Codeforces Round #223 (Div. 2)--A. Sereja and Dima

Sereja and Dima
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Sereja and Dima play a game. The rules of the game are very simple. The players have n cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Sample test(s)
input

4
4 1 2 10

output

12 5

input

7
1 2 3 4 5 6 7

output

16 12

Note
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
문제풀이 사고방식: 바둑론처럼 보이지만 사실은 욕심이다.매번 가장 왼쪽 또는 가장 오른쪽만 뽑을 수 있기 때문에 두 개의 바늘 i=n-1, j=0을 켜고 두 사람이 번갈아 뽑는다. 매번 a[i]>a[j]가 있으면 i--;그렇지 않으면 j++;i==j까지 끝납니다.
AC 코드:

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

int a[1005];

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0; i<n; i++)
            cin>>a[i];
        int ans = 0, anss = 0;
        int t = 0;
        for(int i=n-1, j=0; i>=j; ){                //     i,j
            if(a[i] > a[j]){
                if(t%2) anss += a[i];
                else ans += a[i];
                i --;
            }
            else{
                if(t%2) anss += a[j];
                else ans += a[j];
                j ++;
            }
            t ++;
        }
        cout<<ans<<" "<<anss<<endl;
    }
    return 0;
}