Codeforces Round #223 (Div. 2)A. Sereja and Dima_html/css_WEB-ITnose

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Sereja and Dima

time limit per test

memory limit per test

input

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.

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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

44 1 2 10

output

12 5

input

71 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.

解题思路：看似博弈论，其实就是个贪心。由于每次只能取最左边的或最右边的，则只需开两个指针i = n-1，j = 0，两人轮流取，每次若a[i] > a[j] , 则 i --；否则 j ++;直到i == j时，结束。

AC代码：

#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 0x7fffffffint 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; ){                //开两个指针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;}