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.
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.
Example input:
1.00
3.71
0.04
5.19
0.00
Example output:
3 card(s)
61 card(s)
1 card(s)
273 card(s)
Solution:
// 声明：本代码仅供学习之用，请不要作为个人的成绩提交。
// http://blog.csdn.net/mskia
// email: [email protected]
#include <iostream.h>
#include <stdlib.h>
int main( void ) {
float fz = 2.0;
short int i;
float temp , n ;
while( cin >> n && n != 0 ) {
fz = 2.0;
temp = 0;
i = 0;
while ( temp <= n ) {
++i ;
temp += 1 / fz ;
++fz;
}
cout << i << " card(s)" << endl;
}
return 0;
}
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