codeforces 1051C - Vasya and Multisets (思维)

题目链接：https://codeforces.com/problemset/problem/1051/C

统计出现一次的数的奇偶性
再统计出现三次以上的数是否出现过，补齐即可

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;

const int maxn = 110;

int n;
int a[maxn],bas[maxn];

ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }

int main(){
	n = read();
	int ma = 0;
	for(int i=1;i<=n;++i) a[i] = read(), ++bas[a[i]], ma = max(ma,a[i]);
	
	int cnt1 = 0, cnt2 = 0;
	for(int i=1;i<=ma;++i){
		if(bas[i]==1) ++cnt1;
		if(bas[i]>=3) ++cnt2;
		//if(bas[i]>3) ++cnt3;
	}
	
	if(cnt1%2 == 0){
		printf("YES\n");
		int rec = 0;
		for(int i=1;i<=n;++i){
			if(bas[a[i]] == 1){
				if(rec<cnt1/2) printf("A");
				else printf("B");
				++rec;
			}else {
				printf("B");
			}
		}
	}else{
		if(cnt2>=1){
			printf("YES\n");
			int rec = 0, f = 0;
			for(int i=1;i<=n;++i){
				if(bas[a[i]] == 1){
					if(rec<cnt1/2) printf("A");
					else printf("B");
					++rec;
				}
				else if(bas[a[i]] >=3 ){
					if(!f){
						printf("A");
						f=1;
					}else printf("B");
				}else{
					printf("B");
				}
			}
		}else{
			printf("NO\n");
		}
	}
	return 0;
}