题目链接:

A. Gabriel and Caterpillar

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 byb 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 2pm.

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.

Examples
input
10 30
2 1
output
1
input
10 13
1 1
output
0
input
10 19
1 2
output
-1
input
1 50
5 4
output
1
Note

In the first example at 10 pm of the first day the caterpillar gets the height 26. At 10 am of the next day it slips down to the height 14. And finally at 6 pm of the same day the caterpillar gets the apple.

Note that in the last example the caterpillar was slipping down under the ground and getting the apple on the next day.

AC代码:

#include <bits/stdc++.h>
using namespace std;
int main()
{
int h1,h2,a,b;
scanf("%d%d",&h1,&h2);
scanf("%d%d",&a,&b);
h1+=*a;
if(h1<h2&&a<=b)cout<<"-1"<<endl;
else
{
int ans=;
while()
{
if(h1>=h2)
{
cout<<ans<<endl;
break;
}
h1-=*b;
h1+=*a;
ans++;
} } return ;
}

codeforces 652A A. Gabriel and Caterpillar(水题)的更多相关文章

  1. Educational Codeforces Round 7 B. The Time 水题

    B. The Time 题目连接: http://www.codeforces.com/contest/622/problem/B Description You are given the curr ...

  2. Educational Codeforces Round 7 A. Infinite Sequence 水题

    A. Infinite Sequence 题目连接: http://www.codeforces.com/contest/622/problem/A Description Consider the ...

  3. Codeforces Testing Round #12 A. Divisibility 水题

    A. Divisibility Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/597/probl ...

  4. Codeforces Beta Round #37 A. Towers 水题

    A. Towers 题目连接: http://www.codeforces.com/contest/37/problem/A Description Little Vasya has received ...

  5. codeforces 677A A. Vanya and Fence(水题)

    题目链接: A. Vanya and Fence time limit per test 1 second memory limit per test 256 megabytes input stan ...

  6. CodeForces 690C1 Brain Network (easy) (水题,判断树)

    题意:给定 n 条边,判断是不是树. 析:水题,判断是不是树,首先是有没有环,这个可以用并查集来判断,然后就是边数等于顶点数减1. 代码如下: #include <bits/stdc++.h&g ...

  7. Codeforces - 1194B - Yet Another Crosses Problem - 水题

    https://codeforc.es/contest/1194/problem/B 好像也没什么思维,就是一个水题,不过蛮有趣的.意思是找缺黑色最少的行列十字.用O(n)的空间预处理掉一维,然后用O ...

  8. Codeforces 1082B Vova and Trophies 模拟,水题,坑 B

    Codeforces 1082B Vova and Trophies https://vjudge.net/problem/CodeForces-1082B 题目: Vova has won nn t ...

  9. CodeForces 686A Free Ice Cream (水题模拟)

    题意:给定初始数量的冰激凌,然后n个操作,如果是“+”,那么数量就会增加,如果是“-”,如果现有的数量大于等于要减的数量,那么就减掉,如果小于, 那么孩子就会离家.问你最后剩下多少冰激凌,和出走的孩子 ...

随机推荐

  1. JavaScript-Confirm用法

    function checkMobileBind() {    if(confirm('您尚未绑定手机,是否前往绑定?'))    {        window.location='http://m ...

  2. linux c编程:信号(二) alarm和pause函数

    使用alarm函数可以设置一个定时器,在将来的某个时刻该定时器超时.当定时器超时后,产生SIGALRM信号.如果忽略或不捕捉此信号,则其默认动作是终止调用该alarm函数的进程 #include< ...

  3. android 半透明弹窗

    <style name="edit_AlertDialog_style" parent="@android:style/Theme.Dialog"> ...

  4. linux nginx完全卸载

        Nginx虽然好用,但是一旦关键配置文件被修改,想要卸载重装却是相当困难.本人因为采用apt-get方式安装后又源码安装了Nginx,结果出现冲 突,卸载不了,安装不上,很是蛋疼.主要的问题还 ...

  5. SQL的优化1

    1.对查询进行优化,应尽量避免全表扫描,首先应考虑在 where 及 order by 涉及的列上建立索引. 2.应尽量避免在 where 子句中对字段进行 null 值判断,否则将导致引擎放弃使用索 ...

  6. Spring笔记:IOC基础

    Spring笔记:IOC基础 引入IOC 在Java基础中,我们往往使用常见关键字来完成服务对象的创建.举个例子我们有很多U盘,有金士顿的(KingstonUSBDisk)的.闪迪的(SanUSBDi ...

  7. ELKK 日志处理

    http://blog.csdn.net/u010022051/article/details/54342357 在ELKK的架构中,各个框架的角色分工如下: ElasticSearch1.7.2:数 ...

  8. Data Structure Linked List: Write a function to get the intersection point of two Linked Lists.

    http://www.geeksforgeeks.org/write-a-function-to-get-the-intersection-point-of-two-linked-lists/ 第一第 ...

  9. P3214 [HNOI2011]卡农

    题目 P3214 [HNOI2011]卡农 在被一题容斥\(dp\)完虐之后,打算做一做集合容斥这类的题了 第一次深感HNOI的毒瘤(题做得太少了!!) 做法 求\([1,n]\)组成的集合中选\(m ...

  10. linux下搭建java开发环境

    1 下载jdk包 这里下载.gz格式的,通过ftp上传到服务器 2 解压到指定目录,如/usr/java/ tar -xvf XXX.tar.gz 解压后会在/usr/java下生成一个目录,如jdk ...