Codeforces 919 A. Supermarket

这场cf有点意思，hack场，C题等于1的特判hack很多人(我hack成功3个人，上分了，哈哈哈，咳咳。。。) 

D题好像是树形dp，E题好像是中国剩余定理，F题好像还是dp，具体的不清楚，最近dp的题目好多，一会滚去学dp。

写A,B,C的题解。

A. Supermarket

 

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

We often go to supermarkets to buy some fruits or vegetables, and on the tag there prints the price for a kilo. But in some supermarkets, when asked how much the items are, the clerk will say that a yuan for b kilos (You don't need to care about what "yuan" is), the same as a / b yuan for a kilo.

Now imagine you'd like to buy m kilos of apples. You've asked n supermarkets and got the prices. Find the minimum cost for those apples.

You can assume that there are enough apples in all supermarkets.

Input

The first line contains two positive integers n and m (1 ≤ n ≤ 5 000, 1 ≤ m ≤ 100), denoting that there are n supermarkets and you want to buy m kilos of apples.

The following n lines describe the information of the supermarkets. Each line contains two positive integers a, b (1 ≤ a, b ≤ 100), denoting that in this supermarket, you are supposed to pay a yuan for b kilos of apples.

Output

The only line, denoting the minimum cost for m kilos of apples. Please make sure that the absolute or relative error between your answer and the correct answer won't exceed 10 - 6.

Formally, let your answer be x, and the jury's answer be y. Your answer is considered correct if .

Examples

input

3 5
1 2
3 4
1 3

output

1.66666667

input

2 1
99 100
98 99

output

0.98989899

Note

In the first sample, you are supposed to buy 5 kilos of apples in supermarket 3. The cost is 5 / 3 yuan.

In the second sample, you are supposed to buy 1 kilo of apples in supermarket 2. The cost is 98 / 99 yuan.

大水题。

代码:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
using namespace std;
const int inf=0x3f3f3f3f;
int main(){
    double a,b,ans;
    double minn=inf;
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++){
        cin>>a>>b;
        if(minn>a/b) minn=1.0*a/b;
    }
    ans=minn*(double)m;
    printf("%.8lf",ans);
}

 

滚去学dp，学会了补后面的题|ू･ω･` )