【codeforces】706B—Interesting drink

B. Interesting drink

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 xi 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 mi 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 xi (1 ≤ xi ≤ 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 mi (1 ≤ mi ≤ 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.

Example

input

53 10 8 6 114110311

output

0415

Note

On the first day, Vasiliy won't be able to buy a drink in any of the shops.

On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.

On the third day, Vasiliy can buy a drink only in the shop number 1.

Finally, on the last day Vasiliy can buy a drink in any shop.

用二分和树状数组均可。

二分做法：

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;const int N = 1e5+5;int a[N];int main() {int n;while(scanf("%d",&n)!=EOF) {for(int i=0; i<n; i++) {scanf("%d",&a[i]);}int q,num;scanf("%d",&q);sort(a,a+n);while(q--) {scanf("%d",&num);int ans=upper_bound(a,a+n,num)-a;printf("%d\n",ans);}}return 0;}

树状数组做法：

#include<cstdio>#include<iostream>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int N = 100000+10;int C[N];int lowbit(int x) {return x&(-x);}void add(int t,int x,int nn) {while(t<=nn) {C[t]=C[t]+x;t=t+lowbit(t);}}int Sum(int h) {int s=0;while(h>0) {s=s+C[h];h-=lowbit(h);}return s;}int main() {int n;int nn=100000;while(scanf("%d",&n)!=EOF) {memset(C,0,sizeof(C));for(int i=0; i<n; i++) {int t;scanf("%d",&t);add(t,1,nn);}int q;scanf("%d",&q);while(q--) {int h;scanf("%d",&h);if(h>100000) h=100000;printf("%d\n",Sum(h));}}return 0;}

题目地址：http://codeforces.com/problemset/problem/706/B