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  • Educational Codeforces Round 10 A. Gabriel and Caterpillar 模拟

    A. Gabriel and Caterpillar

    题目连接:

    http://www.codeforces.com/contest/652/problem/A

    Description

    The 9-th grade student Gabriel noticed a caterpillar on a tree when walking around in a forest after the classes. The caterpillar was on the height h1 cm from the ground. On the height h2 cm (h2 > h1) on the same tree hung an apple and the caterpillar was crawling to the apple.

    Gabriel is interested when the caterpillar gets the apple. He noted that the caterpillar goes up by a cm per hour by day and slips down by b cm per hour by night.

    In how many days Gabriel should return to the forest to see the caterpillar get the apple. You can consider that the day starts at 10 am and finishes at 10 pm. Gabriel's classes finish at 2 pm. You can consider that Gabriel noticed the caterpillar just after the classes at 2 pm.

    Note that the forest is magic so the caterpillar can slip down under the ground and then lift to the apple.

    Input

    The first line contains two integers h1, h2 (1 ≤ h1 < h2 ≤ 105) — the heights of the position of the caterpillar and the apple in centimeters.

    The second line contains two integers a, b (1 ≤ a, b ≤ 105) — the distance the caterpillar goes up by day and slips down by night, in centimeters per hour.

    Output

    Print the only integer k — the number of days Gabriel should wait to return to the forest and see the caterpillar getting the apple.

    If the caterpillar can't get the apple print the only integer  - 1

    Sample Input

    10 30
    2 1

    Sample Output

    1

    Hint

    题意

    有一个人,白天每小时爬a米,晚上每小时掉下来b米,可以掉到地下去

    你从第0天的下午两点看到这个人在h1,他要到h2去

    早上10点到晚上十点算白天,其他算晚上

    问你他第几天到达h2

    如果不能到达输出-1

    题解:

    数学智障很慌,直接暴力模拟好了,啪啪啪就好了

    代码

    #include<bits/stdc++.h>
    using namespace std;
    
    long long h1,h2,a,b;
    int main()
    {
        cin>>h1>>h2>>a>>b;
        int now = 14;
        int day = 0;
        int t = 0;
        while(t<1e7)
        {
            t++;
            if(now>=10&&now<=21)h1+=a;
            else h1-=b;
            if(h1>=h2)
            {
                cout<<day<<endl;
                return 0;
            }
            now=(now+1)%24;
            if(now==0)day++;
        }
        cout<<-1<<endl;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5327835.html
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