CF 246C Beauty Pageant

Question:
N (1 <= n <= 50) beauty, charm values are different.
There are k (k <= n * (n + 1)/2) days of beauty contests. The beauty values sent to the contests every day are different.
How to arrange this k-day.

Question:
The point is that the charm values are inconsistent with k <= n * (n + 1)/2. What is the condition for Mao k?
First, the charm values are different. Therefore, if the beauty charm values are sorted in descending order.
When I am dispatched, follow the above dispatching rules:
A [1], a [2],..., a [I-1], a [I]
A [1], a [2],..., a [I-1], a [I + 1]
A [1], a [2],..., a [I-1], a [I + 2]
...
A [1], a [2],..., a [I-1], a [n]
Because they are different, these charm values are definitely different.
Assume that when I + 1 people are dispatched, the dispatch method is as follows:
A [1], a [2],..., a [I], a [I + 1]
A [1], a [2],..., a [I], a [I + 2]
A [1], a [2],..., a [I], a [I + 3]
...
A [1], a [2],..., a [I], a [n] www.2cto.com
Similar to moderate I people, and because they are different, these charm values differ from those of moderate I people.
From this point, I found that there are n-I + 1 dispatching method and n-(I + 1) + 1 dispatching method for I + 1.
Therefore, sum (1 ~ N) = n * (n + 1)/2, which is the maximum limit of k. Therefore, this dispatching method can meet the requirements of the questions.

Code:
[Cpp]
# Include <stdlib. h>
# Include <string. h>
# Include <stdio. h>
# Include <ctype. h>
# Include <math. h>
# Include <stack>
# Include <queue>
# Include <map>
# Include <set>
# Include <vector>
# Include <string>
# Include <iostream>
# Include <algorithm>
Using namespace std;
 
# Define ll _ int64
# Define ls rt <1
# Define rs ls | 1
# Define lson l, mid, ls
# Define rson mid + 1, r, rs
# Define middle (l + r)> 1
# Define eps (1e-8)
# Define type int
# Define clr_all (x, c) memset (x, c, sizeof (x ))
# Define clr (x, c, n) memset (x, c, sizeof (x [0]) * (n + 1 ))
# Define MOD 1000000007.
# Define inf 0x3f3f3f
# Define pi acos (-1.0)
# Define _ max (x, y) (x)> (y ))? (X) :( y ))
# Define _ min (x, y) (x) <(y ))? (X) :( y ))
# Define _ abs (x) <0? (-(X) :( x ))
# Define getmin (x, y) (x = (x <0 | (y) <x )? (Y): x)
# Define getmax (x, y) (x = (y)> x )? (Y): x)
Template <class T> void _ swap (T & x, T & y) {T t = x; x = y; y = t ;}
Int TS, cas = 1;
Const int M = 50 + 5;
Int n, kk, a [M];
 
Void run (){
Int I, j, k;
For (I = 1; I <= n; I ++)
Scanf ("% d", & a [I]);
Sort (a + 1, a + n + 1 );
For (I = 1; I <= n & kk; I ++ ){
For (j = 1; j <= n-I + 1 & kk; j ++ ){
Printf ("% d", I );
For (k = n; k> n-I + 1; k --)
Printf ("% d", a [k]);
Printf ("% d \ n", a [j]);
Kk --;
}
}
}
 
Void preSof (){
}
 
Int main (){
// Freopen ("input.txt", "r", stdin );
// Freopen ("output.txt", "w", stdout );
PreSof ();
// Run ();
While (~ Scanf ("% d", & n, & kk) run ();
// For (scanf ("% d", & TS); cas <= TS; cas ++) run ();
Return 0;
}