# HackerEarth Special Sum problem solution

In this HackerEarth Special Sum problem-solution Special Sum of the number N is defined as follows:

def foo(n):

{

ret = 0

for i = 1 to n:
{

if gcd(n,i) is 1:

ret += 1
}
return ret
}

def SpecialSum(N):

{

ret=0

for i = 1 to N:

{

if i divides N:

ret += foo(i)
}

return ret
}

we have given a N print SpecialSum(N).

## HackerEarth Special Sum problem solution.

`#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<iostream>#include<vector>#include<cassert>#include<sstream>#include<map>#include<set>#include<stack>#include<queue>#include<algorithm>using namespace std;#define pb push_back#define mp make_pair#define clr(x) x.clear()#define sz(x) ((int)(x).size())#define F first#define S second#define REP(i,a,b) for(i=a;i<b;i++)#define rep(i,b) for(i=0;i<b;i++)#define rep1(i,b) for(i=1;i<=b;i++)#define pdn(n) printf("%d\n",n)#define sl(n) scanf("%lld",&n)#define sd(n) scanf("%d",&n)#define pn printf("\n")typedef pair<int,int> PII;typedef vector<PII> VPII;typedef vector<int> VI;typedef vector<VI> VVI;typedef long long LL;#define MOD 1000000007LL mpow(LL a, LL n) {LL ret=1;LL b=a;while(n) {if(n&1) ret=(ret*b)%MOD;b=(b*b)%MOD;n>>=1;}return (LL)ret;}int main(){    int t;    sd(t);    while(t--)    {        LL n;        cin >> n;        cout << n << endl;    }    return 0;}`