zoukankan      html  css  js  c++  java
  • codeforces 1051C

    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
    ");
    		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
    ");
    			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
    ");
    		}
    	}
    	return 0;
    }
    
  • 相关阅读:
    Kendo
    过河
    数组分组(简单dp)
    Codeforces Round #604 (Div. 2)(A-E)
    HDU1253
    HDU1026
    linux常用命令(二) --目录操作
    linux常用命令(一)--ls
    hdu 1072
    Codeforces Round #597 (Div. 2)(A-D)
  • 原文地址:https://www.cnblogs.com/tuchen/p/13824858.html
Copyright © 2011-2022 走看看