A - New Year

題目鏈接：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 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

題目鏈接：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 ---

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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?

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

轉載于:https://www.cnblogs.com/shixinzei/p/8342893.html

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