Codeforces 919 A. Supermarket-白红宇

Codeforces 919 A. Supermarket

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发布时间：2019-06-13

本文共 2369 字，大约阅读时间需要 7 分钟。

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。

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.

大水题。

代码:

1 #include
         
           2 #include
          
            3 #include
           
             4 #include
            
              5 #include
             
               6 #include
              
                7 #include
               
                 8 #include
                
                  9 #include
                 
                  10 #include
                  
                   11 #include
                   12 #include
                    
                     13 using namespace std;14 const int inf=0x3f3f3f3f;15 int main(){16 double a,b,ans;17 double minn=inf;18 int n,m;19 cin>>n>>m;20 for(int i=0;i
                     
                      >a>>b;22 if(minn>a/b) minn=1.0*a/b;23 }24 ans=minn*(double)m;25 printf("%.8lf",ans);26 }

滚去学dp，学会了补后面的题|ू･ω･` )

转载于:https://www.cnblogs.com/ZERO-/p/8401284.html

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