Codeforces 789A Anastasia and pebbles(数学,思维题)
Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park.
She has only two pockets. She can put at most k pebbles in each pocket at the same time. There are n different pebble types in the park, and there are wi pebbles of the i-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day.
Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket.
The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 104) — number of pebbles of each type.
The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles.
3 2
2 3 4
3
5 4
3 1 8 9 7
5
In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day.
Optimal sequence of actions in the second sample case:
- In the first day Anastasia collects 8 pebbles of the third type.
- In the second day she collects 8 pebbles of the fourth type.
- In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type.
- In the fourth day she collects 7 pebbles of the fifth type.
- In the fifth day she collects 1 pebble of the second type.
题目链接:http://codeforces.com/contest/789/problem/A
思路:开始用暴力直接搜,然后太复杂了,然后WA了,看了下别人的题解,发现好像有个这样的计算公式:
设每种石子的数量分别为x(用循环来做),每个口袋可以装的石子数为k
计算每种石子装满一个口袋需要的天数sum=(k+x-1)/k(容易看出它这样做是为了向下取整);
每次将天数进行累加,最终得到的天数sum=(sum+1)/2(同样是向上取整);
下面给出AC代码:
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 int n,k,x; 6 while(scanf("%d%d",&n,&k)!=EOF) 7 { 8 int sum=0; 9 while(n--) 10 { 11 scanf("%d",&x); 12 sum+=(x+k-1)/k; 13 } 14 printf("%d\n",(sum+1)/2); 15 } 16 return 0; 17 }
作 者:Angel_Kitty
出 处:https://www.cnblogs.com/ECJTUACM-873284962/
关于作者:阿里云ACE,目前主要研究方向是Web安全漏洞以及反序列化。如有问题或建议,请多多赐教!
版权声明:本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文链接。
特此声明:所有评论和私信都会在第一时间回复。也欢迎园子的大大们指正错误,共同进步。或者直接私信我
声援博主:如果您觉得文章对您有帮助,可以点击文章右下角【推荐】一下。您的鼓励是作者坚持原创和持续写作的最大动力!
欢迎大家关注我的微信公众号IT老实人(IThonest),如果您觉得文章对您有很大的帮助,您可以考虑赏博主一杯咖啡以资鼓励,您的肯定将是我最大的动力。thx.
我的公众号是IT老实人(IThonest),一个有故事的公众号,欢迎大家来这里讨论,共同进步,不断学习才能不断进步。扫下面的二维码或者收藏下面的二维码关注吧(长按下面的二维码图片、并选择识别图中的二维码),个人QQ和微信的二维码也已给出,扫描下面👇的二维码一起来讨论吧!!!
欢迎大家关注我的Github,一些文章的备份和平常做的一些项目会存放在这里。