LightOj 1278 - Sum of Consecutive Integers

Problem Link: LightOj 1278 - Sum of Consecutive Integers

 // Accepted
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <string>
#include <vector>
#include <cmath>
#include <cctype>
#include <sstream>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <algorithm>
#define sf scanf
#define pf printf
#define sfint scanf ("%d %d",&a,&b)
#define sfl scanf ("%ld %ld",&a,&b)
#define sfll scanf ("%lld %lld",&a,&b)
#define sfd scanf ("%lf %lf",&a,&b)
#define sff scanf ("%f %f",&a,&b)
#define lp1(i,n) for(i=0;i<n;i++)
#define lp2(i,n) for(i=1;i<=n;i++)
#define LL long long
#define L long
#define mem(c,v) memset(c,v,sizeof(c))
#define ui unsigned int
#define cp(a) cout<<" "<<a<<" "<<endl
#define ull unsigned long long int
#define nl puts("")
#define sq(x) ((x)*(x))
#define mx7 20000100
#define mx6 1500000
#define mx5 10000005
#define inf 1<<30                                           //infinity value
#define eps 1e-9
#define mx (65540)
#define mod 1000000007
#define pb push_back
#define pi acos(-1.0)
#define sz size()
#define gc getchar ()


using namespace std;

int setbit(int n, int pos){n=n|(1<<pos); return n;}
bool checkbit(int n, int pos){n=n&(1<<pos); return n;}

bool flag[mx5];
int prm[(mx5/2)+1],plen=0;

void seieve()
{
    mem(flag,false);

    flag[0]=flag[1]=true;

    int qrt=(int)sqrt(double(mx5));

    for(int i=4;i<=mx5;i+=2)flag[i]=true;

    for(int i=3;i<=qrt;i+=2)
    {
        if(!flag[i])
        {
            for(LL j=i*i;j<=mx5;j+=i)
            {
                flag[j]=true;
            }
        }
    }

    for(int i=2;i<=mx5;i++)
    {
        if(!flag[i])
        {
            prm[plen++]=i;
        }
    }
}

LL smi(LL n)
{
    LL m=1;

    for(int i=0;i<plen and sq(prm[i])<=n;i++)
    {
        if(!(n%prm[i]))
        {
            int s=1;

            while(!(n%prm[i]))
            {
                s+=1;
                n/=prm[i];
                if(n==1 or n==0) break;
            }

            if(prm[i]&1)
            {
                m*=s;
            }
        }
    }

    if(n>1 and n&1)
    {
        m=m<<1;
    }

    m--;

    return m;
}

int main()
{
    seieve();
    int t,tc=0;

    sf("%d",&t);

    while(t--)
    {
        LL n;
        sf("%lld",&n);

        pf("Case %d: %lld\n",++tc,smi(n));
    }

    return 0;
}

//Critical TestCase::
//5
//80248409469439
//43660801947259
//75303949780302
//12764781726378
//99999999999973

Accepted Output