CodeForces 417C Football

题目链接：CF 417C

题面：

C. Football

【time limit per test】1 second                   【memory limit per test】  256 megabytes

One day, at the "Russian Code Cup" event it was decided to play football as an out of competition event. All participants was divided into n teams and played several matches, two teams could not play against each other more than once.

The appointed Judge was the most experienced member — Pavel. But since he was the wisest of all, he soon got bored of the game and fell asleep. Waking up, he discovered that the tournament is over and the teams want to know the results of all the matches.

Pavel didn't want anyone to discover about him sleeping and not keeping an eye on the results, so he decided to recover the results of all games. To do this, he asked all the teams and learned that the real winner was friendship, that is, each team beat the other teams exactly k times. Help Pavel come up with chronology of the tournir that meets all the conditions, or otherwise report that there is no such table.

Input

The first line contains two integers — n and k (1 ≤ n, k ≤ 1000).

Output

In the first line print an integer m — number of the played games. The following m lines should contain the information about all the matches, one match per line. The i-th line should contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i). The numbers a_i and b_i mean, that in the i-th match the team with number a_i won against the team with number b_i. You can assume, that the teams are numbered from 1 to n.

If a tournir that meets the conditions of the problem does not exist, then print -1.

Sample test(s)

Input

3 1

Output

3
1 2
2 3
3 1

题意：
    进行一场足球比赛，给定n，k代表共n个人。每两个人之间最多只能进行1场比赛。问是否有可能使得每个人赢k场，可以就输出需要进行的场数，和谁赢谁的序列，不可以则输出-1。

解题：
    简单的构造题，先判断下是否可以。一共可以进行的场次数为（n-1）*n/2。若可以，则让每个人赢其后的k个人，循环接上，保证不会重复。需要注意的是，因为数据量很大，cout会超时，用printf。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
int main()
{
	int n,k,total,tmp;
	scanf("%d%d",&n,&k);
	total=n*(n-1)/2;
	if(total<n*k)
		printf("-1\n");
	else
	{
	  printf("%d\n",n*k);
      for(int i=1;i<=n;i++)
	  {
		  for(int j=1;j<=k;j++)
		  {
			  tmp=i+j;
			  if(tmp>n)
				  tmp-=n;
			  printf("%d %d\n",i,tmp);
		  }
	  }

	}
	return 0;
}