【codeforces】706B—Interesting drink-蒲公英云

【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 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.

Example

input

5
3 10 8 6 11
4
1
10
3
11

output

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;
}