寒假1

You are given an array $a$

You have to color this array in $k$

Each element of the array should be colored in some color;
For each $i$
For each $i$

Obviously, such coloring might be impossible. In this case, print "NO". Otherwise print "YES" and any coloring (i.e. numbers $c_{1}, c_{2}, \dots c_{n}$

Input

The first line of the input contains two integers $n$

The second line of the input contains $n$

Output

If there is no answer, print "NO". Otherwise print "YES" and any coloring (i.e. numbers $c_{1}, c_{2}, \dots c_{n}$

题意：首先给你n个数的一个序列，以及k种颜色，你需要做到序列中的每个数都必须染色，且每种颜色都必须用到，而且每种颜色所涂的数中不能有相同的数。如果能做到，输出这个序列的涂色方案（不唯一）

思路：首先我们判断是否能够涂成功。我们计算这个序列中相同的数出现的最大次数。如果这个最大次数大于k，则失败，否则成功。

　　　对于成功的，我们需要求出方案。我的方法比较复杂，想的有点多。我先统计了每个数的出现次数，然后我们遍历这个序列，每个数的颜色就是他们出现的次数，然后出现次数减一。这里就满足每种颜色所涂的数中没有相同的数，但是没要保证每种颜色都出现。这时我们对于这个方案进行改善。我们需要将出现多次的颜色向后推，推到没有出现的颜色，然后将这个颜色改掉，即满足出现每种颜色。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,m;
    cin>>n>>m;
    int num[5005];
    memset(num,0,sizeof(num));
    int a[5005];
    int maps[5005];
    int k[5005];
    map<int,int>ss;
    memset(maps,0,sizeof(maps));
    for(int i=0;i<n;i++){
        cin>>a[i];
        num[a[i]]++;
    }
    int maxx=-1;
    for(int i=0;i<=5000;i++) maxx=max(maxx,num[i]);
    if(maxx>m) cout<<"NO"<<endl;
    else{
        cout<<"YES"<<endl;
        for(int i=0;i<n;i++){
            k[i]=num[a[i]];
            num[a[i]]--;
        }
        for(int i=0;i<n;i++){
            if(!ss[k[i]]) ss[k[i]]++;
            else {
                int t=k[i]+1;
                while(t<=m){
                    if(!ss[t]){
                        k[i]=t;
                        ss[t]++;
                        break;
                    }
                    t++;
                }
            }
        }
        for(int i=0;i<n;i++) cout<<k[i]<<' ';
    }
    return 0;
}

C - Doors Breaking and Repairing

You are policeman and you are playing a game with Slavik. The game is turn-based and each turn consists of two phases. During the first phase you make your move and during the second phase Slavik makes his move.

There are $n$

During your move you can try to break one of the doors. If you choose door $i$

During Slavik's move he tries to repair one of the doors. If he chooses door $i$

The game lasts $10^{100}$

Your goal is to maximize the number of doors with durability equal to $0$

Input

The first line of the input contains three integers $n$

The second line of the input contains $n$

Output

Print one integer — the number of doors with durability equal to $0$

$0$

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,a,b,c;
    int num[105];
    int t=0;
    cin>>n>>a>>b;
    int ans=0;
    for(int i=0;i<n;i++){
        cin>>c;
        if(c<=a) ans++;
        else num[t++]=c;
    }
    if(a>b) cout<<n<<endl;
    else{
        if(ans%2==1) ans++;
        cout<<ans/2<<endl;
    }
    return 0;
}

D - Balanced Ternary String

You are given a string $s$

Your task is to replace minimum number of characters in this string with other characters to obtain a balanced ternary string (balanced ternary string is a ternary string such that the number of characters '0' in this string is equal to the number of characters '1', and the number of characters '1' (and '0' obviously) is equal to the number of characters '2').

Among all possible balanced ternary strings you have to obtain the lexicographically (alphabetically) smallest.

Note that you can neither remove characters from the string nor add characters to the string. Also note that you can replace the given characters only with characters '0', '1' and '2'.

It is guaranteed that the answer exists.

Input

The first line of the input contains one integer $n$

The second line contains the string $s$

Output

Print one string — the lexicographically (alphabetically) smallest balanced ternary string which can be obtained from the given one with minimum number of replacements.

Because $n$

$n$

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,m;
    string s;
    cin>>n;
    cin>>s;
    map<char,int> num;
    for(int i=0; i<s.length(); i++)
    {
        num[s[i]]++;
    }
    int a=num['0']-n/3;
    int b=num['1']-n/3;
    int c=num['2']-n/3;
    for(int i=0;i<n;i++){
        if(s[i]=='2'&&c>0){
            if(a<0){
                a++;
                c--;
                s[i]='0';
            }
            else if(b<0){
                b++;
                c--;
                s[i]='1';
            }
        }
        if(s[i]=='1'&&b>0){
            if(a<0){
                b--;
                a++;
                s[i]='0';
            }
        }
    }
    for(int i=n-1;i>=0;i--){
        if(s[i]=='0'&&a>0){
            if(c<0){
                c++;
                a--;
                s[i]='2';
            }
            else if(b<0){
                b++;
                a--;
                s[i]='1';
            }
        }
        if(s[i]=='1'&&b>0){
            if(c<0){
                c++;
                b--;
                s[i]='2';
            }
        }
    }
    cout<<s<<endl;
    return 0;
}