Codeforces 919A （Supermarket）

最新推荐文章于 2018-09-09 14:34:22 发布

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最新推荐文章于 2018-09-09 14:34:22 发布

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分类专栏： codefroces 文章标签： codeforces

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题目时空门

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 这里写图片描述 .

Examples
inputCopy
3 5
1 2
3 4
1 3
output
1.66666667
inputCopy
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.

题意：

这个人想去超市买苹果，有n个超市，要买m斤苹果
接下来有n行，每行都是苹果的钱但都不是单价，代表B斤A元，可由此得单件是 A/B 但注意这里要改为浮点数更为精确。

思路：

直接找最小单价然后乘上要买的斤数即可。这里我当时写了一个返回值为double 的判断函数竟然自动给我进位了，估计是c++与c的问题，然后直接写为if就过了可怜。。emmm

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>

using namespace std;
const int N=1e5+10;


int main()
{
   // ios::sync_with_stdio(false);
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        double a,b;
        double minn=N*1.0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&a,&b);
            if(minn>a/b)//这个地方 最初写了一个 判断函数结果答案不对 我也搞不懂 改成这个if判断就a了
                minn=a/b;
        }
        printf("%.8lf\n",m*minn);

    }
    return 0;
}