-- 作者:jollyzlj
-- 发布时间:4/30/2007 3:14:00 PM
-- 求助:一个很短的程序,请高手帮帮忙!
题目是这样的:Hangover Time Limit:1000MS Memory Limit:10000K Total Submit:20182 Accepted:9081 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). Input 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. Output 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) Source Mid-Central USA 2001 我写的是: #include<stdio.h> main() { float c,sum=0.5;int j=1; scanf("%f",&c); while(sum<c) { j++; sum+=1.0/(j+1); } printf("\n%d",j); getch(); } 不知道为什么总是不能通过,说是超时。真搞不懂。
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