Codeforces Round #460 (Div.2) - A - Supermarket

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最新推荐文章于 2024-07-08 02:24:29 发布

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本文介绍了一种算法，用于在多个超市中寻找购买指定数量苹果的最低成本方案。通过比较不同超市提供的价格（以元为单位的价格购买特定公斤数的苹果），算法能够确定哪个超市提供最优惠的单价。

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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 $\text{[math]}$ .

   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.

找到单价最少的超市，然后单价*要买的数量

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int n,m;
int a1,a2,b1,b2;
int main()
{
    scanf("%d%d",&n,&m);
    scanf("%d%d",&a1,&b1);
    for(int i=1;i<n;i++){
        scanf("%d%d",&a2,&b2);
        if(a1*b2>a2*b1) a1 = a2, b1 = b2;
    }
    double ans = 1.0 * a1 * m / b1;
    printf("%.8lf\n",ans);
    return 0;
}