Codeforces758C Unfair Poll

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本文解析了一道关于教师提问模式的算法题，探讨了如何计算特定学生被提问的最大次数、最小次数及指定学生的被提问次数。通过分析不同场景下的提问规律，给出了详细的解答过程。

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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:

the maximum number of questions a particular pupil is asked,
the minimum number of questions a particular pupil is asked,
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 n, m, k, x and y (1 ≤ n, m ≤ 100, 1 ≤ k ≤ 1018, 1 ≤ x ≤ n, 1 ≤ y ≤ m).

Output

Print three integers:

the maximum number of questions a particular pupil is asked,
the minimum number of questions a particular pupil is asked,
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

Note

The order of asking pupils in the first test:

the pupil from the first row who seats at the first table, it means it is Sergei;
the pupil from the first row who seats at the second table;
the pupil from the first row who seats at the third table;
the pupil from the first row who seats at the first table, it means it is Sergei;
the pupil from the first row who seats at the second table;
the pupil from the first row who seats at the third table;
the pupil from the first row who seats at the first table, it means it is Sergei;
the pupil from the first row who seats at the second table;

The order of asking pupils in the second test:

the pupil from the first row who seats at the first table;
the pupil from the first row who seats at the second table;
the pupil from the second row who seats at the first table;
the pupil from the second row who seats at the second table;
the pupil from the third row who seats at the first table;
the pupil from the third row who seats at the second table;
the pupil from the fourth row who seats at the first table;
the pupil from the fourth row who seats at the second table, it means it is Sergei;
the pupil from the third row who seats at the first table;

——————————————————————————————————————

题意：老师点名字按顺序点，排与排之间按1,2……n-1，n，n-1……2,1,2这样，每排按顺序点，输入n排m列，点k次，主人公位置（x，y），问最多的人点几次，最少的人点几次，主人公点几次？

思路：博主用了最笨的方法分类讨论，特判n=1和n=2；可惜考虑不到位终测挂了，气

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

using namespace std;


int main()
{
    long long m,n,k,x,y;
    while(~scanf("%I64d %I64d %I64d %I64d %I64d",&n,&m,&k,&x,&y))
    {   
             if(k==m&&n==1)
        {
            printf("1 1 1\n");
            continue;
        }
        if(k<=m)
        {
            if(x==1&&y<=k)
                printf("1 0 1\n");
            else
                printf("1 0 0\n");
            continue;
        }

        if(n==1)
        {
            long long  ttt=k/m;
            long long tt=k%m;
            if(y<=tt)
            {
                if(tt==0)
                    printf("%I64d %I64d %I64d\n",ttt,ttt,ttt);
                else
                     printf("%I64d %I64d %I64d\n",ttt+1,ttt,ttt+1);
            }
            else
            {
                if(tt==0)
                    printf("%I64d %I64d %I64d\n",ttt,ttt,ttt);
                else
                     printf("%I64d %I64d %I64d\n",ttt+1,ttt,ttt);
            }
            continue;
        }
        if(n==2)
        {
             long long  ttt=k/(m*2);
            long long tt=k%(m*2);
            long long ttc=(x-1)*m+y;
            if(tt==0)
                  printf("%I64d %I64d %I64d\n",ttt,ttt,ttt);
            else
                {
                if(tt<ttc)
                  printf("%I64d %I64d %I64d\n",ttt+1,ttt,ttt);
            else
                  printf("%I64d %I64d %I64d\n",ttt+1,ttt,ttt+1);
                }
            continue;
        }
        k-=m;
        long long int cnt=k/(m*(n-1));
       long long  int md=k%(m*(n-1));
       long long  int zheng=(x-1)*m+y-m;
        long long int fan=(n-x)*m+y-m;
        if(cnt%2==1)
        {
            if(x==1)
            {
                if(md==0)
                    printf("%I64d %I64d %I64d\n",cnt,cnt/2+1,cnt/2+1);
                else
                {
                    if(fan>md)
                     printf("%I64d %I64d %I64d\n",cnt+1,cnt/2+1,cnt/2+1);
                     else
                     printf("%I64d %I64d %I64d\n",cnt+1,cnt/2+1,cnt/2+2);

                }

            }
            else if(x==n)
            {
                if(md==0)
                    printf("%I64d %I64d %I64d\n",cnt,cnt/2+1,cnt/2+1);
                else
                {
                     printf("%I64d %I64d %I64d\n",cnt+1,cnt/2+1,cnt/2+1);
                }
            }
            else if(fan>md)
            {
                if(md==0)
                    printf("%I64d %I64d %I64d\n",cnt,cnt/2+1,cnt);
                else
                    printf("%I64d %I64d %I64d\n",cnt+1,cnt/2+1,cnt);
            }
            else
            {
                printf("%I64d %I64d %I64d\n",cnt+1,cnt/2+1,cnt+1);
            }
        }
        else
        {
            if(x==1)
            {
                if(md==0)
                    printf("%I64d %I64d %I64d\n",cnt,cnt/2,cnt/2+1);
                else
                {
                     printf("%I64d %I64d %I64d\n",cnt+1,cnt/2,cnt/2+1);
                }

            }
            else if(x==n)
            {
                 if(md==0)
                    printf("%I64d %I64d %I64d\n",cnt,cnt/2,cnt/2);
                else
                {
                    if(zheng>md)
                     printf("%I64d %I64d %I64d\n",cnt+1,cnt/2,cnt/2);
                     else
                        printf("%I64d %I64d %I64d\n",cnt+1,cnt/2,cnt/2+1);
                }
            }
            else if(zheng>md)
            {
                if(md==0)
                    printf("%I64d %I64d %I64d\n",cnt,cnt/2,cnt);
                else
                    printf("%I64d %I64d %I64d\n",cnt+1,cnt/2,cnt);
            }
            else
            {
                printf("%I64d %I64d %I64d\n",cnt+1,cnt/2,cnt+1);
            }
        }

    }

    return 0;
}