HackerEarth The furious five problem solution

In this HackerEarth The furious five problem-solution You are given an integer N.

Write a program to find a minimum number P such that 1 <= X <= P, Sigma F(X) >= N ( where F(X) represents the number of times X can be divided by 5 ).

HackerEarth The furious five problem solution.

#include <bits/stdc++.h>
#define ll long long

ll minm(ll a, ll b)
{
    return (a < b) ? a : b;
}

ll get(ll num)
{
    ll ret = 0;
    while(num != 0) {
        ret += num/5;
        num /= 5;
    }
    return ret;
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--) {
        int n;
        ll low, mid, high;
        scanf("%d", &n);
        low = 0, high = 1e18;
        ll ans = high;
        while(low <= high) {
            mid = (low+high)/2;
            ll val = get(mid);
            if(val >= n) {
                high = mid-1;
                ans = minm(ans, mid);
            } else {
                low = mid+1;
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}

Second solution

#include<bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(i= a ; i < b ; ++i)
#define rep(i,n) FOR(i,0,n)
#define INF INT_MAX
#define pb push_back
#define si(n) scanf("%d",&n)
#define pln(n) printf("%lld\n",n)
#define pl(n) printf("%lld ",n)
#define sl(n) scanf("%lld",&n)
#define mod (int)(1e9 + 7)
#define ll long long int

inline ll helper(ll n)
{
    ll ans=0,calc;
    ans+=n;
    calc=5;
    while(n/calc!=0)
    {
        ans+=(n/calc);
        calc*=5;
    }
    return ans;
}
int main()
{
    int t,i;
    ll calc,s=0,n,cnt,low,high,mid,c1,c2;
    si(t);
    assert(t>=1 && t<=100000);
    while(t--)
    {
        sl(n);
        assert(n>=1 && n<=1000000000);
        low=0;
        high=mod;
        while(low<=high)
        {
            mid=(low+high)/2;
            c1=helper(mid);
            c2=helper(mid+1);
            if(c1<n && c2>=n)
            {
                printf("%lld\n",5*(mid+1));
                break;
            }
            else if(n>c1)
                low=mid+1;
            else
                high=mid-1;
        }
    }
    return 0;
}