Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem D: Tour de France

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.

Sample Input

2 4
40 50
12 14 16 19
0

Output for Sample Input

1.19

---

Ondrej Lhotak and Gordon V. Cormack

模擬處理即可。

#include <stdio.h>
#include <algorithm>
using namespace std;
int main() {
    int n, m, q;
    int i, j;
    double f[100], r[100], d[10000];
    while(scanf("%d %d", &n, &m) == 2) {
        q = 0;
        for(i = 0; i < n; i++)
            scanf("%lf", &f[i]);
        for(i = 0; i < m; i++)
            scanf("%lf", &r[i]);
        for(i = 0; i < n; i++)
            for(j = 0; j < m; j++)
                d[q++] = r[j]/f[i];
        sort(d, d+q);
        double mx = 0;
        for(i = 1; i < q; i++)
            mx = max(mx, d[i]/d[i-1]);
        printf("%.2lf\n", mx);
    }
    return 0;
}