HangOver 今天药忘吃喽~ 2022-08-05 05:16 110阅读 0赞 # HangOver # **Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10608 Accepted Submission(s): 4521** Problem Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below. ![1056-1.gif][] The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits. For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples. Sample Input 1.00 3.71 0.04 5.19 0.00 Sample Output 3 card(s) 61 card(s) 1 card(s) 273 card(s) #include<cstdio> #include<iostream> using namespace std; int main() { double n; double sum; while(scanf("%lf",&n)&&n!=0.00) { // printf("%.2lf\n",n); sum=0.00; int i=2,j; while(sum<n) { sum+=1.0/(i++); }//注意这个测试数据0.04 1题目是最少! j=i-2; printf("%d card(s)\n",j); } } [1056-1.gif]: /images/20220805/c1398be3ab644c3d86174956396ab8ea.png
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