Valera and Tubes

C. Valera and Tubes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Valera has got a rectangle table consisting of n rows and m columns. Valera numbered the table rows starting from one, from top to bottom and the columns – starting from one, from left to right. We will represent cell that is on the intersection of row x and column y by a pair of integers (x, y).

Valera wants to place exactly k tubes on his rectangle table. A tube is such sequence of table cells (x1, y1), (x2, y2), ..., (xr, yr), that:

• r ≥ 2;
• for any integer i (1 ≤ i ≤ r - 1) the following equation |xi - xi + 1| + |yi - yi + 1| = 1 holds;
• each table cell, which belongs to the tube, must occur exactly once in the sequence.

Valera thinks that the tubes are arranged in a fancy manner if the following conditions are fulfilled:

• no pair of tubes has common cells;
• each cell of the table belongs to some tube.

Help Valera to arrange k tubes on his rectangle table in a fancy manner.

Input

The first line contains three space-separated integers n, m, k (2 ≤ n, m ≤ 300; 2 ≤ 2k ≤ n·m) — the number of rows, the number of columns and the number of tubes, correspondingly.

Output

Print k lines. In the i-th line print the description of the i-th tube: first print integer ri (the number of tube cells), then print 2ri integersxi1, yi1, xi2, yi2, ..., xiri, yiri (the sequence of table cells).

If there are multiple solutions, you can print any of them. It is guaranteed that at least one solution exists.

Sample test(s)

input

3 3 3

output

3 1 1 1 2 1 3
3 2 1 2 2 2 3
3 3 1 3 2 3 3

input

2 3 1

output

6 1 1 1 2 1 3 2 3 2 2 2 1

Note

Picture for the first sample:

#include <cstdio>
#include <iostream>
#include <algorithm>
#define MAXN 100000
#define ll long long
using namespace std;

int n, m, k, cnt, col, row;
int x[MAXN], y[MAXN];

int main(void)
{
    int i,j,n,m,k;
    cin>>n>>m>>k;
    col=1;row=1;cnt=0;
    int  dir=0;
    while(cnt<n*m)
    {
      //  printf("%d %d %d..
",row,col,cnt);
        if(dir==0)
        {
            if(col<=m)
            {
                x[cnt]=row;
                y[cnt]=col;
               // printf("[%d %d]
",x[cnt],y[cnt]);
                col++;
            }
            else
            {
                x[cnt]=row+1;
                y[cnt]=col-1;
                row++;col-=2;
                dir=1;
            }
        }
        else if(dir==1)
        {
            if(col>=1)
            {
                x[cnt]=row;
                y[cnt]=col;
                col--;
            }
            else if(col==0)
            {
                x[cnt]=row+1;
                y[cnt]=col+1;
                row++;col=2;
                dir=0;
            }
        }
        cnt++;
    }
    cnt=0;
    for(i=0;i<k-1;i++)
    {
        printf("2 ");
        printf("%d %d ",x[cnt],y[cnt]);
        cnt++;
        printf("%d %d
",x[cnt],y[cnt]);
        cnt++;
    }
    printf("%d ",n*m-cnt);
    for(;cnt<n*m;cnt++)
        printf("%d %d ",x[cnt],y[cnt]);
    cout<<endl;
    return 0;
}

另外 有个比較奇怪的现象，假设把cnt++放在printf里面、測试例如以下：

<pre name="code" class="cpp">#include <cstdio>
#include <iostream>
#include <algorithm>
#define MAXN 100000
#define ll long long
using namespace std;

int n, m, k, cnt, col, row;
int x[MAXN], y[MAXN];

int main(void)
{
    int i,j,n,m,k;
    cin>>n>>m>>k;
    col=1;row=1;cnt=0;
    int  dir=0;
    while(cnt<n*m)
    {
      //  printf("%d %d %d..
",row,col,cnt);
        if(dir==0)
        {
            if(col<=m)
            {
                x[cnt]=row;
                y[cnt]=col;
               // printf("[%d %d]
",x[cnt],y[cnt]);
                col++;
            }
            else
            {
                x[cnt]=row+1;
                y[cnt]=col-1;
                row++;col-=2;
                dir=1;
            }
        }
        else if(dir==1)
        {
            if(col>=1)
            {
                x[cnt]=row;
                y[cnt]=col;
                col--;
            }
            else if(col==0)
            {
                x[cnt]=row+1;
                y[cnt]=col+1;
                row++;col=2;
                dir=0;
            }
        }
        cnt++;
    }
    cnt=0;
     for(;cnt<n*m;cnt++)
        printf("%d %d %d
",x[cnt],y[cnt],cnt);
    cnt=0;
    for(i=0;i<k-1;i++)
    {
        printf("2 ");
        printf("[%d] %d [%d] %d [%d] ",cnt,x[cnt],cnt,y[cnt++],cnt);
        printf("[%d] %d [%d] %d [%d]
",cnt,x[cnt],cnt,y[cnt++],cnt);

    }
    printf("%d ",n*m-cnt);
    for(;cnt<n*m;cnt++)
        printf("%d %d ",x[cnt],y[cnt]);
    cout<<endl;
    return 0;
}

这样就非常easy懂了。由于printf是从右向左编译的，cnt++在,处实现自增。