zoukankan      html  css  js  c++  java
  • Codeforces Round #419 (Div. 2)

    A. Karen and Morning
    time limit per test
    2 seconds
    memory limit per test
    512 megabytes
    input
    standard input
    output
    standard output

    Karen is getting ready for a new school day!

    It is currently hh:mm, given in a 24-hour format. As you know, Karen loves palindromes, and she believes that it is good luck to wake up when the time is a palindrome.

    What is the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome?

    Remember that a palindrome is a string that reads the same forwards and backwards. For instance, 05:39 is not a palindrome, because 05:39 backwards is 93:50. On the other hand, 05:50 is a palindrome, because 05:50 backwards is 05:50.

    Input

    The first and only line of input contains a single string in the format hh:mm (00 ≤  hh  ≤ 23, 00 ≤  mm  ≤ 59).

    Output

    Output a single integer on a line by itself, the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome.

    Examples
    input
    05:39
    output
    11
    input
    13:31
    output
    0
    input
    23:59
    output
    1
    Note

    In the first test case, the minimum number of minutes Karen should sleep for is 11. She can wake up at 05:50, when the time is a palindrome.

    In the second test case, Karen can wake up immediately, as the current time, 13:31, is already a palindrome.

    In the third test case, the minimum number of minutes Karen should sleep for is 1 minute. She can wake up at 00:00, when the time is a palindrome.

    要明白palindrome是回文串的意思,看着这个单词懵B了小一会,然后就是分钟个位数字相同和点钟十位数字相同,分钟十位数字相同和点钟个位数字相同,所以就是个很简单的暴力啦

    #include<bits/stdc++.h>
    using namespace std;
    int main(){
        int h,m;
        scanf("%d:%d",&h,&m);
        int t=h*60+m;
        for(int i=h;i<=24;i++)
            for(int j=0;j<60;j++)
            if(60*i+j>=t&&i%24%10==j/10&&j%10==i%24/10)
            return 0*printf("%d",60*i+j-t);
        return 0;
    }
    B. Karen and Coffee
    time limit per test
    2.5 seconds
    memory limit per test
    512 megabytes
    input
    standard input
    output
    standard output

    To stay woke and attentive during classes, Karen needs some coffee!

    Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".

    She knows n coffee recipes. The i-th recipe suggests that coffee should be brewed between li and ri degrees, inclusive, to achieve the optimal taste.

    Karen thinks that a temperature is admissible if at least k recipes recommend it.

    Karen has a rather fickle mind, and so she asks q questions. In each question, given that she only wants to prepare coffee with a temperature between a and b, inclusive, can you tell her how many admissible integer temperatures fall within the range?

    Input

    The first line of input contains three integers, nk (1 ≤ k ≤ n ≤ 200000), and q (1 ≤ q ≤ 200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.

    The next n lines describe the recipes. Specifically, the i-th line among these contains two integers li and ri(1 ≤ li ≤ ri ≤ 200000), describing that the i-th recipe suggests that the coffee be brewed between li and ri degrees, inclusive.

    The next q lines describe the questions. Each of these lines contains a and b, (1 ≤ a ≤ b ≤ 200000), describing that she wants to know the number of admissible integer temperatures between a and b degrees, inclusive.

    Output

    For each question, output a single integer on a line by itself, the number of admissible integer temperatures between a and bdegrees, inclusive.

    Examples
    input
    3 2 4
    91 94
    92 97
    97 99
    92 94
    93 97
    95 96
    90 100
    output
    3
    3
    0
    4
    input
    2 1 1
    1 1
    200000 200000
    90 100
    output
    0
    Note

    In the first test case, Karen knows 3 recipes.

    1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive.
    2. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive.
    3. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.

    A temperature is admissible if at least 2 recipes recommend it.

    She asks 4 questions.

    In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.

    In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.

    In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.

    In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.

    In the second test case, Karen knows 2 recipes.

    1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree.
    2. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.

    A temperature is admissible if at least 1 recipe recommends it.

    In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.

