A

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

题意：有一只蜗牛在井底，问你他是否能够离开井。基础题
H：井的高度
U：每天白天爬的高度
D：每天晚上下滑的高度
F：蜗牛的疲劳期，每天少爬U*F/100

注意：
1、蜗牛每天爬的距离都比前一天少
2、蜗牛每天爬的距离总是要大于0
3、函数类型最好用double
4、当蜗牛爬行的总路程大于井的高度时就成功了，当蜗牛下滑之后的高度小于0就失败

心得：感觉自己对for循环和break的一些用法有些模糊不清了，恩，还是有收获的
AC代码：

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
double a,b,c,d;
double sum,s;
int i;
while(scanf("%lf %lf %lf %lf",&a,&b,&c,&d),a)
{
s=b;
i=0;
sum=0;
//if(a==0) break;
while(1)
{
i++;
if(s>0) sum=sum+s;
if(sum>a) break;
sum=sum-c;
if(sum<0) break;
s=s-(d/100)*b;
}
if(sum>a) printf("success on day %d ",i);
else printf("failure on day %d ",i) ;

}
return 0;
}