| 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 feetwhile the sun is up, but slides down 1 foot at night while sleeping.The snail has a fatigue factorof 10%, which means that on each successive day the snail climbs10% 
3 = 0.3 feet less thanit did the previous day. (The distance lost to fatigue is always 10% of thefirst day's climbingdistance.) On what day does the snail leave the well, i.e., what is the firstday during whichthe snail's height exceeds 6 feet? (A day consists of a period of sunlightfollowed by a period ofdarkness.) As you can see from the following table, the snail leaves the wellduring 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 parametersof the problem, thesnail will eventually either leave the well or slide back to the bottom ofthe well. (In other words,the snail's height will exceed the height of the well or become negative.)You must find out whichhappens first and on what day.
Input
The input file contains one or more test cases, each on a line by itself.Each line contains fourintegers 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 theheight of the well infeet, U is the distance in feet that the snail can climb during the day, D is the distance in feetthat the snail slides down during the night, and F is the fatigue factorexpressed as a percentage.The snail never climbs a negative distance. If the fatigue factor dropsthe snail's climbing distancebelow zero, the snail does not climb at all that day. Regardless of how farthe 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: 小时候常见的蜗牛爬墙, 我在描述一下
一只蜗牛白天能向上爬, 但是晚上睡觉会滑落一点
然后由于有疲劳的原因, 每天能向上爬的距离越来越少
如第一组数据, 6英尺的墙, 第一天能向上爬3英尺, 但是晚上会滑落1英尺
至于10, 是说百分之10, 第一天是爬3英尺没错, 但是第二天只有 (3 - 3*10%) = 2.7英尺了, 第三天就只有2.4英尺了...
注意点:(容易WA的地方)
首先是临界, 如第一组数据, 爬到6英尺的地方不算爬出, 要大于6才行, 等于不行!
落到墙底也是, 要<0, 等于0不算!
另外每次白天的爬行距离也要判定, 必须是大于零的, 不会说向上爬行的距离最后变的往下爬~
AC代码:
#include<stdio.h>
int main() {
double h, u, d, f;
while(scanf("%lf%lf%lf%lf", &h, &u, &d, &f) != EOF) {
if(h == 0)
break;
int day = 1;
double init_h = 0;
double down = u * (f / 100);
double day_down;
while(init_h < h) {
day_down = (day-1) * down;
if((u - day_down) > 0)
init_h = (u - day_down) + init_h;
if(init_h > h) {
printf("success on day %d\n", day);
break;
}
init_h = init_h - d;
if(init_h < 0) {
printf("failure on day %d\n", day);
break;
}
day++;
}
}
return 0;
}
