codechef+Name Reduction

Name Reduction

Problem code: NAME1

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In an attempt to reduce the growing population, Archer was asked to come up with a plan.Archer being as intelligent as he is, came up with the following plan:

If N children, with names C₁, C₂, ..., C_N, are born to parents with namesA and B, and you consider C to be theconcatenation of all the names of the children, i.e.C = C₁ + C₂ + ... + C_N (where + is concatenation operator), then C should be a substring of one of the permutations ofA + B.

You are given the task to verify whether the names parents propose to give their children are in fact permissible by Archer's plan or not.

Input

The first line contains an integer T, the number of test cases.T test cases follow. Each test case stats with a line containing two space separated stringsA and B, denoting the names of the parents. The next line contains a single integerN denoting the number of children A and B are planning to have. Following this are N lines, thei'th line containing C_i, the proposed name for thei'th child.

Output

For each test case output a single line containing "YES" if the names are permissible by Archer's plan, otherwise print"NO". (quotes are meant for clarity, please don't print them)

Constraints

• 1 ≤ T ≤ 100
• 1 ≤ N ≤ 1000
• The lengths of all the strings including A, B, and all C_i will be in the range [1, 40000], both inclusive. All these strings will contain only lowercase English letters.
• The combined lengths of all names of children will not exceed the combined length of the names of their parents.

Example

Input:3tom marvoloriddle2lordvoldemortcheap up1heapcupbruce wayne2batmanOutput:YESYESNO

Explanation:

Let Y denote the concatenation of names of all the children, andX denote the concatenation of the names of the parents.

Case 1: Here X = "tommarvoloriddle", and Y = "lordvoldemort". Consider Z = "iamlordvoldemort". It is not difficult to see thatZ is a permutation of X and Y is a substring ofZ. Hence Y is a substring of a permutation of X, so the answer is "YES".

Case 2: Here X = "cheapup", and Y = "heapcup". SinceY in itself is a permutation of X, and as every string is a substring of itself,Y is a substring of X and also a permutation ofX. Hence "YES".

Case 3: Here X = "brucewayne", and Y = "batman". As "t" is not present inX, "t" wont be present in any permutation of X, hence the answer is "NO".

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/*题目大意：给你双亲的姓名，然后给你n个孩纸的姓名，问你双亲合起来的字符串能够、否经过变形包含孩纸的字符串和；思路：将双亲和孩纸字符串按字典序排序，然后遍历即可*/#include<iostream>#include<string>#include<algorithm>using namespace std;int main(){int t;cin>>t;while(t--){string a,b,c;cin>>a>>b;a=a+b;sort(a.begin(),a.end());int i,n;cin>>n;for(i=0;i<n;i++){cin>>b;c+=b;}sort(c.begin(),c.end());int k=0;for(i=0;i<a.length();i++)if(a[i]==c[k])k++;if(k==c.length())cout<<"YES"<<endl;elsecout<<"NO"<<endl;//cout<<a<<" "<<c<<endl;}return 0;}