+s[(i+2)%3]-1;
    cout<<i<<endl;
    return 0;
}

代码摘抄于别人，写的挺巧妙的，利用求余实现了循环，实现了循环也就实现了排序