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.
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.
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
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:输入H,U,D,F。H表示墙的高度,U表示白天能爬多少,D表示晚上下降多少,F表示疲劳程序(表示每天爬的高度都会减少U*F%),U最少为0,。。要求输出,在第几天爬出,或者在第几天掉回原点。
直接模拟,不过注意,高度大于H才能出墙,高度小于0才是落回原点。。
#include <stdio.h> #include <string.h> double h, u, d, f; int day; double sum; int main() { while (scanf("%lf%lf%lf%lf", &h, &u, &d, &f) != EOF && h) { day = 0; sum = 0; double up = u; while (1) { sum += up - d; up -= u * f / 100; if (up < 0) up = 0; day ++; if (sum < 0) { printf("failure on day %d\n", day); break; } if (sum + d > h) { printf("success on day %d\n", day); break; } } } return 0; }