The Snail
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1461 Accepted Submission(s): 1069
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
题目陷阱：
疲劳度是在第一天的基础上减去的相应的百分比，并且每天上升的高度，必须要是double型，不能变为int，int就变成了wa。
代码如下：
#include<stdio.h>
int main()
{
int h,d,f,i;
double u;
while(~scanf("%d",&h),h)
{
scanf("%lf%d%d",&u,&d,&f);
double sum=0,p,U=u*(1.0*f/100);
for(i=0;;i++)
{
sum+=u-d;//加上的是每天的净高度
if(sum>=h)
{
p=1;
break;
}
else if(sum<0)
{
p=0;
break;
}
u-=U;
}
p?printf("success on day %d\n",i):printf("failure on day %d\n",i+1);//此处的i+1是因为判断条件是加上的每天的净高度，当天白天并没有滑下去，所以又加上了一。
}
return 0;
}