B. Painting Pebbles
time limit per test
memory limit per test
input
output
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
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
Input
n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
n positive integers a1, a2, ..., an (1 ≤ ai) denoting number of pebbles in each of the piles.
Output
NO" (without quotes) .
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
贪心，每种颜色尽可能放
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,k;
int a[MAXN];
int f[MAXN][MAXN];
int main()
{
// freopen("Painting.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n>>k;
For(i,n) {cin>>a[i];f[i][0]=a[i];}
int mi=a[1],ma=a[1];
For(i,n) ma=max(ma,a[i]),mi=min(mi,a[i]);
if (ma-mi>k)
{
cout<<"NO"<<endl;
return 0;
}
cout<<"YES"<<endl;
For(i,n)
{
printf("1");
Fork(j,2,a[i])
{
if (j<=mi) printf(" 1");
else printf(" %d",j-mi);
}
printf("\n");
}
return 0;
}
