codeforces 439C Devu and Partitioning of the Array（简单题）

题目链接

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, ..., 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. The i^th of them should contain the content of the i^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 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

题意：n个数字，把他们分到K个集合，使得有P个集合中数的和为偶数，另外k-p个集合的数的和为奇数。如果有解，并任意一种输出方案。

题解：由于只要求和的奇偶性，所以我们只用考虑数字的奇偶性，先提取出所有的奇数和偶数，然后先将要求和为奇数的集合放一个奇数，放完后把和为偶数的集合放一个偶数或两个奇数。放完以后，如果剩余的奇数有奇数个，则无解。否则剩余的偶数随便放，剩余的奇数两个为一组随便放即可。

代码如下：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<string.h>
#include<string>
#include<stdlib.h>
typedef __int64 LL;
typedef unsigned __int64 LLU;
const int nn=110000;
const int inf=0x3fffffff;
const LL inf64=(LL)inf*inf;
using namespace std;
int n,k,p;
queue<int>odd,even;
vector<int>ve[nn];
int main()
{
    int i,x;
    while(scanf("%d%d%d",&n,&k,&p)!=EOF)
    {
        for(i=1;i<=k;i++)
        {
            ve[i].clear();
        }
        while(odd.size())
            odd.pop();
        while(even.size())
            even.pop();
        for(i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(x%2)
            {
                odd.push(x);
            }
            else
                even.push(x);
        }
        if(odd.size()<k-p)
        {
            puts("NO");
            continue;
        }
        for(i=1;i<=k-p;i++)
        {
            ve[i].push_back(odd.front());
            odd.pop();
        }
        if(odd.size()%2==1)
        {
            puts("NO");
            continue;
        }
        for(i=k-p+1;i<=k;i++)
        {
            if(odd.size())
            {
                ve[i].push_back(odd.front());
                odd.pop();
                ve[i].push_back(odd.front());
                odd.pop();
            }
            else if(even.size())
            {
                ve[i].push_back(even.front());
                even.pop();
            }
            else
                break;
        }
        if(i<=k)
        {
            puts("NO");
            continue;
        }
        else
        {
            while(even.size())
            {
                ve[1].push_back(even.front());
                even.pop();
            }
            while(odd.size())
            {
                ve[1].push_back(odd.front());
                odd.pop();
            }
            puts("YES");
            for(i=1;i<=k;i++)
            {
                int lv=ve[i].size();
                printf("%d",lv);
                for(int j=0;j<lv;j++)
                {
                    printf(" %d",ve[i][j]);
                }
                puts("");
            }
        }
    }
    return 0;
}