AtCoder Beginner Contest 084（AB）

A - New Year

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

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.

Input

Input is given from Standard Input in the following format:

Output

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

Sample Input 1

Copy

Sample Output 1

Copy

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

Sample Input 2

Copy

Sample Output 2

Copy

 1 #include <iostream>
 2 using namespace std;
 3 int main()
 4 {
 5     int n;
 6     while(cin>>n){
 7         cout<<24-n+24<<endl;
 8     }
 9     return 0;
10 }

View Code

B - Postal Code

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

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.

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.

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.

Sample Input 2

Copy

1 1
---

Sample Output 2

Copy

No

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

Sample Input 3

Copy

1 2
7444

Sample Output 3

Copy

No

 1 #include <iostream>
 2 using namespace std;
 3 int main()
 4 {
 5     int a,b;
 6     while(cin>>a>>b){
 7         string s;
 8         cin>>s;
 9         int l=s.length();
10         int flag=1;
11         for(int i=0;i<l;i++){
12             if(i==a&&s[i]!='-'){
13                 flag=0;
14                 break;
15             }
16             if(i!=a&&!(s[i]<='9'&&s[i]>='0')){
17                 flag=0;
18                 break;
19             }
20         }
21         if(flag) cout<<"Yes"<<endl;
22         else cout<<"No"<<endl;
23     }
24     return 0;
25 }

View Code

转载于:https://www.cnblogs.com/shixinzei/p/8342893.html