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 qconsecutive 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
5 3 10 8 6 11 4 1 10 3 11
Output
0 4 1 5
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.
如果直接暴力去做，这道题会之间超限，所以应该选择用二分的方法。
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int a[10000000];
int begin(int a[],int n,int number)
{
int left=1;
int right=n;
int l=0;
while(left<=right)
{
int mid=(left+right)/2;
if(a[mid]<=number)
{
l=mid;
left=mid+1;
}
else
right=mid-1;
}
return l;
}
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;++i)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
int m,x;
scanf("%d",&m);
while(m--)
{
scanf("%d",&x);
printf("%d\n",begin(a,n,x));
}
return 0;
}
不过这道题用到C++里面的upper_bound（）会更简单一些。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int a[1000000];
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;++i)
scanf("%d",&a[i]);
sort(a+1,a+1+n);
int m;
scanf("%d",&m);
while(m--)
{
int qqq;
scanf("%d",&qqq);
int putss=upper_bound(a+1,a+1+n,qqq)-a;
printf("%d\n",putss-1);
}
return 0;
}