emp room play min output rep exce have spa

A. Supermarket 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.

找最便宜的超市買東西即可。

#include <bits/stdc++.h>

using namespace std;

int main()
{

    int n;
    double m;
    scanf("%d%lf",&n,&m);

    double minx = 0x3f3f3f3f;
    for(int i = 0; i < n; i++) {
        double a,b;
        scanf("%lf%lf",&a,&b);
        minx = min(minx,a/b);
    }

    printf("%lf\n",m*minx);


    return 0;
}

View Code

B. Perfect Number

We consider a positive integer perfect, if and only if the sum of its digits is exactly 10. Given a positive integer k, your task is to find the k-th smallest perfect positive integer.

Input

A single line with a positive integer k (1 ≤ k ≤ 10 000).

Output

A single number, denoting the k-th smallest perfect integer.

Examples input

1

output

19

input

2

output

28

Note

The first perfect integer is 19 and the second one is 28.

暴力打表

#include <bits/stdc++.h>

using namespace std;

bool calc(int x) {
    int sum = 0;
    while(x) {
        sum+=(x%10);
        x/=10;
    }
    return sum==10;
}

int main()
{
    vector<int> ans;
    for(int i = 19; i < 20000000; i++) {
        if(calc(i))
            ans.push_back(i);
    }
    //printf("%d\n",ans.size());
    int k;
    scanf("%d",&k);
    printf("%d\n",ans[k-1]);

    return 0;
}

View Code

C. Seat Arrangements

Suppose that you are in a campus and have to go for classes day by day. As you may see, when you hurry to a classroom, you surprisingly find that many seats there are already occupied. Today you and your friends went for class, and found out that some of the seats were occupied.

The classroom contains n rows of seats and there are m seats in each row. Then the classroom can be represented as an n × m matrix. The character ‘.‘ represents an empty seat, while ‘*‘ means that the seat is occupied. You need to find k consecutive empty seats in the same row or column and arrange those seats for you and your friends. Your task is to find the number of ways to arrange the seats. Two ways are considered different if sets of places that students occupy differs.

Input

The first line contains three positive integers n, m, k (1 ≤ n, m, k ≤ 2 000), where n, m represent the sizes of the classroom and k is the number of consecutive seats you need to find.

Each of the next n lines contains m characters ‘.‘ or ‘*‘. They form a matrix representing the classroom, ‘.‘ denotes an empty seat, and ‘*‘ denotes an occupied seat.

Output

A single number, denoting the number of ways to find k empty seats in the same row or column.

Examples input

2 3 2
**.
...

output

3

input

1 2 2
..

output

1

input

3 3 4
.*.
*.*
.*.

output

0

Note

In the first sample, there are three ways to arrange those seats. You can take the following seats for your arrangement.

• (1, 3), (2, 3)
• (2, 2), (2, 3)
• (2, 1), (2, 2)

k連坐

做的時候把我做傻了，一步三坑。以至於我算重復了。

1 ，行數只有一行

2 ，k = 1

搞得我換了兩種寫法。第二種好看一點。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 2005;
char maps[maxn][maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    int n,m,K;
    scanf("%d%d%d",&n,&m,&K);
    int ans = 0;
    for(int i = 0; i < n; i++) scanf("%s",maps[i]);
        if(K == 1) {
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(maps[i][j]==‘.‘)
                    ans++;
            }
        }
        printf("%d\n",ans);
        return 0;
    }

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++)
        {
            if(maps[i][j]==‘.‘) {
                int cnt = 0;
                int k;
                for(k = j; k < m; k++) {
                    if(maps[i][k]==‘.‘)
                        cnt++;
                    else break;
                }
                if(cnt>=K) {
                    ans = ans + cnt-K+1;
                }
                j = k;
            }
        }
    }

    if(n!=1) {
        for(int j = 0; j < m; j++) {
            for(int i = 0; i < n; i++)
            {
                if(maps[i][j]==‘.‘) {
                    int cnt = 0;
                    int k;
                    for(k = i; k < n; k++) {
                        if(maps[k][j]==‘.‘)
                            cnt++;
                        else break;
                    }
                    if(cnt>=K)
                        ans=ans + cnt - K +1;
                    i = k;
                }
            }
        }
    }

    printf("%d\n",ans);
    return 0;
}

View Code

#include <bits/stdc++.h>

using namespace std;

const int maxn = 2005;
char maps[maxn][maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    int n,m,K;
    scanf("%d%d%d",&n,&m,&K);

    for(int i = 0; i < n; i++) scanf("%s",maps[i]);
    
    if(K == 1) {
        int ans = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(maps[i][j]==‘.‘)
                    ans++;
            }
        }
        printf("%d\n",ans);
        return 0;
    }
    

    int ans = 0;
    int cnt = 0;
    int i,j;
    for(i = 0; i < n; i++)
    {
        cnt = 0;
        for(j = 0; j < m; j++)
        {
            if(maps[i][j]==‘*‘||maps[i][j+1]==‘\0‘)
            {
                if(maps[i][j]==‘.‘) cnt++;
                ans += cnt>=K ? cnt - K + 1:0;
                cnt = 0;
                continue;
            }
            cnt ++;
        }
    }
    if(n!=1)
    {
        for(i = 0; i < m; i++)
        {
            cnt = 0;
            for(j = 0; j < n; j++)
            {
                if(maps[j][i]==‘*‘||maps[j+1][i]==‘\0‘)
                {
                    if(maps[j][i]==‘.‘) cnt++;
                    ans += cnt>=K ? cnt - K + 1:0;
                    cnt = 0;
                    continue;
                }
                cnt++;
            }
        }
    }

    printf("%d\n",ans);
    return 0;
}

View Code

Codeforces Round #460 (Div. 2)