C. Devu and Partitioning of the Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which he could not solve, can you please solve it for him?

Given an array consisting of distinct integers. Is it possible to partition the whole array into k disjoint non-empty parts such that p of the parts have even sum (each of them must have even sum) and remaining k - p have odd sum? (note that parts need not to be continuous).

If it is possible to partition the array, also give any possible way of valid partitioning.

Input

The first line will contain three space separated integers n, k, p (1 ≤ k ≤ n ≤ 105; 0 ≤ p ≤ k). The next line will contain n space-separated distinct integers representing the content of array a: a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

In the first line print "YES" (without the quotes) if it is possible to partition the array in the required way. Otherwise print "NO" (without the quotes).

If the required partition exists, print k lines after the first line. The ith of them should contain the content of the ith part. Print the content of the part in the line in the following way: firstly print the number of elements of the part, then print all the elements of the part in arbitrary order. There must be exactly p parts with even sum, each of the remaining k - p parts must have odd sum.

As there can be multiple partitions, you are allowed to print any valid partition.

Sample test(s)

input

5 5 3 2 6 10 5 9

output

YES 1 9 1 5 1 10 1 6 1 2

input

5 5 3 7 14 2 9 5

output

NO

input

5 3 1 1 2 3 7 5

output

YES 3 5 1 3 1 7 1 2

 

#include
            
             using namespace std;int n , k , p ;vector
             
               o , e ;int op , ep ;int m ;int main () {        cin >> n >> k >> p ;        p = k-p ;        for (int i = 0 ; i < n ; i ++) {                int x ;                cin >> x ;                if (x & 1) o.push_back (x) ;                else e.push_back (x) ;        }        n = o.size () ; m = e.size () ;        if (n < p || n-p & 1 || (n-p)/2+m < k-p) {                puts ("NO") ;                return 0 ;        }        puts ("YES") ;        op = 0 , ep = 0 ;        for (; op < p-1 ;) printf ("1 %d\n" , o[op++]) ;        for (int i = 0 ; i < k-p-1 ; i ++) {                if (ep < m) printf ("1 %d\n" , e[ep++]) ;                else printf ("2 %d %d\n" , o[op++] , o[op++]) ;        }        if (p && k-p) printf ("1 %d\n" , o[op++]) ;        printf ("%d " , n-op + m-ep) ;        while (op < n) printf ("%d " , o[op++]) ;        while (ep < m) printf ("%d " , e[ep++]) ; puts ("") ;        return 0 ;}
             
            

　　这个构造题，跟14年牡丹江的一道题的构造思路有些相似的地方。

为了防止复杂化思路，你在进行的每一步操作都应该尽可能 -----符合题设的条件

并不使问题进一步复杂化。

这道题还需要的一个技巧是，因为要把所有的元素输出，所以用了两个变量op，ep来控制长度。