树状数组--离散化

- 二分

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit  Status

Description

Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to x_i coins.

Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent m_i coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of shops in the city that sell Vasiliy's favourite drink.

The second line contains n integers x_i (1 ≤ x_i ≤ 100 000) — prices of the bottles of the drink in the i-th shop.

The third line contains a single integer q (1 ≤ q ≤ 100 000) — the number of days Vasiliy plans to buy the drink.

Then follow q lines each containing one integer m_i (1 ≤ m_i ≤ 10⁹) — the number of coins Vasiliy can spent on the i-th day.

Output

Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.

Sample Input

Input

5
3 10 8 6 11
4
1
10
3
11

Output

0
4
1
5
有n个商店，每个商店都有这种饮品，并且给你每个商店对应的价格，
然后有Q个询问，表示你有那么些钱，可以在几个商店选择购买。
思路：
1、简单树状数组，将输入进来的数都加入树中。
2、简单查询，对应Mi如果超过了Xi的极限数据的时候，我们直接离散化一下，
将Mi设定为最大的Xi然后去查询即可。
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#include<cstdio>
#include<cstring>
int n,tree[1000001];
int lowbit(int x)
{
	return x&(-x);
}
void add(int x,int y)
{
	while(x<=n)
	{
		tree[x]+=y;
		x+=lowbit(x);
	}
}
int sum(int n)
{
	int s=0;
	while(n>0)
	{
		s+=tree[n];
		n-=lowbit(n);
	}
	return s;
}

int main()
{
	n=1000000;
	int m,a,b;
	while(~scanf("%d",&m))
	{
		for(int i=1;i<=m;i++)
		{
			scanf("%d",&a);
			add(a,1);
		}
		int p;
		scanf("%d",&p);
		for(int i=1;i<=p;i++)
		{
			scanf("%d",&b);
			if(b>n)
			b=n-1;
			printf("%d\n",sum(b));
		}
	}
	return 0;
}