HDOJ 1302-The Snail
来源:互联网 发布:时代互联域名 编辑:程序博客网 时间:2024/05/17 00:08
The Snail
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2072 Accepted Submission(s): 1464
Problem Description
A snail is at the bottom of a 6-foot well and wants to climb to the top. The snail can climb 3 feet while the sun is up, but slides down 1 foot at night while sleeping. The snail has a fatigue factor of 10%, which means that on each successive day the snail climbs 10% * 3 = 0.3 feet less than it did the previous day. (The distance lost to fatigue is always 10% of the first day's climbing distance.) On what day does the snail leave the well, i.e., what is the first day during which the snail's height exceeds 6 feet? (A day consists of a period of sunlight followed by a period of darkness.) As you can see from the following table, the snail leaves the well during the third day.
Day Initial Height Distance Climbed Height After Climbing Height After Sliding
1 0 3 3 2
2 2 2.7 4.7 3.7
3 3.7 2.4 6.1 -
Your job is to solve this problem in general. Depending on the parameters of the problem, the snail will eventually either leave the well or slide back to the bottom of the well. (In other words, the snail's height will exceed the height of the well or become negative.) You must find out which happens first and on what day.
Day Initial Height Distance Climbed Height After Climbing Height After Sliding
1 0 3 3 2
2 2 2.7 4.7 3.7
3 3.7 2.4 6.1 -
Your job is to solve this problem in general. Depending on the parameters of the problem, the snail will eventually either leave the well or slide back to the bottom of the well. (In other words, the snail's height will exceed the height of the well or become negative.) You must find out which happens first and on what day.
Input
The input file contains one or more test cases, each on a line by itself. Each line contains four integers H, U, D, and F, separated by a single space. If H = 0 it signals the end of the input; otherwise, all four numbers will be between 1 and 100, inclusive. H is the height of the well in feet, U is the distance in feet that the snail can climb during the day, D is the distance in feet that the snail slides down during the night, and F is the fatigue factor expressed as a percentage. The snail never climbs a negative distance. If the fatigue factor drops the snail's climbing distance below zero, the snail does not climb at all that day. Regardless of how far the snail climbed, it always slides D feet at night.
Output
For each test case, output a line indicating whether the snail succeeded (left the well) or failed (slid back to the bottom) and on what day. Format the output exactly as shown in the example.
Sample Input
6 3 1 1010 2 1 5050 5 3 1450 6 4 150 6 3 11 1 1 10 0 0 0
Sample Output
success on day 3failure on day 4failure on day 7failure on day 68success on day 20failure on day 2
Source
Mid-Central USA 1997
解题思路:
H-总高 U-初始速度 D-下滑距离 F-速度下降的概率 先计算出每次速度下降的大小为U*F 然后统计出天数,当天起点高度,爬升距离,爬升后的高度,下滑后的高度 然后下滑后的高度小于0就失败 爬升后的高度大于总高度就成功。
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){double h,u,d,f;double sh,ph,suh,xih,t;int day=0;while(~scanf("%lf%lf%lf%lf",&h,&u,&d,&f)){if(!h)break;f=f/100;t=u*f;sh=ph=suh=xih=0;day=0;int flag=1;while(suh<=h){ph=u;suh=sh+ph;xih=suh-d;sh=xih;u-=t;if(u<=0){u=t=0;}day++;if(xih<0){flag=0;break;}}if(flag)printf("success on day %d\n",day); else printf("failure on day %d\n",day);} return 0;}
0 0
- hdoj 1302 The Snail
- HDOJ The Snail 1302
- hdoj 1302 The Snail
- HDOJ 1302-The Snail
- HDOJ 1302 The Snail
- hdoj.1302 The Snail 20140819
- HDOJ 1302 The Snail(水题)
- hdoj The Snail
- HDOJ 题目1302 The Snail(模拟,水题)
- HDOJ 1302(UVa 573) The Snail(蜗牛爬井)
- HDU 1302 The Snail
- hdu 1302 The Snail
- hdu 1302 The Snail
- hdu -1302 The Snail
- HDU 1302 The Snail
- HDU 1302 The Snail
- 1302 The Snail【模拟】
- hdu 1302 The Snail
- debug current instruction pointer
- 时差,不同国家之间的大概时差
- 学习过程中的快思考与慢思考
- Scalaz(21)-类型例证:Liskov and Leibniz - type evidence
- 一些移动开发的前端框架分享
- HDOJ 1302-The Snail
- Fld selectn for mvmt type 551 / acct 400001 differs for Customer goods movement (015) Message no. M7
- java类加载的先后顺序
- SQL Server审计功能入门:SQL Server审核 (SQL Server Audit)
- 【Hibernate3】(3)数据库生成策略
- 文档的编码格式问题
- 你真的理解大数据吗?
- 下拉刷新
- 属性动画要素(简记)