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Hangover
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.
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 |
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int main() 6 { 7 double cnt[400]; 8 double c; 9 cnt[1]=0.5; 10 for(int i=2;i<=400;i++){ 11 cnt[i]=cnt[i-1]+1.0/(i+1); 12 } 13 while(cin>>c&&c){ 14 for(int i=1;i<=400;i++){ 15 if(cnt[i]>=c) { 16 cout<<i<<" card(s)"<<endl; 17 break; 18 } 19 } 20 } 21 return 0; 22 }
原文：http://www.cnblogs.com/shenyw/p/5161455.html