Codeforces Global Round 15 |
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Finished |
The numbers $$$1, \, 2, \, \dots, \, n \cdot k$$$ are colored with $$$n$$$ colors. These colors are indexed by $$$1, \, 2, \, \dots, \, n$$$. For each $$$1 \le i \le n$$$, there are exactly $$$k$$$ numbers colored with color $$$i$$$.
Let $$$[a, \, b]$$$ denote the interval of integers between $$$a$$$ and $$$b$$$ inclusive, that is, the set $$$\{a, \, a + 1, \, \dots, \, b\}$$$. You must choose $$$n$$$ intervals $$$[a_1, \, b_1], \, [a_2, \, b_2], \, \dots, [a_n, \, b_n]$$$ such that:
One can show that such a family of intervals always exists under the given constraints.
The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of colors and the number of occurrences of each color.
The second line contains $$$n \cdot k$$$ integers $$$c_1, \, c_2, \, \dots, \, c_{nk}$$$ ($$$1 \le c_j \le n$$$), where $$$c_j$$$ is the color of number $$$j$$$. It is guaranteed that, for each $$$1 \le i \le n$$$, it holds $$$c_j = i$$$ for exactly $$$k$$$ distinct indices $$$j$$$.
Output $$$n$$$ lines. The $$$i$$$-th line should contain the two integers $$$a_i$$$ and $$$b_i$$$.
If there are multiple valid choices of the intervals, output any.
4 3 2 4 3 1 1 4 2 3 2 1 3 4
4 5 1 7 8 11 6 12
1 2 1 1
1 2
3 3 3 1 2 3 2 1 2 1 3
6 8 3 7 1 4
2 3 2 1 1 1 2 2
2 3 5 6
In the first sample, each number can be contained in at most $$$\left\lceil \frac{4}{3 - 1} \right\rceil = 2$$$ intervals. The output is described by the following picture:
In the second sample, the only interval to be chosen is forced to be $$$[1, \, 2]$$$, and each number is indeed contained in at most $$$\left\lceil \frac{1}{2 - 1} \right\rceil = 1$$$ interval.
In the third sample, each number can be contained in at most $$$\left\lceil \frac{3}{3 - 1} \right\rceil = 2$$$ intervals. The output is described by the following picture:
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