AtCoder Beginner Contest 084（AB）

A - New Year

题目链接：https://abc084.contest.atcoder.jp/tasks/abc084_a

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Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?

Constraints

• 1≤M≤23
• M is an integer.

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Input

Input is given from Standard Input in the following format:

M

Output

If we have x hours until New Year at M o'clock on 30th, December, print x.

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Sample Input 1

Copy

21

Sample Output 1

Copy

27

We have 27 hours until New Year at 21 o'clock on 30th, December.

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Sample Input 2

Copy

12

Sample Output 2

Copy

36

 #include <iostream>
 using namespace std;
 int main()
 {
     int n;
     while(cin>>n){
         cout<<-n+<<endl;
     }
     return ;
 }

B - Postal Code

题目链接：https://abc084.contest.atcoder.jp/tasks/abc084_b

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Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen -, and the other characters are digits from 0 through 9.

You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

Constraints

• 1≤A,B≤5
• |S|=A+B+1
• S consists of - and digits from 0 through 9.

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Input

Input is given from Standard Input in the following format:

A B
S

Output

Print Yes if S follows the postal code format in AtCoder Kingdom; print No otherwise.

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Sample Input 1

Copy

3 4
269-6650

Sample Output 1

Copy

Yes

The (A+1)-th character of S is -, and the other characters are digits from 0 through 9, so it follows the format.

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Sample Input 2

Copy

1 1
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Sample Output 2

Copy

No

S contains unnecessary -s other than the (A+1)-th character, so it does not follow the format.

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Sample Input 3

Copy

1 2
7444

Sample Output 3

Copy

No

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 #include <iostream>
 using namespace std;
 int main()
 {
     int a,b;
     while(cin>>a>>b){
         string s;
         cin>>s;
         int l=s.length();
         int flag=;
         for(int i=;i<l;i++){
             if(i==a&&s[i]!='-'){
                 flag=;
                 break;
             }
             if(i!=a&&!(s[i]<=''&&s[i]>='')){
                 flag=;
                 break;
             }
         }
         if(flag) cout<<"Yes"<<endl;
         else cout<<"No"<<endl;
     }
     return ;
 }