Toda 2-白红宇

Toda 2

阅读量：4931 次

发布时间：2019-06-11

本文共 3498 字，大约阅读时间需要 11 分钟。

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 5 members).

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.

Examples

Input

Copy

5 4 5 1 7 4

Output

Copy

1 8 01010 00011 01010 10010 00011 11000 00011 11000

Input

Copy

2 1 2

Output

Copy

0 2 11 11

Input

Copy

3 1 1 1

Output

Copy

1 0 题解：数据比较小，每次操作只用两个数。还要确定最终的状态。

1 #pragma warning(disable:4996) 2 #include 3 #include
         
           4 #include
          
            5 #include
           
             6 #include
            
              7 #include
             
               8 #include
              
                9 using namespace std;10 11 typedef pair
               
                 P;12 13 const int maxn = 105;14 15 P a[maxn];16 int n;17 int G[maxn*maxn][maxn];18 19 int main()20 {21 while (scanf("%d", &n) != EOF) {22 memset(G, 0, sizeof(G));23 for (int i = 1; i <= n; i++) {24 int tp;25 scanf("%d", &tp);26 a[i] = P(tp, i);27 }28 int cnt = 0;29 while (true) {30 sort(a + 1, a + n + 1);31 if (a[n].first == a[1].first) break;32 cnt++;33 int k = 0;34 for (int i = 2; i <= n; i++) if (a[i].first > a[1].first) k++;35 if (2 <= k && k <= 5 && a[n].first - 1 == a[1].first) { //最终的状态！！！！！36 for (int i = n - k + 1; i <= n; i++) {37 a[i].first--;38 G[cnt][a[i].second] = 1;39 }40 break;41 }42 else {43 if (a[n].first) a[n].first--;44 if (a[n - 1].first) a[n - 1].first--;45 G[cnt][a[n].second] = G[cnt][a[n - 1].second] = 1;46 }47 }48 printf("%d\n", a[1].first);49 printf("%d\n", cnt);50 for (int i = 1; i <= cnt; i++) {51 for (int j = 1; j <= n; j++) {52 printf("%d", G[i][j]);53 }54 printf("\n");55 }56 }57 return 0;58 }

转载于:https://www.cnblogs.com/zgglj-com/p/9022282.html

你可能感兴趣的文章