zoukankan      html  css  js  c++  java
  • CF758C Unfair Poll

    题意:

    On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.

    Seating in the class looks like a rectangle, where n rows with m pupils in each.

    The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...

    The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.

    During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:

    1. the maximum number of questions a particular pupil is asked,
    2. the minimum number of questions a particular pupil is asked,
    3. how many times the teacher asked Sergei.

    If there is only one row in the class, then the teacher always asks children from this row.

    Input

    The first and the only line contains five integers nmkx and y (1 ≤ n, m ≤ 100, 1 ≤ k ≤ 1018, 1 ≤ x ≤ n, 1 ≤ y ≤ m).

    Output

    Print three integers:

    1. the maximum number of questions a particular pupil is asked,
    2. the minimum number of questions a particular pupil is asked,
    3. how many times the teacher asked Sergei.
    Examples
    input
    1 3 8 1 1
    output
    3 2 3
    input
    4 2 9 4 2
    output
    2 1 1
    input
    5 5 25 4 3
    output
    1 1 1
    input
    100 100 1000000000000000000 100 100
    output
    101010101010101 50505050505051 50505050505051

    思路:

    模拟。

    实现:

     1 #include <iostream>
     2 #include <cstdio>
     3 using namespace std;
     4 typedef long long ll;
     5 ll n, m, k, x, y, maxn, minn, t;
     6 int main()
     7 {
     8     cin >> n >> m >> k >> x >> y;
     9     if (n == 1)
    10     {
    11         maxn = k / m + (k % m ? 1 : 0);
    12         minn = k / m;
    13         t = y > k % m ? minn : maxn;
    14         cout << maxn << " " << minn << " " << t << endl;
    15     }
    16     else
    17     {
    18         ll t = (2 * n - 2) * m;
    19         ll tmp = k / t;
    20         ll rem = k % t;
    21         maxn = n > 2 ? 2 * tmp : tmp, minn = tmp;
    22         t = (x == 1 || x == n) ? minn : maxn;
    23         if (rem)
    24         {
    25             if (rem > n * m)
    26             {
    27                 maxn += 2;
    28                 minn++;
    29                 ll p = (rem - n * m) / m;
    30                 ll q = (rem - n * m) % m;
    31                 ll nx = n - 1 - p;
    32                 ll ny = q;
    33                 if (ny == 0)
    34                 {
    35                     nx++;
    36                     ny = m;
    37                 }
    38                 if (x < nx || x == nx && y > ny || x == n)
    39                 {
    40                     t++;
    41                 }
    42                 else
    43                 {
    44                     t += 2;
    45                 }
    46             }
    47             else if (rem == n * m)
    48             {
    49                 maxn++;
    50                 minn++;
    51                 t++;
    52             }
    53             else
    54             {
    55                 if (rem > m)
    56                     maxn++;
    57                 else if (maxn == minn)
    58                     maxn++;
    59                 ll nx = rem / m;
    60                 ll ny = rem % m;
    61                 if (ny)
    62                 {
    63                     nx++;
    64                 }
    65                 else
    66                 {
    67                     ny = m;
    68                 }
    69                 if (!(x > nx || x == nx && y > ny))
    70                 {
    71                     t++;
    72                 }
    73             }
    74         }
    75         cout << maxn << " " << minn << " " << t << endl;
    76     }
    77     return 0;
    78 }
  • 相关阅读:
    2018北京网络赛 G The Mole /// 分块暴力 点线距离
    POJ 3525 /// 半平面交 模板
    买不到的数目 /// 结论公式 oj26316
    Number Sequence /// oj21456
    Round Numbers /// 组合计数 oj21455
    POJ 1265 /// 皮克定理+多边形边上整点数+多边形面积
    fread 快速读入 (神奇挂!)
    HDU6395(分段+矩阵快速幂)
    如何得出保留某位小数,不进行四舍五入
    Codeforces1114 D. Flood Fill (DP)(整个区间染成同色)
  • 原文地址:https://www.cnblogs.com/wangyiming/p/6387636.html
Copyright © 2011-2022 走看看