ACM--字符串--CSU--1550-- Simple String

中南大学OJ题目地址：http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1550

1550: Simple String

Time Limit: 1 Sec   Memory Limit: 256 MB
Submit: 471   Solved: 206
[ Submit][ Status][ Web Board]

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

AABB
BBCC
AACC
AAAA
BBBB
AAAA

Sample Output

YES
NO

=================================傲娇的分割线===============================

题意：

题目的意思是给你两个长度为2*N的字符串，A,B,然后给定一个2*N长的字符串C，C字符串分别由A,B中的N个字符串组成，判断给定的两个字符串能否组成C。

分析：

由题意，我们就可以列出9种情况：判断c中的某个字符串是否符合题意。

因为A,B中的某个字母最多拿出N个去组成C，（当A,B字符串中的某个字母如果大于N个时，也最多只能拿出N个出来）这样才能满足组成C字符串的条件。

	A[I]<N	A[I]=N	A[I]>N
B[I]<N	C[I]<=A+B	C[I]<=A+B	C[I]<=N+B
B[I]=N	C[I]<=A+B	C[I]<=A+B	C[I]<=B+N
B[I]>N	C[I]<=A+N	C[I]<=A+N	C[I]<=N+N

#include <iostream>
#include <string.h>
using namespace std;
int A[26],B[26],C[26];
char str[100001];
bool flag[26];
int main(){
    int i;
    while(cin>>str){
       int len=strlen(str);
       //初始化数组
       memset(A,0,sizeof(A));
       memset(B,0,sizeof(B));
       memset(C,0,sizeof(C));
       //将数据读入A数组
       for(i=0;i<len;i++){
         A[str[i]-'A']++;
       }
       cin>>str;
       //将数据读入B数组
        for(i=0;i<len;i++){
         B[str[i]-'A']++;
       }
       cin>>str;
       //将数据读入C数组
        for(i=0;i<len;i++){
         C[str[i]-'A']++;
       }
       memset(flag,true,sizeof(flag));
       for(i=0;i<26;i++){
          if(A[i]>len/2){//当A中的某个字母的数量大于N的时候只能取N个
            A[i]=len/2;
          }
          if(B[i]>len/2){//当B中的某个字母的数量大于N的时候只能取N个
            B[i]=len/2;
          }
          if(C[i]<=A[i]+B[i]){
            flag[i]=false;
          }
       }
       //循环判断是不是全部都是false
       for(i=0;i<26;i++){
          if(flag[i]==true){
            break;
          }
       }
       if(i<26){
          cout<<"NO"<<endl;
       }else{
          cout<<"YES"<<endl;
       }

    }
  return 0;
}

参考博客：http://blog.csdn.net/whjkm/article/details/45510167