codeforces 251 div2 C. Devu and Partitioning of the Array 模拟-白红宇

codeforces 251 div2 C. Devu and Partitioning of the Array 模拟

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发布时间：2019-06-09

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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.

Examples

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

题意：给你n个数，分成m堆，这些堆的和为偶数的堆为p堆；

思路：分析清楚情况就是，偶数永远是偶数，但是两个奇数可以组成一个偶数；

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                       using namespace std;#define ll __int64#define mod 1000000007#define inf 999999999//#pragma comment(linker, "/STACK:102400000,102400000")int scan(){ int res = 0 , ch ; while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) ) { if( ch == EOF ) return 1 << 30 ; } res = ch - '0' ; while( ( ch = getchar() ) >= '0' && ch <= '9' ) res = res * 10 + ( ch - '0' ) ; return res ;}int even[100010],ji;int odd[100010],ou;int main(){ int x,y,z,i,t; int n,k,p; scanf("%d%d%d",&n,&k,&p); for(i=0;i

转载于:https://www.cnblogs.com/jhz033/p/5475078.html

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