     给你n个区间,k次查询,如果一个值覆盖超过k次即是一个特别好的温度,前缀和处理下,类似于扫描线处理下这么多区间,不然会超时的

    #include<bits/stdc++.h>
    using namespace std;const int N=200005;
    int n,a[N],b[N],c[N],q,k;
    int main(){
        scanf("%d%d%d",&n,&k,&q);
        for(int i=0;i<n;i++){
            int l,r;
            scanf("%d%d",&l,&r);
            a[l]++;
            b[r]++;
        }
        int f=0;
        for(int i=1;i<200005;i++){
            f+=a[i];
            c[i]=c[i-1]+(f>=k);
            f-=b[i];
        }
        while(q--){
            int l,r;
            scanf("%d%d",&l,&r);
            printf("%d
    ",c[r]-c[l-1]);
        }
        return 0;
    }
    C. Karen and Game
    time limit per test
    2 seconds
    memory limit per test
    512 megabytes
    input
    standard input
    output
    standard output

    On the way to school, Karen became fixated on the puzzle game on her phone!

    The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

    One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

    To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

    Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

    Input

    The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

    The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

    Output

    If there is an error and it is actually not possible to beat the level, output a single integer -1.

    Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

    The next k lines should each contain one of the following, describing the moves in the order they must be done:

    • row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row".
    • col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column".

    If there are multiple optimal solutions, output any one of them.

    Examples
    input
    3 5
    2 2 2 3 2
    0 0 0 1 0
    1 1 1 2 1
    output
    4
    row 1
    row 1
    col 4
    row 3
    input
    3 3
    0 0 0
    0 1 0
    0 0 0
    output
    -1
    input
    3 3
    1 1 1
    1 1 1
    1 1 1
    output
    3
    row 1
    row 2
    row 3
    Note

    In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

    In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

    In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

    Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

    C题就是让你选行或列+1,加到你要得到的这个矩阵,一直暴力错误,等会再做下

    这题有个特别坑的地方,它要这个操作的最小值,所以要分情况讨论

    #include<bits/stdc++.h>
    using namespace std;
    int n, m;
    int a[105][105];
    int ans;
    vector<int> r, c;
    void dorow()
    {
        for(int i=0;i<n;i++)
        {
            int val=501;
            for(int j=0;j<m;j++)
            val=min(val,a[i][j]);
            ans+=val;
            for(int j=0;j<m;j++)
            a[i][j]-=val;
            for(int k=0;k<val;k++)
            r.push_back(i);
        }
    }
    
    void docol()
    {
        for(int i=0;i<m;i++)
        {
            int val=501;
            for(int j=0;j<n;j++)
            val=min(val,a[j][i]);
            ans+=val;
            for(int j=0;j<n;j++) a[j][i]-=val;
            for(int k=0;k<val;k++) c.push_back(i);
        }
    }
    
    int main()
    {
        cin>>n>>m;
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        cin>>a[i][j];
        if(n<m) dorow(),docol();
        else docol(),dorow();
        bool ok=1;
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        if (a[i][j]) ok=0;
        if (!ok) cout<<-1;
        else
        {
            cout<<ans<<"
    ";
            for(int i=0;i<r.size();i++)
            cout<<"row "<<r[i]+1<<"
    ";
            for(int i=0;i<c.size();i++)
            cout<<"col "<<c[i]+1<<"
    ";
        }
    }

    还好升了几名。还是小学生,等下次再翻身吧

  • 相关阅读:
    loadrunner获取Http信息头中指定值作为参数
    soapUI使用-DataSource获取oracle库中的参数
    [转]vim编辑器---批量注释与反注释
    String() 函数把对象的值转换为字符串。
    自定义滚动条mCustomScrollbar
    css实现强制不换行/自动换行/强制换行
    在网页中添加新浪微博“加关注”按钮
    移动前端调试方案(Android + Chrome 实现远程调试)
    font-family
    移动端touch事件滚动
  • 原文地址:https://www.cnblogs.com/BobHuang/p/7043404.html
Copyright © 2011-2022 走看看