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.

2 440 5012 14 16 190

1.19

, 2007.7.14

山地自行车，有前轮和后轮，前轮有fn种驱动齿轮，后轮有rn种驱动齿轮，每种驱动齿轮上和链条结合的齿数不一样，前轮为fj(j=1,2,...fn)，后轮为ri(i=1,2...rn)，现在前后两个齿轮相结合工作的齿轮比例为dk = ri / fj,然后走到相邻的d1<d2，且没有d3使得d1<d3<d2，定义这时的spread为d2/d1。问fn种前驱齿轮和种后驱齿轮形成的spread的最大值为多少？

读入所有的ri、fj，然后计算出所有的dk，再对d进行从小到大排序，找到d[i+1] / d[i]的最大值。

注意点：

1）是多case题目，以0结束，注意初始化。

2）结果保留两位小数。

代码（1AC）：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;int rear[15];int front[15];float dratio[300];int main(void){    int f, r;    int i, j, k;    float max, tmp;    while (scanf("%d", &f), f != 0){        memset(dratio, 0, sizeof(dratio));        memset(rear, 0, sizeof(rear));        memset(front, 0, sizeof(front));        scanf("%d", &r);        for (i = 0; i < f; i++){            scanf("%d", &front[i]);        }        for (i = k = 0, max = -1; i < r; i++){            scanf("%d", &rear[i]);            for (j = 0; j < f; j++){                dratio[k++] = (float)rear[i] / (float)front[j];            }        }        sort(dratio, dratio+k);        max = -1;        for (i = 0; i < k - 1 ; i++){            tmp = dratio[i + 1] / dratio[i];            if (tmp > max){                max = tmp;            }        }        printf("%.2f\n", max);    }    return 0;}