poj3300

最新推荐文章于 2025-08-09 20:51:19 发布

转载最新推荐文章于 2025-08-09 20:51:19 发布 · 110 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/eric-blog/archive/2011/06/01/2067441.html

文章标签：

#javascript #ViewUI

Tour de France

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3418		Accepted: 1655

Description

A racing bicycle is driven by a chain connecting two sprockets. Sprockets are grouped into two clusters: the front cluster (typically consisting of 2 or 3 sprockets) and the rear cluster (typically consisting of between 5 and 10 sprockets). At any time the chain connects one of the front sprockets to one of the rear sprockets. The drive ratio -- the ratio of the angular velocity of the pedals to that of the wheels -- is n : m where n is the number of teeth on the rear sprocket and m is the number of teeth on the front sprocket. Two drive ratios d₁< d₂ are adjacent if there is no other drive ratio d₁< d₃< d₂. The spread between a pair of drive ratios d₁< d₂ is their quotient: d₂ ⁄ d₁. You are to compute the maximum spread between two adjacent drive ratios achieved by a particular pair of front and rear clusters.

Input

Input consists of several test cases, followed by a line containing 0. Each test case is specified by the following input:

f: the number of sprockets in the front cluster;
r: the number of sprockets in the rear cluster;
f integers, each giving the number of teeth on one of the gears in the front cluster;
r integers, each giving the number of teeth on one of the gears in the rear cluster.

You may assume that no cluster has more than 10 sprockets and that no gear has fewer than 10 or more than 100 teeth.

Output

For each test case, output the maximum spread rounded to two decimal places.

Sample Input

2 4
40 50
12 14 16 19
0

Sample Output

1.19

Source

Waterloo Local Contest, 2007.7.14

题目很难看懂，看了半天才发现是自行车变速器的问题。水过，补习英语去了。。。

#include <cstdio>
#include <algorithm>

using namespace std;

float ratio[100];
int f,r;
float front[10],rear[10];
float max_ratio;
int index;

int main()
{
    while(scanf("%d",&f)!=EOF && f)
    {
        scanf("%d",&r);
        for(int i=0; i<f; i++)
            scanf("%f",&front[i]);
        for(int i=0; i<r; i++)
            scanf("%f",&rear[i]);
        index=0;
        for(int i=0; i<r; i++)
        {
            for(int j=0; j<f; j++)
            {
                ratio[index++]=rear[i]/front[j];
            }
        }
        sort(ratio,ratio+index);
        for(int i=0; i<index-1; i++)
            ratio[i]=ratio[i+1]/ratio[i];
        max_ratio=0.0;
        for(int i=0; i<index-1; i++)
        {
            if(max_ratio<ratio[i])
                max_ratio=ratio[i];
        }
        printf("%.2f\n",max_ratio);

    }
    return 0;
}

转载于:https://www.cnblogs.com/eric-blog/archive/2011/06/01/2067441.html