CF 439C（251C题）Devu and Partitioning of the Array

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 intok disjoint non-empty parts such thatp of the parts have even sum (each of them must
have even sum) and remainingk -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 containn space-separated distinct integers representing
the content of arraya:
a₁, a₂, ..., a_n (1 ≤ a_i ≤ 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. Thei^th of them should contain the content of thei^th
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 exactlyp parts with even
sum, each of the remainingk -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

题目大意：一组数组分成k部分，p组和为偶数的部分，k-p组和为奇数的部分。可以输出YES。否则输出NO

第3组的案例来讲，

1， 2 是偶的，

剩余的

3 。5 1 3

1， 7

和是奇的

推断不难，主要是记录奇数的个数odd，假设odd>=k-p(须要的奇数组)而且剩余的奇数个数是偶数个（以配套凑成和为偶数的组合），同一时候偶数的个数even  +（剩余奇数可以凑成的最多的和为偶数的组合）>=p(即须要的偶数组);

所以一个条件语句就能控制

if(odd>=(k-p)&&(odd-(k-p))%2==0&&(even+(odd-(k-p))/2)>=p)

题目难处在于，后面的组合分配，在代码中具体解释。

AC代码例如以下：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<iomanip>
using namespace std;
#define ll long long
#define M 100005

struct H
{
    int x,y;//x为值。y为奇偶标记

}a[M];

int cmp(H q,H w)
{
    return q.y<w.y;
}

int main()
{
    int n,k,p,i,j,t=0;
    int even=0,odd=0,cont=0;
    cin>>n>>k>>p;
    for(i=0;i<n;i++)
    {
        cin>>a[i].x;

        if(a[i].x%2==0)
            {a[i].y=2;even++;}//标记偶数为2。统计偶数个数
        else {a[i].y=1;odd++;}//标记奇数为1，统计奇数个数
    }
    sort(a,a+n,cmp);//通过对标记值的排序，将数组奇数前置
        if(odd>=(k-p)&&(odd-(k-p))%2==0&&(even+(odd-(k-p))/2)>=p)
        {
            cout<<"YES"<<endl;
            if(p==0)//前面一直错就是卡在这。特殊情况的处理
            {
                for(i=1;i<=k;i++)
                {
                    if(i!=k)
                        {cout<<"1"<<" "<<a[t++].x<<endl;cont++;}//前面k-1都仅仅输出一个数
                        else//i=k时把剩余的输出
                        {
                            cout<<n-cont<<" ";
                        for(j=1;j<=n-cont;j++)
                            cout<<a[t++].x<<" ";
                        cout<<endl;
                        }
                }
                return 0;
            }
            for(i=1;i<=k;i++)
            {
                if(i<=k-p)
                    {cout<<"1"<<" "<<a[t++].x<<endl;cont++;}//首先把奇数全输出
                else if(i>k-p&&i<k)
                {
                    if(a[t].y==1)//当还剩余奇数时。因为条件控制，它们肯定是偶数个。所以每两个奇数凑成偶数组
                    {
                        {cout<<"2"<<" "<<a[t++].x<<" "<<a[t++].x<<endl;cont+=2;}
                    }
                    else
                    {
                        {cout<<"1"<<" "<<a[t++].x<<endl;cont++;}
                    }
                }
                else//把剩余的输出
                    {
                        cout<<n-cont<<" ";
                        for(j=1;j<=n-cont;j++)
                            cout<<a[t++].x<<" ";
                        cout<<endl;
                    }
            }

        }
        else
            cout<<"NO"<<endl;
    return 0;
}