FZU 2219 StarCraft（优先队列贪心）

思路：手推一下样例大概猜到这样一个贪心，n个工作m个工人，那么需要n-m次分裂，每次分裂出来的一个人去做时间最长的工作，剩下的一个继续分裂，直到工人的数量够了，其实就是一个小根堆，每次取两个元素出来，将次小的元素+k再扔回去即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
priority_queue<int,vector<int>,greater<int> >q;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m,k;
scanf("%d%d%d",&n,&m,&k);
int ans = 0;
for(int i = 1;i<=n;i++)
{
int x;
scanf("%d",&x);
q.push(x);
}
while(q.size()>m)
{
q.pop();
int tmp = q.top();
q.pop();
q.push(tmp+k);
}
while(q.size()!=1)
q.pop();
printf("%d\n",q.top());
while(!q.empty())
q.pop();
}
}


Description

ZB loves playing StarCraft and he likes Zerg most!

One day, when ZB was playing SC2, he came up with an idea:

He wants to change the queen's ability, the queen's new ability is to choose a worker at any time, and turn it into an egg, after K units of time, two workers will born from that egg. The ability is not consumed, which means you can use it any time without
cooling down.

Now ZB wants to build N buildings, he has M workers initially, the i-th building costs t[i] units of time, and a worker will die after he builds a building. Now ZB wants to know the minimum time to build all N buildings.

Input

The first line contains an integer T, meaning the number of the cases. 1 <= T <= 50.

For each test case, the first line consists of three integers N, M and K. (1 <= N, M <= 100000, 1 <= K <= 100000).

The second line contains N integers t[1] ... t
(1 <= t[i] <= 100000).

Output

For each test case, output the answer of the question.

Sample Input

2
3 1 1
1 3 5
5 2 2
1 1 1 1 10

Sample Output

6
10

Hint

For the first example, turn the first worker into an egg at time 0, at time 1 there’s two worker. And use one of them to build the third building, turn the other one into an egg, at time 2, you have 2 workers and a worker building the third building. Use
two workers build the first and the second building, they are built at time 3, 5, 6 respectively.