E - Toda 2 CodeForces - 730A(简单思维和处理技巧）

最新推荐文章于 2019-08-09 14:36:57 发布

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#简单思维

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本文介绍了一种策略，用于帮助游戏团队通过特定的比赛输赢机制，使得团队成员之间的初始评级达到尽可能高的统一水平。该策略关注如何有效地减少部分成员的评级，确保最终所有成员的评级相等。

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A group of n friends enjoys playing popular video game Toda 2. There is a rating system describing skill level of each player, initially the rating of the i-th friend is r_i.

The friends decided to take part in the championship as a team. But they should have equal ratings to be allowed to compose a single team consisting of all n friends. So the friends are faced with the problem: how to make all their ratings equal.

One way to change ratings is to willingly lose in some matches. Friends can form a party consisting of two to five (but not more than n) friends and play a match in the game. When the party loses, the rating of each of its members decreases by 1. A rating can't become negative, so r_i = 0 doesn't change after losing.

The friends can take part in multiple matches, each time making a party from any subset of friends (but remember about constraints on party size: from 2 to 5members).

The friends want to make their ratings equal but as high as possible.

Help the friends develop a strategy of losing the matches so that all their ratings become equal and the resulting rating is maximum possible.

Input

The first line contains a single integer n (2 ≤ n ≤ 100) — the number of friends.

The second line contains n non-negative integers r₁, r₂, ..., r_n (0 ≤ r_i ≤ 100), where r_i is the initial rating of the i-th friend.

Output

In the first line, print a single integer R — the final rating of each of the friends.

In the second line, print integer t — the number of matches the friends have to play. Each of the following t lines should contain n characters '0' or '1', where the j-th character of the i-th line is equal to:

'0', if friend j should not play in match i,
'1', if friend j should play in match i.

Each line should contain between two and five characters '1', inclusive.

The value t should not exceed 10⁴, it is guaranteed that such solution exists.

Remember that you shouldn't minimize the value t, but you should maximize R. If there are multiple solutions, print any of them.

Example

Input

5
4 5 1 7 4

Output

Input

2
1 2

Output

Input

3
1 1 1

Output

1
0

题意：给出一串数字，让你经过若干次减一操作，每次可操作的数2到5个，最后让所有数相同。

思路：任何数都可以被2或者3整除，所以，若一个数是奇数的话，让它减三就会变成偶数，然后再两两处理就可以了。

#include<bits/stdc++.h>
using namespace std;
struct node
{
	int x,y;
}a[106];
int n,t,ans[10010][106];
bool cmp(node a,node b)
{
	return a.x<b.x;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		cin>>a[i].x;
		a[i].y=i;
	}
	while(1)
	{
		sort(a+1,a+n+1,cmp);
		if(a[1].x==a[n].x)
		break;
		t++;
		if(n>=3&&a[n].x==a[n-1].x&&a[n-1].x==a[n-2].x&&a[n-2].x!=a[n-3].x)
		{
			ans[t][a[n].y]=1;ans[t][a[n-1].y]=1;ans[t][a[n-2].y]=1;
			a[n].x--,a[n-1].x--,a[n-2].x--;
		}
		else
		{
			ans[t][a[n].y]=1;
			ans[t][a[n-1].y]=1;
			a[n].x--,a[n-1].x--;
			if(a[n-1].x<0)
			a[n-1].x=0;
	    }
	}
	cout<<a[1].x<<endl;
	cout<<t<<endl;
	for(int i=1;i<=t;i++)
	{
		for(int j=1;j<=n;j++)
		{
			printf("%d",ans[i][j]);
		}
		printf("\n");
	}
	return 0;

}