Spoj-VISIBLEBOX Decreasing Number of Visible Box

Shadowman loves to collect box but his roommates woogieman and itman don't like box and so shadowman wants to hide boxes as many as possible. A box can be kept hidden inside of another box if and only if the box in which it will be held is empty and the size of the box is at least twice as large as the size of the box.

Print the minimum number of box that can be shown.

Input

The input set starts with single line integer T (1<=T<=50) the number of test cases. Then following T cases starts with an integer N (1<=N<=100000) denoting the number of box. The next line contains N space separated positive integer. i-th of them contains a numbers A_i(1<=A_i<=100000) size of the i-th box.

Output

Output the the case number and the minimum number of box that can be shown.

Example

Input:
2
4
1 2 4 8
4
1 3 4 5

Output:
Case 1: 1
Case 2: 3

n个盒子要叠起来，大小为k的盒子可以放进大小为2k的空盒子里，问最少剩多少

排个序贪心就好了，维护一个队列表示当前可以塞到其他盒子里的盒子，每次新来一个盒子的时候，把最小的那个塞进去

最后答案就是队列大小了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
#define mod 100007
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,t,w,ans;
int q[100010];
int a[100010];
inline void work(int cur)
{
    n=read();
    for (int i=1;i<=n;i++)a[i]=read();
    sort(a+1,a+n+1);
    t=w=ans=0;
    for (int i=1;i<=n;i++)
    {
        if (t==w)q[++w]=a[i]*2,ans++;
        else if (q[t+1]<=a[i])t++,q[++w]=a[i]*2;
        else q[++w]=a[i]*2,ans++;
    }
    printf("Case %d: %d
",cur,ans);
}
int main()
{
    int T=read(),tt=0;
    while (T--)work(++tt);
}