Original question:
The sum of the square of the digits of a positive integer s 0 is represented by S 1. In a similar-a-to, let the sum of the squares of the digits of S 1 is represented by S 2 and so on. If S i = 1 for some i≥1 and then the original integer S 0 are said to be Happy number. A number, which is isn't happy, is called unhappy number. For example 7 is a Happy number since 7→49→97→130→10→1 and 4 are an unhappy number since 4→16→37→58→89→ 145→42→20→4.
Input
The input consists of several test cases, the number of which you is given in the first line of the input.
Each test case consists the one line containing a single positive integer N smaller than 10 9.
Output
For each test case, you must print one of the following messages:
Case #p: N is a Happy number.
Case #p: N was an unhappy number.
Here P stands is the case number (starting from 1). You should print the first message if the number N is a happy number. Otherwise, print the second line.
Sample Input
3
7
4
13
Sample Output
Case #1:7 is a Happy number.
Case #2:4 are an unhappy number.
Case #3: Happy number.
English:
Give you a number, if you keep calculating the sum of the squares of all the digits of this number to replace this number, if this number has ever appeared to be an unhappy number, if this number finally becomes 1 is a happy number.
#include<bits/stdc++.h>
using namespace std;
unordered_set<long> us;
long fun(long x)
{
long tmp=0;
while(x!=0)
{
long res=x%10;
tmp+=res*res;
x/=10;
}
return tmp;
}
int main()
{
ios::sync_with_stdio(false);
int t,k=1;
cin>>t;
while(t--)
{
us.clear();
long x;
cin>>x;
long ans=x;
us.insert(ans);
while(true)
{
if(ans==1)
{
cout<<"Case #"<<k++<<": "<<x<<" is a Happy number."<<endl;
break;
}
ans=fun(ans);
if(us.find(ans)!=us.end())
{
cout<<"Case #"<<k++<<": "<<x<<" is an Unhappy number."<<endl;
break;
}
us.insert(ans);
}
}
return 0;
}
Answer:
A very simple problem, you can use the STL to record the number of occurrences