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CF - 789A. Anastasia and pebbles - 贪心+模拟

1.题目描述:

A. Anastasia and pebbles time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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.

Input

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.

Output

The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles.

Examples input

3 2
2 3 4
      

output

3
      

input

5 4
3 1 8 9 7
      

output

5
      

Note

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.

2.题意概述:

这个小女孩有n个口袋,每个口袋最大容量是k,小女孩会把每天拾取的鹅卵石放到口袋里面,条件是每个口袋只能放相同种类的鹅卵石。给你每种鹅卵石的个数,问你最少要多少天才能拾取完。

3.解题思路:

最开始想的是模拟,成功地T了一发,仔细一想:对于每个卵石类型,我们可以计算出最少数量的口袋,阿纳斯塔西娅需要收集这种类型的鹅卵石。 很容易注意到这个数字是等于

CF - 789A. Anastasia and pebbles - 贪心+模拟

。 所以问题的答案是

CF - 789A. Anastasia and pebbles - 贪心+模拟

.。 解决方案的复杂度是O(N)。

4.AC代码:

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 100100
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;

int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	long _begin_time = clock();
#endif
	int n, k;
	while(~scanf("%d%d", &n, &k))
	{
		int cnt = 0;
		for (int i = 0; i < n; i++)
		{
			int x;
			scanf("%d", &x);
			while (x > 0)
			{
				x -= k;
				cnt++;
			}
		}
		int ans = ceil(1.0 * cnt / 2);
		printf("%d\n", ans);
	}
#ifndef ONLINE_JUDGE
	long _end_time = clock();
	printf("time = %ld ms.", _end_time - _begin_time);
#endif
	return 0;
}
           

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