Algorithm
problem link- https://www.spoj.com/problems/HPYNOS/
HPYNOS - Happy Numbers I
The process of “breaking” an integer is defined as summing the squares of its digits. For example, the result of breaking the integer 125 is (12 + 22 + 52) = 30. An integer N is happy if after “breaking” it repeatedly the result reaches 1. If the result never reaches 1 no matter how many times the “breaking” is repeated, then N is not a happy number.
TASK
Write a program that given an integer N, determines whether it is a happy number or not.
CONSTRAINTS
2 ≤ N ≤ 2,147,483,647
Input
A single line containing a single integer N.
Output
A single line containing a single integer T which is the number of times the process had to be done to determine that N is happy, or -1 if N is not happy.
Example
Input: 19 Output: 4
1) 19 : 12 + 92 = 82 2) 82 : 82 + 22 = 68 3) 68 : 62 + 82 = 100 4) 100 : 12 + 02 + 02 = 1
The solution is 4 because we discovered that the integer 19 is happy after we repeated the process 4 times.
Input: 204 Output: -1
204 –> 20 –> 4 –> 16 –> 37 –> 58 –> 89 –> 145 –> 42 –> 20 –> 4 –> 16 –> 37 –> 58 –> 89 –> 145 ……
204 is not a happy number because after breaking it several times the results start repeating so we can deduce that if we continue breaking it, the result will never reach 1.
Number of input files is 32.
Don't use pre-calculated values (Don't Cheat)!!!
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <sstream>
#include <map>
using namespace std;
int main(){
bool s[1000];
memset(s,false,sizeof(s));
long long int num=0;
cin>>num;
long long int temp=0;
int count=0;
while(true){
if(temp==1){
cout<<count<<endl;
break;
}
count++;
temp=0;
while(num>0){
temp+=pow(num%10,2);
num/=10;
}
if(s[temp]==true)
{
cout<<"-1"<<endl;
break;
}
else
{
s[temp]=true;
num=temp;
}
}
return 0;
}Input
Output
Demonstration
SPOJ Solution-Happy Numbers-Solution in C, C++, Java, Python