题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1302
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.
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 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
Sample Output
success on day 3 failure on day 4 failure on day 7 failure on day 68 success on day 20 failure on day 2
Source
题意：
给出H（瓶子的高度），U(蜗牛每天白天爬的高度)，D（蜗牛每天晚上下滑的距离），F（蜗牛因为的疲劳度，注意蜗牛每天减少的爬行距离是相同的，都是第一天的爬行距离减少U*（F%））；问蜗牛在第几天能爬出瓶子（则成功），或者又回到瓶底（则失败）。
代码如下：
#include <cstdio>
int main()
{
double H, U, D, F;
while(~scanf("%lf%lf%lf%lf",&H,&U,&D,&F))
{
if(H==0 && U==0 && D==0 && F==0)
break;
double sum = 0;
int k = 0;
int flag = 0;
double dd = (U*F)/100.0;
while(sum <= H)
{
if(sum < 0)
{
flag = 1;
break;
}
k++;
sum+=U;
if(sum > H)
break;
sum-=D;
U-=dd;
}
if(flag)
printf("failure on day %d\n",k);
else
printf("success on day %d\n",k);
}
return 0;
}