CodeForces 701A Cards（水水的玩扑克）

http://codeforces.com/problemset/problem/701/A

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 thei-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.

题意：

好多人一块打扑克牌，存在部分两张牌的和相等，问都是哪两张的和与另外两张的和等值。

思路：

统计总和取平均，最后暴力！宇神说了，会暴力是最 low 的%%%%%%%%%%

Code：

#include<stdio.h> #include<cstring> const int MYDD=1103; int c[MYDD]; bool vis[MYDD];//标记该牌是否还存在 int main() { int n; while(scanf("%d",&n)!=EOF) { int sum=0; for(int j=1; j<=n; j++) { scanf("%d",&c[j]); sum+=c[j]; } memset(vis,false,sizeof(vis)); int a=sum/(n/2);//每个人应该获得的点数 for(int j=1; j<=n; j++) { for(int k=1; k<=n; k++) { if(j==k) continue;//一张牌 if(vis[j]||vis[k]) continue;//存在一张牌已经被使用 if(c[j]+c[k]==a) { printf("%d %d\n",j,k); vis[j]=vis[k]=true; } } } } return 0; }