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 ?
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.
For each the case, if you can get C, please print “YES”. If you cann’t get C, please print “NO”.
AABB
BBCC
AACC
AAAA
BBBB
AAAA
YES
NO
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>
#include <algorithm>
using namespace std;
int A[26], B[26], C[26];
bool flag[26];
char str[200002];
int main()
{
int i, lenth;
while (cin >> str)
{
memset(A, 0, sizeof(A));
memset(B, 0, sizeof(B));
memset(C, 0, sizeof(C));
lenth = strlen(str);
for (i=0; i<lenth; i++)
{
A[str[i]-'A']++;//统计A字符串中某个字母的个数
}
cin >> str;
for (i=0; i<lenth; i++)
{
B[str[i]-'A']++;//统计B字符串中字母的个数
}
cin >> str;
for (i=0; i<lenth; i++)
{
C[str[i]-'A']++;
}
memset(flag, true, sizeof(flag));
for (i=0; i<26; i++)
{
if (A[i] > lenth/2)
{
A[i] = lenth/2; //当A中某个字母>N中，最多也只能提供N个，赋值为N
}
if (B[i] > lenth/2)
{
B[i] = lenth/2;
}
if (C[i] <= A[i] + B[i])//当c字符串满足是，标记为false
{
flag[i] = false;
}
}
for (i=0; i<26; i++)
{
if (flag[i] == true)//c字符串中有一个不满足，跳出循环
{
break;
}
}
if (i<26)
{
cout << "NO" << endl;
}
else
{
cout << "YES" << endl;
}
}
return 0;
}
原文地址：http://blog.csdn.net/whjkm/article/details/45510167
