Codeforces Beta Round #84 (Div. 1) A. Lucky Sum of Digits

Problem link: http://codeforces.com/problemset/problem/109/A

A minimum number must contain less number of digits as possible. So to find the minimum number whose digit sum is equal to given n, we have to look into the case 1 and if it fails we have to move to case 2.If case 2 also fails then such minimum lucky  number cannot be formed.

Case 1:if the number can be formed using digits 7 alone.

Case 2: if the number can be formed with combination of digits 4 and 7 such that the number is minimum as possible

	#include <vector>
	#include <list>
	#include <map>
	#include <set>
	#include <deque>
	#include <queue>
	#include <stack>
	#include <bitset>
	#include <algorithm>
	#include <functional>
	#include <numeric>
	#include <utility>
	#include <sstream>
	#include <iostream>
	#include <iomanip>
	#include <cstdio>
	#include <cmath>
	#include <cstdlib>
	#include <ctime>
	#include <cstring>
	#include <climits>
	#include <stdlib.h>
	#include <stdio.h>
	using namespace std;
	#define REP(i,n) for(int i=0; i<n; i++)
	#define FOR(i,st,end) for(int i=st;i<end;i++)
	#define db(x) cout << (#x) << " = " << x << endl;
	#define mp make_pair
	#define pb push_back
	#define MAX 10000005
	typedef long long int ll;
	int main(){
	int n;
	cin>>n;
	if(n<4){
	cout<<"-1";
	return 0;
	}
	int numSeven=n/7;
	int rem=n%7;
	int numFour=rem/4;
	int toBecomeFour=rem%4;
	if(toBecomeFour!=0){
	if(numSeven>=toBecomeFour){
	numFour+=((7*toBecomeFour)+toBecomeFour)/4;
	numSeven-=toBecomeFour;
	}
	else{
	cout<<"-1";
	return 0;
	}
	}
	REP(i,numFour)
	cout<<"4";
	REP(i,numSeven)
	cout<<"7";
	}

view raw Lucky Sum of Digits.cpp hosted with ❤ by GitHub