首页 > ACM题库 > HDU-杭电 > HDU 1302 The Snail-模拟-[解题报告] C++
2013
12-04

HDU 1302 The Snail-模拟-[解题报告] C++

The Snail

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

6 3 1 10
10 2 1 50
50 5 3 14
50 6 4 1
50 6 3 1
1 1 1 1
0 0 0 0

样例输出:

success on day 3
failure on day 4
failure on day 7
failure on day 68
success on day 20
failure on day 2

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1302

这个题目也不是很复杂。一只蜗牛,白天上升一些,晚上下滑一些,每天会因为没有力气而比上一天少爬,求这个蜗牛几天滑到底或者几天滑到顶。简单模拟即可,如果用数学反而麻烦。另外当这个蜗牛上升比下降还快的时候,其实就已经不行了= =#
思路:直接按着题目意思模拟就行了。。。

#define _CRT_SECURE_NO_WARNINGS
 #include<iostream>
 #include<cstdio>
 #include<cmath>
 using namespace std;
 
 int main(){
     double h,u,d,f;
     while(~scanf("%lf",&h)&&h){
         scanf("%lf%lf%lf",&u,&d,&f);
         int day=0;
         bool flag=true;
         double dist=0,s=u*f/100;
         while(1){
             day++;
             dist+=u;
             if(dist>h)break;
             dist-=d;
             if(dist<0){flag=false;break;};
             u-=s;
             if(u<0)u=0;
         }
         if(flag){
             printf("success on day %d\n",day);
         }else 
             printf("failure on day %d\n",day);
     }
     return 0;
 }