ZCMU—1233

1233: Tour de France

Time Limit: 1 Sec  Memory Limit: 128 MB
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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 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.

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

Input

Please Input Input Here

Output

Please Input Output Here

Sample Input

2 4

40 50

12 14 16 19

0

Sample Output

1.19

【分析】

主要还是题意了吧。。

自行车用链条将两个齿轮,前面的齿轮由2或3根齿条组成,后面的齿轮由5……10根齿条组成, 这个齿轮转动的角速度f是由前后两个齿轮所包含的齿条数决定的,假设前面的齿轮的齿条数为m,后面齿轮的齿条数为n,那么角速度为n/m,求出所有f后将其按升序排序,然后再计算d[i+1]/d[i]的最大值就可以了

【代码】

#include <stdio.h>#include <algorithm>using namespace std;double f[100];double a[100],b[100];int len,n,m;int main(){    while(~scanf("%d",&n) && n)    {        scanf("%d",&m);        for(int i=0;i<n;i++) scanf("%lf",&a[i]);        for(int i=0;i<m;i++) scanf("%lf",&b[i]);        int len=0;        for(int i=0;i<m;i++)            for(int j=0;j<n;j++)            f[len++]=b[i]/a[j];        sort(f,f+len);        double ans=0.0;for(int i=0;i<len-1;i++) ans=max(ans,f[i+1]/f[i]);        printf("%.2f\n",ans);    }    return 0;}