Climbing Worm
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10288 Accepted Submission(s):
6815
Problem Description
An inch worm is at the bottom of a well n inches deep.
It has enough energy to climb u inches every minute, but then has to rest a
minute before climbing again. During the rest, it slips down d inches. The
process of climbing and resting then repeats. How long before the worm climbs
out of the well? We'll always count a portion of a minute as a whole minute and
if the worm just reaches the top of the well at the end of its climbing, we'll
assume the worm makes it out.
Input
There will be multiple problem instances. Each line
will contain 3 positive integers n, u and d. These give the values mentioned in
the paragraph above. Furthermore, you may assume d < u and n < 100. A
value of n = 0 indicates end of output.
Output
Each input instance should generate a single integer on
a line, indicating the number of minutes it takes for the worm to climb out of
the well.
Sample Input
10 2 1
20 3 1
0 0 0
Sample Output
17
19
Source
思路：模拟
代码：
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; int n,u,d; int timee; int main() { while(scanf("%d%d%d",&n,&u,&d),n != 0) { timee = 0; for(int i = n;i >= 0;) { timee ++; i -= u; if(i <= 0) break ; timee ++; i += d; } printf("%d ",timee); } return 0; }