# codeforces 509B（构造，思维）

### 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.

### 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;
}```

0 条评论

• ### codeforces 1384A（构造）

The length of the longest common prefix of two strings s=s1s2…sn and t=t1t2…tm i...

• ### codeforces 628B（数学）

Max wants to buy a new skateboard. He has calculated the amount of money that is...

• ### Code Forces 21C Stripe 2

C. Stripe 2 time limit per test 1 second memory limit per test 64 megabyte...

• ### codeforces 1384A（构造）

The length of the longest common prefix of two strings s=s1s2…sn and t=t1t2…tm i...

• ### codeforces 628B（数学）

Max wants to buy a new skateboard. He has calculated the amount of money that is...

• ### c/c++基础零散补充

一、C语言的指针与数组、结构体里的成员数组和指针、传入传出参数、回调函数、头文件包含 指针的本质都只是一个内存地址，如果是多字节变量，则是其内存首地址（低地址...

• ### PAT 1018 Public Bike Management（Dijkstra 最短路）

1018. Public Bike Management (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16...

• ### HDU 1017 A Mathematical Curiosity【水，坑】

A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit:...

• ### PCIE AER

PCI Express error signaling can occur on the PCI Express link itself or on behal...

### 活动推荐 