CodeForces 441C Valera and Tubes

链接：http://codeforces.com/problemset/problem/441/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 andm 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 rowx 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),..., (x_r, y_r), that:

• r ≥ 2;
• for any integer i (1 ≤ i ≤ r - 1) the following equation |x_i - x_i + 1| + |y_i - y_i + 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 integerr_i (the number of tube cells), then print2r_i integersx_i1, y_i1, x_i2, y_i2, ..., x_iri, y_iri (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 33 2 1 2 2 2 33 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:

Picture for the second sample:

思路：因为只要填满格子，并且要求每个管子最少占据两个格子。那么只要k-1个管子每个管子占据两个格子，剩下的一个管子占据剩下的所有格子。这样得到的一定是正解当中的一个解。

附上AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cmath>#include <iomanip>#include <ctime>#include <climits>#include <cstdlib>#include <cstring>#include <cctype>#include <algorithm>#include <queue>#include <vector>#include <set>#include <map>#include <stack>#include <deque>//#pragma comment(linker, "/STACK:102400000, 102400000")using namespace std;typedef long long ll;const double pi = acos(-1.0);const double e = exp(1.0);const double eps = 1e-8;int n, m, k;int main(){ios::sync_with_stdio(false);while (~scanf("%d%d%d", &n, &m, &k)){int x=1, y=1;bool flag = false;for (int i=1; i<k; i++){printf("2");for (int j=0; j<2; j++){printf(" %d %d", x, y);if (!flag) y++;else y--;if (y > m){flag = !flag;x++; y=m;}if (y < 1){flag = !flag;x++; y=1;}}printf("\n");}printf("%d", n*m-2*k+2);while (x <= n){printf(" %d %d", x, y);if (!flag) y++;else y--;if (y > m){flag = !flag;x++; y=m;}if (y < 1){flag = !flag;x++; y=1;}}printf("\n");}return 0;}