967B Watering System

B. Watering System

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.

Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1,s_2,\dots ,s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i\cdot A}{S}$ liters of water will flow out of it.

What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole?

Input

The first line contains three integers $n$, $A$, $B$ ($1\le n\le 100000$, $1\le B\le A\le {10}^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.

The second line contains $n$ integers $s_1,s_2,\dots ,s_n$ ($1\le s_i\le {10}^4$) — the sizes of the holes.

Output

Print a single integer — the number of holes Arkady should block.

Examples

input

Copy

4 10 3
2 2 2 2

output

Copy

1

input

Copy

4 80 20
3 2 1 4

output

Copy

0

input

Copy

5 10 10
1000 1 1 1 1

output

Copy

4

Note

In the first example Arkady should block at least one hole. After that, $\frac{10\cdot 2}{6}\approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.

In the second example even without blocking any hole, $\frac{80\cdot 3}{10}=24$ liters will flow out of the first hole, that is not less than $20$.

In the third example Arkady has to block all holes except the first to make all water flow out of the first hole.

题意：昨晚相对于dir2来说，这应该是最简单的一道题，题意没有那么难懂，注意细节也好做，难怪那些大佬们一开始都做b题，a大多是后面做的。反而总的过题人数是b过的最多。一个人想要浇花，但是他的浇水系统很糟糕，他有一个水管，上面n个漏洞，他能堵住一些漏洞，然后他把a升水，倒进水管里，然后用第一个漏洞浇花，每个漏洞都有一个尺寸大小。每个漏洞能够流$\frac{s_i\cdot A}{S}$ 这么多水，问他给他的花浇b升水，最少需要堵住多少漏洞。输出最少堵住漏洞的个数。

题解：模拟+贪心  根据题目给的公式，我们可以看出，那个公式几乎是定了的，因为用第一个浇水，si和A固定了，所以尽量要让分母小，也就是总的s最小就行。所以，我们把输入的漏洞尺寸，从第二个开始从小到大排序，如果算出的水不>=b,就依次堵住当前最大的漏洞，ans个数++。满足就直接跳出输出答案。

#include<bits/stdc++.h>
using namespace std;
long long c[100100],n,a,b,ans,sum;
int main()
{
    cin>>n>>a>>b;
    for(int i=0; i<n; i++)
        cin>>c[i],sum+=c[i];
    sort(c+1,c+n);
    for(int i=n-1; i>=1; i--)
    {
        if(a*c[0]*1.0/sum>=b)
            break;
        ans++,sum-=c[i];
    }
    cost<<years;
    return 0;
}