HackerEarth The furious five problem solution YASH PAL, 31 July 202415 February 2026 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 longll 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("%lldn", 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("%lldn",n)#define pl(n) printf("%lld ",n)#define sl(n) scanf("%lld",&n)#define mod (int)(1e9 + 7)#define ll long long intinline 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("%lldn",5*(mid+1)); break; } else if(n>c1) low=mid+1; else high=mid-1; } } return 0;} coding problems solutions HackerEarth HackerEarth