1550: Simple String （做得少的思维题，两个字符串能否组成另外一个字符串问题）...-白红宇

1550: Simple String （做得少的思维题，两个字符串能否组成另外一个字符串问题）...

阅读量：4965 次

发布时间：2019-06-12

本文共 1906 字，大约阅读时间需要 6 分钟。

Time Limit: 1 Sec Memory Limit: 256 Mb Submitted: 682 Solved: 287

Description

Welcome,this is the 2015 3th Multiple Universities Programming Contest ,Changsha ,Hunan Province. In order to let you feel fun, ACgege will give you a simple problem. But is that true? OK, let’s enjoy it.

There are three strings A , B and C. The length of the string A is 2*N, and the length of the string B and C is same to A. You can take N characters from A and take N characters from B. Can you set them to C ?

Input

There are several test cases.

Each test case contains three lines A,B,C. They only contain upper case letter.

0<N<100000

The input will finish with the end of file.

Output

For each the case, if you can get C, please print “YES”. If you cann’t get C, please print “NO”.

Sample Input

AABBBBCCAACCAAAABBBBAAAA

Sample Output

YESNO

Hint

Source

题目意思：

给你三个字符串

a，b，c

长度都是2*n

问你从a和b中各抽出n给字符

能不能组成c

分析：

统计a，b，c中26个字符出现的次数

如果某字符在a中出现次数+在b中出现次数小于该字符在c中出现次数

那么肯定不能组成c

1.A+B 与C的交集必须>=n

2.A与C的交集必须>=n/2，B与C的交集必须>=n/2。

理解上面两句话就可以了

#include
        
         #include
         
          #include
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  
                   #include 
                   #include
                    
                     #include
                     
                      #include
                      
                       typedef long long LL;using namespace std;#define max_v 125int c1[30];int c2[30];int c3[30];void init(){ memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); memset(c3,0,sizeof(c3));}int main(){ string str1,str2,str3; while(cin>>str1) { init(); cin>>str2; cin>>str3; int l=str1.length(); for(int i=0; i
                       
                        l/2) flag=0; if(flag) printf("YES\n"); else printf("NO\n"); } return 0;}/*题目意思：给你三个字符串a，b，c长度都是2*n问你从a和b中各抽出n给字符能不能组成c分析：统计a，b，c中26个字符出现的次数如果某字符在a中出现次数+在b中出现次数小于该字符在c中出现次数那么肯定不能组成c1.A+B 与C的交集必须>=n2.A与C的交集必须>=n/2，B与C的交集必须>=n/2。理解上面两句话就可以了*/

转载于:https://www.cnblogs.com/yinbiao/p/9498622.html

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