time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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;
【题目链接】:http://codeforces.com/contest/758/problem/C
【题解】
画一画;
会发现;
走完第一行之后;
之后每2*(n-1)*m个格子为一个周期;
这一个周期;
固定把第一行和最后一行的格子每个格子递增1;
同时第二行到第n-1行每个格子都递增2;
我是以k<=n*m和k>n*m作为划分的;
对于k<=n*m的情况直接暴力模拟(n*m最大为10000完全可以接受);
对于k>n*m的情况
先把第一行都加上1(先走完第一行);
然后k-=m;
周期的个数就为k/(2*(n-1)*m);
然后对于剩下的k%(2*(n-1)*m)
暴力搞完就好;
特判一下n=1的情就好;
不会很难的;
到了n+1则变成n-1，转个方向;
到了0则变成2,转个方向;
【完整代码】
#include <bits/stdc++.h>
#define LL long long
using namespace std;
int x,y;
LL n,m,k;
LL a[110][110];
int main()
{
//freopen("F:\rush.txt","r",stdin);
cin >> n >> m >> k >> x >> y;
if (k<=n*m)
{
int rest = k;
int nx = 1,ny = 1,fx=1;
while (rest>0)
{
a[nx][ny]++;
rest--;
ny++;
if (ny>m)
{
ny = 1;
nx+=fx;
}
if (nx>n)
{
nx-=2;
fx=-fx;
}
if (nx<1)
{
nx = 2;
if (nx>n)
nx = 1;
fx=-fx;
}
}
}
else
if (k > n*m)
{
if (n>1)
{
for (int i = 1;i <= m;i++)
a[1][i] = 1;
LL temp,rest;
k-=m;
temp = k/(2*(n-1)*m);
rest = k%(2*(n-1)*m);
for (int i = 1;i <= m;i++)
a[1][i]+=temp,a[n][i]+=temp;
for (int i = 2;i <= n-1;i++)
for (int j = 1;j <= m;j++)
a[i][j]+=temp*2;
int nx = 2,ny = 1,fx=1;
if (nx>n)
nx = 1;
while (rest>0)
{
a[nx][ny]++;
rest--;
ny++;
if (ny>m)
{
ny = 1;
nx+=fx;
}
if (nx>n)
{
nx-=2;
fx=-fx;
}
if (nx<1)
{
nx = 2;
fx = -fx;
}
}
}
else
if (n==1)
{
LL temp = k/m;
LL rest = k%m;
for (int i = 1;i <= m;i++)
a[1][i]+=temp;
for (int i = 1;i <= rest;i++)
a[1][i]++;
}
}
LL ma = a[1][1],mi = a[1][1];
for (int i = 1;i <= n;i++)
for (int j = 1;j <= m;j++)
{
ma = max(ma,a[i][j]);
mi = min(mi,a[i][j]);
}
cout << ma << ' ' <<mi << ' '<<a[x][y]<<endl;
return 0;
}