A. Toda 2

A. Toda 2

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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 ri.

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 ri = 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, ..., rn (0 ≤ ri ≤ 100), where ri 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 ncharacters '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.

思路：贪心；

因为可以同时操作2-5个，那么我们只要讨论时操作2个或三个就行了因为2，3能组成其他数，然后每次分类，当当前的数有偶数个，那么每次操作2个，当前为1的时候也操作2个，否则操作3个，并且选出的是最大的，这个用优先队列维护。

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<stdlib.h>
  4 #include<queue>
  5 #include<string.h>
  6 #include<iostream>
  7 #include<math.h>
  8 using namespace std;
  9 typedef struct node
 10 {
 11         int id;
 12         int cost;
 13         bool operator<(const node&cx)const
 14         {
 15                 return cx.cost > cost;
 16         }
 17 } ss;
 18 int ans[1000];
 19 int ab[10005][105];
 20 priority_queue<ss>que;
 21 int flag[1005];
 22 int main(void)
 23 {
 24         int n;
 25         scanf("%d",&n);
 26         int i,j;
 27         for(i = 0; i < n; i++)
 28         {
 29                 scanf("%d",&ans[i]);
 30         }
 31         memset(ab,0,sizeof(ab));
 32         while(!que.empty())
 33                 que.pop();
 34         memset(flag,0,sizeof(flag));
 35         int cn = 0;
 36         for(i = 0; i < n; i++)
 37         {
 38                 if(!flag[ans[i]])
 39                 {
 40                         cn++;
 41                 }
 42                 flag[ans[i]]++;
 43                 ss ak;
 44                 ak.cost = ans[i];
 45                 ak.id = i;
 46                 que.push(ak);
 47         }
 48         int cnt = 0;
 49         while(cn>1)
 50         {
 51                 ss a = que.top();
 52                 que.pop();
 53                 ss b = que.top();
 54                 que.pop();
 55                 if(a.cost!=b.cost)
 56                 {
 57                         if(a.cost>0)
 58                         {
 59                                 flag[a.cost]--;
 60                                 if(flag[a.cost]==0)cn--;
 61                                 if(flag[a.cost-1]==0)cn++;
 62                                 a.cost--;
 63                                 flag[a.cost]++;
 64                         }
 65                         if(b.cost>0)
 66                         {
 67                                 flag[b.cost]--;
 68                                 if(flag[b.cost]==0)cn--;
 69                                 if(flag[b.cost-1]==0)cn++;
 70                                 b.cost--;
 71                                 flag[b.cost]++;
 72                         }
 73                         ab[cnt][a.id]++;
 74                         ab[cnt][b.id]++;
 75                 }
 76                 else if(flag[a.cost]%2==0)
 77                 {
 78                         if(a.cost>0)
 79                         {
 80                                 flag[a.cost]--;
 81                                 if(flag[a.cost]==0)cn--;
 82                                 if(flag[a.cost-1]==0)cn++;
 83                                 a.cost--;
 84                                 flag[a.cost]++;
 85                         }
 86                         if(b.cost>0)
 87                         {
 88                                 flag[b.cost]--;
 89                                 if(flag[b.cost]==0)cn--;
 90                                 if(flag[b.cost-1]==0)cn++;
 91                                 b.cost--;
 92                                 flag[b.cost]++;
 93                         }
 94                             ab[cnt][a.id]++;
 95                         ab[cnt][b.id]++;
 96                 }
 97                 else
 98                 {
 99                         ss c = que.top();
100                         que.pop();
101                           if(a.cost>0)
102                         {
103                                 flag[a.cost]--;
104                                 if(flag[a.cost]==0)cn--;
105                                 if(flag[a.cost-1]==0)cn++;
106                                 a.cost--;
107                                 flag[a.cost]++;
108                         }
109                         if(b.cost>0)
110                         {
111                                 flag[b.cost]--;
112                                 if(flag[b.cost]==0)cn--;
113                                 if(flag[b.cost-1]==0)cn++;
114                                 b.cost--;
115                                 flag[b.cost]++;
116                         }
117                         if(c.cost>0)
118                         {
119                                 flag[c.cost]--;
120                                 if(flag[c.cost]==0)cn--;
121                                 if(flag[c.cost-1]==0)cn++;
122                                 c.cost--;
123                                 flag[c.cost]++;
124                         }ab[cnt][c.id]++;
125                             ab[cnt][a.id]++;
126                         ab[cnt][b.id]++;
127                         que.push(c);
128                 }
129                 cnt++;
130                 que.push(a);
131                 que.push(b);
132         }
133         int x = que.top().cost;
134         printf("%d\n",x);
135         printf("%d\n",cnt);
136         for(i = 0;i < cnt;i++)
137         {
138             for(j = 0;j < n;j++)
139             {
140                 if(j == 0)
141                     printf("%d",ab[i][j]);
142                 else printf("%d",ab[i][j]);
143             }
144             printf("\n");
145         }
146         return 0;
147 }

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