Algorithm
Problem link- https://www.spoj.com/problems/HANGOVER/
HANGOVER - Hangover
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.
Input: 1.00 3.71 0.04 5.19 0.00 Output: 3 card(s) 61 card(s) 1 card(s) 273 card(s)
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
float t;
cin>>t;
while(t!=0.00)
{
int count=2;
float z=0.0;
while(z<t)
{
z=z+(1.0/count);
count=count+1;
}
cout<<(count-2)<<" card(s)"<<endl;
cin>>t;
}
return 0;
}Input
3.71
0.04
5.19
0.00
Output
61 card(s)
1 card(s)
273 card(s)
#2 Code Example with Java Programming
Code -
Java Programming
import java.util.*;
class hangover{
public static void main(String [] args){
Scanner in = new Scanner(System.in);
while(true){
double test = in.nextDouble();
if(test == 0.0){break;}
double hangover = 0.5;
int n = 1;
while(hangover < test){
n++;
hangover += 1.0/(n+1);
}
System.out.println(n + " card(s)");
}
}
}Input
3.71
0.04
5.19
0.00
Output
61 card(s)
1 card(s)
273 card(s)
#3 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
void cards(float a)
{
int i=2;
float sum=0.00;
float x=1.00;
while(sum<a)
{
sum=sum+x/i;
i++;
}
printf("%d card(s)\n",i-2);
}
int main(void) {
// your code goes here
float n;
scanf("%f",&n);
while(n!=0.00)
{
cards(n);
scanf("%f",&n);
}
return 0;
}Input
3.71
0.04
5.19
0.00
Output
61 card(s)
1 card(s)
273 card(s)
Demonstration
SPOJ Solution-Hangover-Solution in C, C++, Java, Python