杭电 2832 水题

最新推荐文章于 2024-08-01 16:24:15 发布

原创最新推荐文章于 2024-08-01 16:24:15 发布 · 783 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#integer #input #output

水题专栏收录该内容

22 篇文章

订阅专栏

探讨解决赛马难题的算法策略，面对不断伸长的跑道，小蜗牛何时能到达终点？通过数学公式和迭代计算揭示答案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

此题应该是有公式的，可惜吾等弱菜实在太弱，程序愣是跑了9500ms才过，，，惭愧。看着别人0msAC的，无颜面对啊。。。题目：

Snail’s trouble

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 416 Accepted Submission(s): 226

Problem Description

Once upon a time, there was a poor snail. Every day, it tried very hard to crawl forward, while there was a keeper who’d like to maltreat this poor little snail. The snail was crawling on a one-meter rubber band at first, and it can move on k cm every minute. But after that, the keeper stretches the rubber band quickly, and it would be elongated one meter, during that the distances ratio between the snail and the two endpoints remain unchanged. In the next minute, little snail tried to keep moving forward again.
“Can I finally get to the endpoint?”
The snail often asked himself such a question because he was afraid he would never succeed. Now, we hope you can tell this poor snail when he would reach the endpoint.

Input

Every line of the input contains an integer number k, indicating that the snail moved forward k cm every minute. (5 <= k <= 100)

Output

Output an integer t, indicating that the snail doesn’t get to the endpoint until t-1 minutes later, while t minutes later, it finally succeed.

Sample Input

10
100

Sample Output

12367
1

ac代码：

#include <iostream>
#include <cstdio>
using namespace std;
double ee=1e-8;
int main(){
  //freopen("1.txt","r",stdin);
  int k;
  while(~scanf("%d",&k)){
    _int64 count=1,len=0,stick=100;
	double leftside=k*1.0;
	double rightside=100;
	double ss;
	while((stick*1.0)-leftside>ee){
	  rightside=stick-leftside;
	  stick+=100;
	  leftside=((stick*1.0)/(stick-100))*leftside+k*1.0;
	  count++;
	}
	printf("%I64d\n",count);
  }
  return 0;
}