A. YES or YES?

B. ICPC Balloons

C. Cypher

D. Double Strings

A. YES or YES?

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a string ss of length 33, consisting of uppercase and lowercase English letters. Check if it is equal to "YES" (without quotes), where each letter can be in any case. For example, "yES", "Yes", "yes" are all allowable.

Input

The first line of the input contains an integer tt (1≤t≤1031≤t≤103) — the number of testcases.

The description of each test consists of one line containing one string ss consisting of three characters. Each character of ss is either an uppercase or lowercase English letter.

Output

For each test case, output "YES" (without quotes) if ss satisfies the condition, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Example

input

10
YES
yES
yes
Yes
YeS
Noo
orZ
yEz
Yas
XES

output

YES
YES
YES
YES
YES
NO
NO
NO
NO
NO

#include<bits/stdc++.h>
using namespace std;
int main()
{int t;cin>>t;while(t--){string s;cin>>s;int tmp1=0,tmp2=0,tmp3=0;for(int i=0;i<s.size();i++){if(s[0]=='Y'||s[0]=='y') tmp1++;if(s[1]=='E'||s[1]=='e') tmp2++;if(s[2]=='S'||s[2]=='s') tmp3++;}if(tmp1!=0&&tmp2!=0&&tmp3!=0)cout<<"YES"<<endl;else cout<<"NO"<<endl;}return 0;
}

B. ICPC Balloons

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In an ICPC contest, balloons are distributed as follows:

Whenever a team solves a problem, that team gets a balloon.
The first team to solve a problem gets an additional balloon.

A contest has 26 problems, labelled AA, BB, CC, ..., ZZ. You are given the order of solved problems in the contest, denoted as a string ss, where the ii-th character indicates that the problem sisi has been solved by some team. No team will solve the same problem twice.

Determine the total number of balloons that the teams recieved. Note that some problems may be solved by none of the teams.

Input

The first line of the input contains an integer tt (1≤t≤1001≤t≤100) — the number of testcases.

The first line of each test case contains an integer nn (1≤n≤501≤n≤50) — the length of the string.

The second line of each test case contains a string ss of length nn consisting of uppercase English letters, denoting the order of solved problems.

Output

For each test case, output a single integer — the total number of balloons that the teams received.

Example

input

6
3
ABA
1
A
3
ORZ
5
BAAAA
4
BKPT
10
CODEFORCES

output

#include<bits/stdc++.h>
using namespace std;
int main()
{int t;cin>>t;while(t--){int n;cin>>n;string s;cin>>s;set<char> st;for(int i=0;i<s.size();i++){st.insert(s[i]);}int tmp=st.size();cout<<tmp*2+n-st.size()<<endl;}return 0;
}

C. Cypher

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Luca has a cypher made up of a sequence of nn wheels, each with a digit aiai written on it. On the ii-th wheel, he made bibi moves. Each move is one of two types:

up move (denoted by UU): it increases the ii-th digit by 11. After applying the up move on 99, it becomes 00.
down move (denoted by DD): it decreases the ii-th digit by 11. After applying the down move on 00, it becomes 99.

Example for n=4n=4. The current sequence is 0 0 0 0.

Luca knows the final sequence of wheels and the moves for each wheel. Help him find the original sequence and crack the cypher.

Input

The first line contains a single integer tt (1≤t≤1001≤t≤100) — the number of test cases.

The first line of each test case contains a single integer nn (1≤n≤1001≤n≤100) — the number of wheels.

The second line contains nn integers aiai (0≤ai≤90≤ai≤9) — the digit shown on the ii-th wheel after all moves have been performed.

Then nn lines follow, the ii-th of which contains the integer bibi (1≤bi≤101≤bi≤10) and bibi characters that are either UU or DD — the number of moves performed on the ii-th wheel, and the moves performed. UU and DD represent an up move and a down move respectively.

Output

For each test case, output nn space-separated digits — the initial sequence of the cypher.

Example

input

3
3
9 3 1
3 DDD
4 UDUU
2 DU
2
0 9
9 DDDDDDDDD
9 UUUUUUUUU
5
0 5 9 8 3
10 UUUUUUUUUU
3 UUD
8 UUDUUDDD
10 UUDUUDUDDU
4 UUUU

output

2 1 1
9 0
0 4 9 6 9

#include<bits/stdc++.h>
using namespace std;
int main()
{int t;cin>>t;while(t--){int n;cin>>n;int a[n];for(int i=0;i<n;i++){cin>>a[i];}int j=0;while(n--){int b;string s;cin>>b>>s;for(int i=0;i<s.size();i++){if(s[i]=='D'){if(a[j]==9) a[j]=0;else  a[j]+=1;;} if(s[i]=='U'){if(a[j]==0) a[j]=9;else a[j]-=1;}}cout<<a[j]<<' ';j++;}cout<<endl;}return 0;
}

D. Double Strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given nn strings s1,s2,…,sns1,s2,…,sn of length at most 88.

For each string sisi, determine if there exist two strings sjsj and sksk such that si=sj+sksi=sj+sk. That is, sisi is the concatenation of sjsj and sksk. Note that jj can be equal to kk.

Recall that the concatenation of strings ss and tt is s+t=s1s2…spt1t2…tqs+t=s1s2…spt1t2…tq, where pp and qq are the lengths of strings ss and tt respectively. For example, concatenation of "code" and "forces" is "codeforces".

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The first line of each test case contains a single integer nn (1≤n≤1051≤n≤105) — the number of strings.

Then nn lines follow, the ii-th of which contains non-empty string sisi of length at most 88, consisting of lowercase English letters. Among the given nn strings, there may be equal (duplicates).

The sum of nn over all test cases doesn't exceed 105105.

Output

For each test case, output a binary string of length nn. The ii-th bit should be 11 if there exist two strings sjsj and sksk where si=sj+sksi=sj+sk, and 00 otherwise. Note that jj can be equal to kk.

Example

input

3
5
abab
ab
abc
abacb
c
3
x
xx
xxx
8
codeforc
es
codes
cod
forc
forces
e
code

output

10100
011
10100101

思路：拆分字符串，判断拆分的字符串在不在已输入的字符里。

#include<bits/stdc++.h>
using namespace std;
int main()
{int t;cin>>t;while(t--){int n;cin>>n;string ans=""; //保存二进制结果vector<string> a(n);  //字符串数组map<string,int> mp;   // map标记for(int i=0;i<n;i++){cin>>a[i];mp[a[i]]=1; }for(int i=0;i<n;i++) //遍历字符串{string s="";int ok=0;for(int j=0;j<a[i].size();j++) //拆分字符串{s+=a[i][j];string c="";for(int k=j+1;k<a[i].size();k++) {c+=a[i][k];}if(mp.count(s)==1 && mp.count(c)==1) //判断在不在已标记的mp里面{ok=1;}}if(ok==1) ans+='1';else ans+='0';}cout<<ans<<endl;}return 0;
}

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Codeforces Round #806 (Div. 4)题解

A. YES or YES?

B. ICPC Balloons

C. Cypher

D. Double Strings

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