Codeforces #273 (Div.2) C.Table Decorations

Problem link: http://codeforces.com/problemset/problem/478/C

Logic:

Case 1:

If (number of maximum colored balloon)/2 is greater than or equal to sum of rest of the 2 colored balloons, then we can decorate each table with 2 balloons from maximum colored balloon and 1 balloon from either of the 2 other colored balloons.

Case 2:

If (number of maximum colored balloon)/2 is lesser than the sum of rest of the 2 colored balloons, then maximum number of tables that we can decorate would be (total number of balloons available)/3.

	#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
	typedef long long int ll;
	int main(){
	ll arr[3];
	REP(i,3){
	cin>>arr[i];
	}
	sort(arr,arr+3);
	if(arr[0]+arr[1]<=arr[2]/2){
	cout<<arr[0]+arr[1];
	}
	else{
	cout<<((arr[0]+arr[1]+arr[2])/3);
	}
	}

view raw TableDecorations.cpp hosted with ❤ by GitHub