题意描述

There are n piles of pebbles on the table, the i-th pile contains a i 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 b i, 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 |b i, c - b j, c| ≤ 1. It isn’t necessary to use all k colors: if color c hasn’t been used in pile i, then b i, c is considered to be zero.

给每一组的石头染色，要求任意两组颜色的数量差不能超过1.

思路

观察样例可以发现，如果每个数和最小的数的差值大于k的话是无解的。对于有解的情况，我们可以先输出min个1，然后再依次输出1，2，3，…，n即可

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=1e5+10;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
int a[N];
void solve(){
    int n,k;cin>>n>>k;
    int MIN=INF;
    rep(i,1,n+1){
        cin>>a[i];
        MIN=min(MIN,a[i]);
    }
    int flag=0;
    rep(i,1,n+1){
        if(a[i]-MIN>k) flag=1;
    }
    if(flag) cout<<"NO"<<endl;
    else{
        cout<<"YES"<<endl;
        rep(i,1,n+1){
            rep(j,1,MIN+1) cout<<1<<' ';
            rep(j,1,a[i]-MIN+1) cout<<j<<' ';
            cout<<endl;
        }
    }
}
int main(){
    IOS;
    solve();
    return 0;
}