codeforces 234G Practice

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Little time is left before Berland annual football championship. Therefore the coach of team "Losewille Rangers" decided to resume the practice, that were indefinitely interrupted for uncertain reasons. Overall there are n players in "Losewille Rangers". Each player on the team has a number — a unique integer from 1 to n. To prepare for the championship, the coach Mr. Floppe decided to spend some number of practices.

Mr. Floppe spent some long nights of his holiday planning how to conduct the practices. He came to a very complex practice system. Each practice consists of one game, all n players of the team take part in the game. The players are sorted into two teams in some way. In this case, the teams may have different numbers of players, but each team must have at least one player.

The coach wants to be sure that after the series of the practice sessions each pair of players had at least one practice, when they played in different teams. As the players' energy is limited, the coach wants to achieve the goal in the least number of practices.

Help him to schedule the practices.

Input

A single input line contains integer n (2 ≤ n ≤ 1000).

Output

In the first line print m — the minimum number of practices the coach will have to schedule. Then print the descriptions of the practices inm lines.

In the i-th of those lines print f_i — the number of players in the first team during the i-th practice (1 ≤ f_i < n), and f_i numbers from 1 to n— the numbers of players in the first team. The rest of the players will play in the second team during this practice. Separate numbers on a line with spaces. Print the numbers of the players in any order. If there are multiple optimal solutions, print any of them.

Sample test(s)

input
2

output
1
1 1

input
3

output
2
2 1 2
1 1

题目大意:将编号为1~n的n个人分成两组（每组至少1人），求至少分多少次组，能满足任意取两个人，这两个人必存在被分到过不同的组。

模拟：

#include<cstdio>
const int N=1002;
int a[12][N],cnt[12];
void dfs(int l,int r,int d)
{
	int i,m=(l+r)>>1;
	for(i=l;i<=m;i++)a[d][cnt[d]++]=i;
	if(l<r)
	{
		dfs(l,m,d+1);
		dfs(m+1,r,d+1);
	}
}
int main()
{
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	int n,i,j,ans;
	while(~scanf("%d",&n))
	{
		for(ans=1;1<<ans<n;ans++);
		for(i=1;i<=ans;i++)cnt[i]=0;
		printf("%d\n",ans);
		dfs(1,n,1);
		for(i=1;i<=ans;i++)
		{
			printf("%d",cnt[i]);
			for(j=0;j<cnt[i];j++)printf(" %d",a[i][j]);
			puts("");
		}
	}
	return 0;
}

位运算：

#include<cstdio>
int main()
{
        freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	int n,i,j,ans,cnt;
	while(~scanf("%d",&n))
	{
		for(ans=1;1<<ans<n;ans++);
		printf("%d\n",ans);
		for(i=0;i<ans;i++)
		{
			cnt=0;
			for(j=0;j<n;j++)if(1<<i&j)cnt++;
			printf("%d",cnt);
			for(j=0;j<n;j++)if(1<<i&j)printf(" %d",j+1);
			puts("");
		}
	}
	return 0;
}