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".
InputThe 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.
OutputPrint 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 5 3 10 8 6 11 4 1 10 3 11 output 0 4 1 5 NoteOn 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.
题解：排序后直接用二分公式即可#include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define CLR(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f int a[100010]; int main() { int n; while(~scanf("%d",&n)) { CLR(a,0); for(int i=1;i<=n;i++) scanf("%d",&a[i]); sort(a+1,a+1+n); int q; scanf("%d",&q); while(q--) { int t; scanf("%d",&t); int ans=upper_bound(a+1,a+1+n,t)-a-1; printf("%d\n",ans); } } return 0; }
