codeforces 701A Cards

最新推荐文章于 2019-09-26 08:29:49 发布

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本文介绍了一个涉及卡片游戏的算法问题，目标是在确保每位玩家获得两张卡片的前提下，使得所有玩家手中的卡片数值之和相等。文章提供了一种通过排序并配对最大与最小值来实现这一目标的有效贪心算法。

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There are n cards (n is even) in the deck. Each card has a positive integer written on it. n / 2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.

Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.

Input

The first line of the input contains integer n (2 ≤ n ≤ 100) — the number of cards in the deck. It is guaranteed that n is even.

The second line contains the sequence of n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is equal to the number written on the i-th card.

Output

Print n / 2 pairs of integers, the i-th pair denote the cards that should be given to the i-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.

It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.

        Examples
      

          input
        
6
1 5 7 4 4 3

          output
        
1 3
6 2
4 5

          input
        
4
10 10 10 10

          output
        
1 2
3 4

Note

In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.

In the second sample, all values ai are equal. Thus, any distribution is acceptable.

简单的贪心

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
    int x,y;
}s[110];
int comp(node a,node b)
{
    return a.x<b.x;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&s[i].x);
        s[i].y=i;
    }
    sort(s,s+n,comp);
    for(int i=0;i<n/2;i++)
        printf("%d %d\n",s[i].y+1,s[n-i-1].y+1);
    return 0;
}