UVA 573 The Snail

The Snail

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 Dfeet
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

直接模拟即可，青蛙出井问题，判断青蛙在第几天跳出井或再次抵达井底。

注意题中有说当疲劳度对应调高小于零时，青蛙停在原处。

因为没注意这个，w了三次！

代码如下：

#include <stdio.h>
int main(void)
{
int h,d,f,day;
double percent,u,height;

while(scanf("%d%lf%d%d",&h,&u,&d,&f)!=EOF&&h)
{
day=1;
height=0.0;
percent=u*f/100.0;
for(;;day++)
{
if(u>=0)
height+=u;
if(height>h)
{
printf("success on day %d\n",day);
break;
}
height-=d;
if(height<0)
{
printf("failure on day %d\n",day);
break;
}
u-=percent;
}
}
return 0;
}