B. Painting Pebbles

B. Painting Pebbles
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.

In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.

Input

The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.

Output

If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .

Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.

Sample test(s)
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <string>
 7 #include <vector>
 8 #include <set>
 9 #include <map>
10 #include <queue>
11 #include <stack>
12 #include <sstream>
13 #include <cctype>
14 #include <cassert>
15 #include <typeinfo>
16 #include <utility>    //std::move()
17 using namespace std;
18 const int INF = 0x7fffffff;
19 const double EXP = 1e-8;
20 const int MS = 101;
21 typedef long long LL;
22 
23 int a[MS];
24 int main()
25 {
26     int n, k, i, j;
27     cin >> n >> k;
28     int mini = INF, maxi = -INF;
29     for (int i = 0; i < n; i++)
30     {
31         cin >> a[i];
32         if (mini>a[i])
33             mini = a[i];
34         if (maxi < a[i])
35             maxi= a[i];
36     }
37     if ((maxi / k - 1)*k+ maxi%k > mini)
38     {
39         cout << "NO" << endl;
40     }
41     else
42     {
43         cout << "YES" << endl;
44         for (i = 0; i < n; i++)
45         {
46             for (j = 0; j < a[i]; j++)
47             {
48                 if (j)
49                     cout << " ";
50                 cout << j%k + 1;
51             }
52             cout << endl;
53         }
54     }
55     return 0;
56 }

 

posted @ 2015-02-01 00:07  daydaycode  阅读(171)  评论(0)  编辑  收藏  举报