Cards(Codeforces 701A)(Codeforces Round #364Div.2 A)

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本文介绍了一种确保每名玩家获得两张卡片且卡片数值之和相等的游戏算法。通过示例展示了解决方案，并提供了两种AC代码实现方式，一种通过直接匹配数值求和，另一种通过排序简化问题。

701A

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.

这题还算简单，自己成功A掉。

AC代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[105];
int visited[105];
int main()
{
    int n;
    scanf("%d",&n);
    int sum=0;
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        sum+=a[i];
    }
    memset(visited,0,sizeof(visited));
    int res=sum*2/n;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n&&j!=i;j++){
            if(a[i]+a[j]==res&&!visited[i]&&!visited[j]){
                printf("%d %d\n",i,j);
                visited[i]=1;
                visited[j]=1;
            }
        }
    }
    return 0;
}

附上大牛的代码:(思路是排序)

#include<bits/stdc++.h>
using namespace std;
struct p{
	int i,v;
};
int com(p a,p b){
	return a.v<b.v;
}
int main(){
	int n;
	cin>>n;
	p a[n];
	for(int i=0;i<n;i++)
	{	cin>>a[i].v;
		a[i].i=i+1;
	}
	sort(a,a+n,com);
	for(int i=0;i<n/2;i++){
		cout<<a[i].i<<' '<<a[n-1-i].i<<'\n';
	}
}