codeforce789A

/*哇比赛时候vj炸了，去codeforce交题就有一种万分期待的感觉虽然题不难直接1A但是心里还是很舒服啊*/

/*这个题的思路是通过三个计数器计数，将大于k的与小于k的分为两组并统计数量cnt1，cnt2，还有一个总的计数器cnt计最终答案，通过对2*k取余各种操作完成详情看代码*/

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e5+10;int num[maxn];bool cmp (int a, int b){return a > b;}int main(){int n, k;while(scanf("%d%d", &n, &k) != EOF){memset(num, 0, sizeof(num));int cnt_s = 0;int cnt_b = 0;int cnt = 0;for(int i = 0 ; i < n; i++){scanf("%d", &num[i]);if(num[i] <= k) cnt_s++;}cnt_b = n - cnt_s;sort(num, num+n, cmp);for(int i = 0 ;i < cnt_b; i++){if(num[i] % (2*k) == 0) cnt = cnt + num[i]/(2*k);else{if(num[i]%(2*k) > k) cnt = cnt + (num[i]/(2*k) + 1);if(num[i]%(2*k) <= k) {cnt = cnt + (num[i]/(2*k)); cnt_s++;}   //主要是这里的cnt_s++,意思是这种情况将余的部分当做小于k的并++}}//cout << cnt_s;if(cnt_s%2 == 0) cnt = cnt + cnt_s/2;else cnt = cnt + cnt_s/2 + 1;cout << cnt << endl;}return 0;}