codeforces #364a Cards

cf的a题没什么好说到，100的量级，每个人给2张牌，使每个人手中的牌点数相等。保证有一种分配方案。

对每个人，先计算出手中的牌的点数，然后循环两遍拿牌就可以。

 

A. Cards

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

#include<iostream>
#include<stdint.h>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<math.h>

using namespace std;
int main()
{
    int n;
    scanf("%d",&n);
    int a[105];
    int sum=0;
    for(int i=0; i<n; i++)
    {
        scanf("%d",&a[i]);
        sum+=a[i];
    }
    int vis[105];
    for(int i=0; i<=105; i++) vis[i]=0;
    int tt=sum/(n/2);
    for(int i=0; i<n/2; i++)
    {
        int tmp=tt;
        for(int j=0; j<n; j++)
        {
            if(tmp>=a[j]&&!vis[j])
            {
                printf("%d ",j+1);
                tmp-=a[j];
                vis[j]=1;
                break;
            }
        }
        for(int j=0; j<n; j++)
        {
            if(tmp==a[j]&&!vis[j])
            {
                printf("%d
",j+1);
                vis[j]=1;
                break;
            }
        }

    }
    return 0;
}