CodeForces - 758C Unfair Poll （模拟+暴力+思维）

题目链接：http://codeforces.com/problemset/problem/758/C点击打开链接

C. Unfair Poll

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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
bea6
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;

100*100不算大 细节可以暴力找

只要把整个覆盖的轮数算出来就好 再加上细节就能得到答案

#include <bits/stdc++.h>
using namespace std;
long long int mmap[111][111];
int main()
{
long long int n,m,k,x,y;
long long int numnum;
cin >> n >> m >> k >> x >> y;
if(n==1)
numnum=1;
else
numnum=2;
long long int round=k/((n*m)+(n-numnum)*m);
long long int rounds=k%((n*m)+(n-numnum)*m);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
if(i==1||i==n)
mmap[i][j]+=round;
else
mmap[i][j]+=(round*2);
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
mmap[i][j]+=rounds/(n*m);
int flag=rounds/(n*m);
rounds=rounds%(n*m);

if(!flag)
{
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
if(rounds)
{
mmap[i][j]++;
rounds--;

}
}
else
{
for(int i=n-1;i>=1;i--)
for(int j=1;j<=m;j++)
if(rounds)
{
mmap[i][j]++;
rounds--;
}
}
long long int maxn=mmap[1][1];
long long int minn=mmap[1][1];
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
maxn=max(maxn,mmap[i][j]);
minn=min(minn,mmap[i][j]);
}

printf("%lld %lld %lld",maxn,minn,mmap[x][y]);
}