Vasya and Multisets

Vasya has a multiset ss consisting of nn integer numbers. Vasya calls some number xx nice if it appears in the multiset exactly once. For example, multiset {1,1,2,3,3,3,4}{1,1,2,3,3,3,4} contains nice numbers 22 and 44.

Vasya wants to split multiset ss into two multisets aa and b(one of which may be empty) in such a way that the quantity of nice numbers in multiset aa would be the same as the quantity of nice numbers in multiset bb (the quantity of numbers to appear exactly once in multiset aa and the quantity of numbers to appear exactly once in multiset bb).

Input

The first line contains a single integer n (2n100)n (2≤n≤100).

The second line contains nn integers s1,s2,sn (1si100)s1,s2,…sn (1≤si≤100) — the multiset ss.

Output

If there exists no split of ss to satisfy the given requirements, then print "NO" in the first line.

Otherwise print "YES" in the first line.

The second line should contain a string, consisting of nn characters. ii-th character should be equal to 'A' if the ii-th element of multiset ss goes to multiset aa and 'B' if if the ii-th element of multiset ss goes to multiset bb. Elements are numbered from 11 to nn in the order they are given in the input.

If there exist multiple solutions, then print any of them.

Examples
input
Copy
4
3 5 7 1
output
Copy
YES
BABA
input
Copy
3
3 5 1
output
Copy
NO

写的垃圾,疯狂扣细节
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
//#include <xfunctional>
#define ll long long
#define mod 1000000007
using namespace std;
int dir[4][2] = { { 0,1 },{ 0,-1 },{ -1,0 },{ 1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;

int main()
{
    int n, cnt = 0,pos=0;
    cin >> n;
    vector<int> s(101,0),a(n+1);
    for (int i = 1; i <= n; i++)
    {
        int t;
        cin >> t;
        a[i] = t;
        s[t]++;
        if (s[t] > 2)
        {
            pos = t;
        }
    }
    for (int i = 1; i <= 100; i++)
    {
        if (s[i] == 1)
            cnt++;
    }
    if (cnt % 2 == 0)
    {
        cout << "YES" << endl;
        int flag = 1;
        for (int i = 1; i <= n; i++)
        {
            if (flag && s[a[i]]==1)
            {
                cout << "A";
                flag = 0;
            }
            else if (flag==0 && s[a[i]] == 1)
            {
                cout << "B";
                flag = 1;
            }
            else if(s[a[i]]>1)
            {
                cout << "B";
            }
        }
    }
    else if (cnt % 2 && pos)
    {
        int flag = 1,firstone=1,special=1;
        cout << "YES" << endl;
        for (int i = 1; i <= n; i++)
        {
            if (a[i] == pos)
            {
                if (special)
                {
                    cout << "A";
                    special = 0;
                }
                else
                {
                    cout << "B";
                }
                continue;
            }
            if (s[a[i]] == 1)
            {
                if (firstone)
                {
                    cout << "B";
                    firstone = 0;
                    continue;
                }
                if (flag && s[a[i]] == 1)
                {
                    cout << "A";
                    flag = 0;
                }
                else if (flag == 0 && s[a[i]] == 1)
                {
                    cout << "B";
                    flag = 1;
                }
            }    
            else if(s[a[i]]>1)
            {
                    cout << "B";
            }
        }
    }
    else
        cout << "NO";
    return 0;
}
原文地址:https://www.cnblogs.com/dealer/p/12439117.html