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  • Codeforces Round #370 (Div. 2) C. Memory and De-Evolution 水题

    C. Memory and De-Evolution

    题目连接:

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

    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 ≤ 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.

    Sample Input

    6 3

    Sample Output

    4

    Hint

    题意

    给你一个边长为x的正三角形,你每秒钟可以修改一条边的边长,使得他变成边长为y的正三角形。

    你需要每时每刻都保证他是一个三角形。

    问你最少需要多少秒。

    题解:

    从小的变成大的,就可以贪心去变了。

    代码

    #include<bits/stdc++.h>
    using namespace std;
    int a,b,ans;
    void solve(int x,int y,int z)
    {
        if(z==a)return;
        ans++;
        z=min(a,x-1+y);
        solve(z,x,y);
    }
    int main()
    {
        cin>>a>>b;
        solve(b,b,b);
        cout<<ans<<endl;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5867867.html
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