HDU 2832-Snail’s trouble（分治）

原创于 2017-07-06 10:12:40 发布 · 288 阅读

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探讨一只蜗牛在一米长的跑道上爬行的问题，跑道每分钟增加一米但蜗牛与两端的比例保持不变，求解蜗牛达到终点的时间。

Snail’s trouble

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 79 Accepted Submission(s): 54

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

Source

2009 Multi-University Training Contest 2 - Host by TJU

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gaojie

题意：有一只蜗牛在一个一米长的跑道上，每分钟爬k厘米，跑道每分钟增加1米，但是蜗牛与跑道两端的比率还是不变的。求蜗牛多少时间到达终点。

设总跑道长度为1，第 n 分钟走过的长度为（k／100）*（1 / n），对，没错，所以可以求解了

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
long long ans[105];
int main() {
    for (int i = 5; i <= 100; ++i) {
        double s = 100.0 / i;
        double temp = 0;
        for (int j = 1;; ++j) {
            temp += 1.0 / j;
            if (temp - s > -(1e-11)) {
                ans[i] = j;
                break;
            }
        }
    }
    int n;
    while (cin >> n) {
        cout << ans[n] << endl;
    }
}