AtCoder Beginner Contest 115 题解
题目链接:https://abc115.contest.atcoder.jp/
A Christmas Eve Eve Eve
题目:
Time limit : 2sec / Memory limit : 1024MB
Score : 100 points
Problem Statement
In some other world, today is December D-th.
Write a program that prints Christmas if D=25, Christmas Eve if D=24, Christmas Eve Eve if D=23 and Christmas Eve Eve Eve if D=22.
Constraints
- 22≤D≤25
- D is an integer.
Input
Input is given from Standard Input in the following format:
D
Output
Print the specified string (case-sensitive).
Sample Input 1
25
Sample Output 1
Christmas
Sample Input 2
22
Sample Output 2
Christmas Eve Eve Eve
Be sure to print spaces between the words.
题解:
没什么好说的。
1 #include <cstdio> 2 3 using namespace std; 4 5 int main() 6 { 7 int d; 8 scanf("%d", &d); 9 if(d == 25) 10 printf("Christmas "); 11 else if(d == 24) 12 printf("Christmas Eve "); 13 else if(d == 23) 14 printf("Christmas Eve Eve "); 15 else if(d == 22) 16 printf("Christmas Eve Eve Eve "); 17 return 0; 18 }
B Christmas Eve Eve
题目:
Time limit : 2sec / Memory limit : 1024MB
Score : 200 points
Problem Statement
In some other world, today is the day before Christmas Eve.
Mr. Takaha is buying N items at a department store. The regular price of the i-th item (1≤i≤N) is pi yen (the currency of Japan).
He has a discount coupon, and can buy one item with the highest price for half the regular price. The remaining N−1 items cost their regular prices. What is the total amount he will pay?
Constraints
- 2≤N≤10
- 100≤pi≤10000
- pi is an even number.
Input
Input is given from Standard Input in the following format:
N
p1
p2
:
pN
Output
Print the total amount Mr. Takaha will pay.
Sample Input 1
3
4980
7980
6980
Sample Output 1
15950
The 7980-yen item gets the discount and the total is 4980+7980⁄2+6980=15950 yen.
Note that outputs such as 15950.0 will be judged as Wrong Answer.
Sample Input 2
4
4320
4320
4320
4320
Sample Output 2
15120
Only one of the four items gets the discount and the total is 4320⁄2+4320+4320+4320=15120 yen.
题解:
没什么好说的。
1 #include <cstdio> 2 #include <iostream> 3 4 using namespace std; 5 6 int main() 7 { 8 int n; 9 scanf("%d", &n); 10 int p[15], maxp = 0, sum = 0; 11 for(int i = 0; i < n; ++i) 12 { 13 scanf("%d", &p[i]); 14 maxp = max(maxp, p[i]); 15 sum += p[i]; 16 } 17 sum = sum - maxp + maxp / 2; 18 printf("%d ", sum); 19 return 0; 20 }