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".
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.
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.
5
3 10 8 6 11
4
1
10
3
11
0
4
1
5
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.
题目链接：http://codeforces.com/problemset/problem/706/B
分析：二分的简单应用吧！在区间内查找上限下限，然后根据这些求出即可！
下面给出AC代码：
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 int a[100010]; 6 int n,i,m,num; 7 while(scanf("%d",&n)!=EOF) 8 { 9 for(i=1;i<=n;i++) 10 scanf("%d",&a[i]); 11 sort(a+1,a+1+n); 12 scanf("%d",&m); 13 while(m--) 14 { 15 scanf("%d",&num); 16 if(num<a[1]) 17 printf("0\n"); 18 else if(num>=a[n]) 19 printf("%d\n",n); 20 else 21 { 22 int left=1,right=n; 23 int mid,ans; 24 while(left<=right) 25 { 26 mid=(left+right)/2; 27 if(a[mid]<=num) 28 { 29 ans=mid; 30 left=mid+1; 31 } 32 else right=mid-1; 33 } 34 printf("%d\n",ans); 35 } 36 } 37 } 38 return 0; 39 }
