Unfair Poll

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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

思路：直接分情况讨论的。

难点就是特定点的，这个地方来了个模拟，方便计算。

注意循环节，这里n<=2是一种情况，普通的循环。

当n大于2的时候，循环节变成了1....n....n-1。

#include <bits/stdc++.h>using namespace std;const int MAXN=1e5+7;long long n,m,k,x,y;int main(){    long long MAX,MIN,ans;    scanf("%I64d%I64d%I64d%I64d%I64d",&n,&m,&k,&x,&y);    if(n<=2||k<n*m)    {        long long num=n*m;        MAX=MIN=k/num;        if(k%num)MAX++;        long long pos=(x-1)*m+y;        if(pos>k%num)ans=MIN;        else ans=MAX;    }    else    {        long long num=(2*n-2)*m;        MIN=k/num;        MAX=MIN*2;        long long p=k%num;        //ans        long long pos=(x-1)*m+y;        if(pos>p||p==0)        {            if(x>1&&x<n)ans=MAX;            else ans=MIN;        }        else        {                if(pos>0&&pos<=m)ans=MIN+1;                if(pos>m&&pos<=m*(n-1))                {                    if(p<=m*n)ans=MAX+1;                    else                    {                        long long l=pos-m;                        long long r=p-m*n;                        int t1=2,t2=n-1;                        while(l>m)                        {                            l-=m;                            t1++;                        }                        while(r>m)                        {                            r-=m;                            t2--;                        }                        if(t1>t2)ans=MAX+2;                        else if(t1==t2)                        {                            if(r>=l)ans=MAX+2;                            else ans=MAX+1;                        }                        else ans=MAX+1;                    }                }                else ans=MIN+1;        }        //MAX,MIN        if(p>0&&p<=m)        {        }        else if(p>m&&p<=m*(n-1))        {            MAX++;        }        else if(p>m*(n-1)&&p<m*n)        {            MAX++;        }        else if(p==m*n)        {            MAX++;            MIN++;        }        else if(p>m*n)        {            MIN++;            MAX+=2;        }    }    printf("%I64d %I64d %I64d\n",MAX,MIN,ans);    return 0;}

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