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

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

    题目:

    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 ab, 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: 

    .

     思路:倒着贪心就好,每次把最小的边变换成最大的。。

    代码:

    #include <bits/stdc++.h>
    
    using namespace std;
    
    #define MP make_pair
    #define PB push_back
    typedef long long LL;
    typedef pair<int, int> PII;
    const double eps = 1e-8;
    const double pi = acos(-1.0);
    const int K = 1e6 + 7;
    int a[4];
    int main(void)
    {
        int ans=0,st,se;
        cin>>st>>se;
        a[1]=a[2]=a[3]=se;
        while(1)
        {
            sort(a+1,a+1+3);
            if(a[1]<st)
                a[1]=min(a[2]+a[3]-1,st);
            else
                break;
            //printf("%d:%d %d %d
    ",ans+1,a[1],a[2],a[3]);
            ans++;
        }
        cout<<ans<<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/weeping/p/5869433.html
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