1 second
256 megabytes
standard input
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 kgiven 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 iand
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.
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.
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.
4 4 1 2 3 4
YES 1 1 4 1 2 4 1 2 3 4
5 2 3 2 4 1 3
NO
5 4 3 2 4 3 5
YES 1 2 3 1 3 1 2 3 4 1 3 4 1 1 2 3 4
# include<cstdio> # include<iostream> # include<algorithm> # include<cmath> # include<cstring> using namespace std; int a1[105]; int main(void) { int n,k; int p; cin>>n>>k; for( int i = 0;i < n;i++ ) { cin>>a1[i]; } int _min = a1[0]; int _max = a1[0]; for ( int i = 0;i < n;i++ ) { if ( a1[i] < _min ) { _min = a1[i]; } if ( a1[i] > _max ) { _max = a1[i]; } } if( _max-_min > k ) { printf("NO\n"); } else{ printf("YES\n"); for( int i = 0;i < n;i++ ) { printf("1"); for( int j = 2;j <= a1[i];j++ ) { p = j; while( p > k ) p = p - k; printf(" %d",p); } printf("\n"); } } return 0; }