Problem Greg's Workout

Link : http://codeforces.com/problemset/problem/255/A

A. Greg's Workout

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was nintegers a₁, a₂, ..., a_n. These numbers mean that Greg needs to do exactly n exercises today. Besides, Greg should repeat the i-th in order exercise a_i times.

Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the n-th exercise.

Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.

Input

The first line contains integer n (1 ≤ n ≤ 20). The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 25) — the number of times Greg repeats the exercises.

Output

Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.

It is guaranteed that the input is such that the answer to the problem is unambiguous.

Sample test(s)

input

2
2 8

output

biceps

input

3
5 1 10

output

back

input

7
3 3 2 7 9 6 8

output

chest

Note

In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.

In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.

In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.

Thuật toán: Đơn giản là cài đặt thôi.

Code:

/* 
    Coder : Nguyen Duc Tam 
    template for contest
*/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
#include<vector>
#include<utility>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<stack>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<iomanip>
#include<ctime>
#include<cctype>

using namespace std;

/*
    define for statement loop
*/
#define REP(i, start, end, step) for(int i = start; i < end; i += step)
#define DOWN(i, start, end, step) for(int i = start; i > end; i -= step)
#define FOR(it,c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
#define ALL(c) (c).begin(), (c).end()
#define SZ(x) ((int)(x).size())

/*
    define operator bit
*/
#define L(x,i) ((x) << (i))
#define R(x,i) ((x) >> (i))
#define AND(a,b) ((a) & (b))
#define OR(a,b) ((a) | (b))
#define XOR(a,b) ((a) ^ (b))
#define NOT(a) (~(a))
#define SB(x,i) (OR((x), L(1, (i))))    // x | 1 << i 
#define CB(x,i) (AND((x),NOT(L(1,(i))))) // x & ~(1 << i)
#define TB(x,i) (AND((x), L(1,(i))))    // x & (1 << i) 

/*
    define init data    
*/
#define FILL(a,val) memset(a,val,sizeof(a));
#define INIT(a,l,r,val) REP(i,l,r,1) (a)[i] = val;

/*
    define convert data
*/
#define DIG(c) (int)((c) - '0')
#define CHR(c) (char)((c) + '0')
#define LOW(c) (char)((c) + 32)
#define UPP(c) (char)((c) - 32)

/*
    define math function    
*/
#define sqr(a) (a) * (a)
#define abs(a) (a < 0 ? -(a) : (a))

/*
    define find element in container very fast.
*/
#define LO(c,x) lower_bound(ALL(c),x)   
#define UP(c,x) upper_bound(ALL(c),x)
#define IOF(c,x) distance((c).begin(), LO((c),x))
#define IOL(c,x) distance((c).begin(), UP((c),x))
#define IN(c,x) binary_search(ALL(c),x)
#define ER(c,x) equal_range(ALL(c),x)

/*
    define geo
*/
#define DIS(p,q) sqrt(sqr(p.x - q.x) + sqr(p.y - q.y)) // khoang cach Euclide
#define DIM(p,q) abs(p.x - q.x) + abs(p.y - q.y) // khoang cach mahatan


/*
    define constant
*/

#define EPS 1e-7
#define OO 1000000005
#define N 100005

#define X first
#define Y second

struct Point
{
    double x, y;
    Point(){}
    Point(double x,double y):x(x), y(y){}   
};

struct Node
{
    int value;
    Node* next;
    Node(){}
    Node(int value, Node* next):value(value),next(next){}
};

struct Edge
{
    int src, dst, weight;
    Edge(){}
    Edge(int src, int dst, int weight):src(src), dst(dst), weight(weight){} 
    
};

struct Compare
{
    
    bool operator() (const int i, const int j)
    {
        return i > j;   
    }

    /*
    bool operator() (const Point& p, const Point& q)
    {
        return (p.x != q.x ? p.x < q.x : p.y < q.y);
    }   
    */
    /*
    bool operator()(const Edge& e, const Edge& f)
    {
        return  (e.weight != f.weight ? e.weight > f.weight : e.src != f.src ? 
            e.src < f.src : e.dst < f.dst); // inverse weight
    }
    */
};

    
const int DAY[13] = {-1,31,29,31,30,31,30,31,31,30,31,30,31};

const int dx4[4] = {-1, 0, 1, 0}; //U - R - D - L
const int dy4[4] = {0, 1, 0, -1};

const int dx8[8] = {-1, -1, 0, 1, 1, 1, 0, -1}; //U - UR - R - DR - D - DL - L - UL 
const int dy8[8] = {0, 1, 1, 1, 0, -1, -1, -1};

typedef pair<int,int> II;
typedef pair<II,int> III;
typedef long long LL;
typedef unsigned long long ULL;
typedef unsigned char UC; 

vector<int>::iterator it, low, up; // using lower_bound, upper_bound
pair<vector<int>::iterator, vector<int>::iterator > bound;  // using function equal_range(first, last, x)
Compare comp;
Point p[2];

int n, x, ans[3], res ;

int main()
{
    #define Off  false
    if(Off)
    {
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    }
    cin >> n;
    REP(i,0,n,1)
    {
        cin >> x;
        ans[i % 3] += x;
    }
    res = max(ans[0],max(ans[1],ans[2]));
    if(res == ans[0])
        cout << "chest" << endl;
    else if(res == ans[1])
        cout << "biceps" << endl;
        else
            cout << "back" << endl;
    return 0;
}