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 nkp (1 ≤ k ≤ n ≤ 105; 0 ≤ p ≤ k).
The next line will contain n space-separated distinct integers representing the content of array aa1, 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

题意:给出n个数。要分成k份。每份有若干个数。可是仅仅须要关注该份的和为奇数还是偶数,要求偶数堆的个数为p。输出方案。

分析:先输出k-p-1组奇数,然后输出p-1组偶数;假设 p!=0&&(k-p)!=0,再输出一个奇数,这时奇数的已经构造完毕。最后把剩下的所有输出为一组就可以。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN = 1e5 + 100;
int a[MAXN], b[MAXN];
int main()
{
int n, p, k, i, odd = 0, even = 0, c;
scanf("%d%d%d",&n,&k,&p);
for(i = 0; i < n; i++)
{
scanf("%d",&c);
if(c&1)
a[odd++] = c;
else
b[even++] = c;
}
if(odd < k-p || (odd - k + p) % 2 == 1 || even + (odd - k + p) / 2 < p)
printf("NO\n");
else
{
printf("YES\n");
int tmp = k - p;
for(i = 0; i < tmp - 1; i++)
printf("1 %d\n",a[--odd]);
for(i = 0; i < p - 1; i++)
{
if(even)
printf("1 %d\n", b[--even]);
else
{
printf("2 %d %d\n", a[odd-1], a[odd-2]);
odd -= 2;
}
}
if(tmp && p)
printf("1 %d\n", a[--odd]);
printf("%d", odd+even);
while(odd)
printf(" %d",a[--odd]);
while(even)
printf(" %d\n", b[--even]);
printf("\n");
}
return 0;
}

CodeForce 439C Devu and Partitioning of the Array(模拟)的更多相关文章

  1. Codeforces 439C Devu and Partitioning of the Array(模拟)

    题目链接:Codeforces 439C Devu and Partitioning of the Array 题目大意:给出n个数,要分成k份,每份有若干个数,可是仅仅须要关注该份的和为奇数还是偶数 ...

  2. CF 439C Devu and Partitioning of the Array

    题目链接: 传送门 Devu and Partitioning of the Array time limit per test:1 second     memory limit per test: ...

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

    C. Devu and Partitioning of the Array time limit per test 1 second memory limit per test 256 megabyt ...

  4. codeforces 439C Devu and Partitioning of the Array(烦死人的多情况的模拟)

    题目 //这是一道有n多情况的烦死人的让我错了n遍的模拟题 #include<iostream> #include<algorithm> #include<stdio.h ...

  5. 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 ...

  6. Codeforces Round #251 (Div. 2) C. Devu and Partitioning of the Array

    注意p的边界情况,p为0,或者 p为k 奇数+偶数 = 奇数 奇数+奇数 = 偶数 #include <iostream> #include <vector> #include ...

  7. codeforces 439D Devu and Partitioning of the Array(有深度的模拟)

    题目 //参考了网上的代码 注意答案可能超过32位 //要达成目标,就是要所有数列a的都比数列b的要小或者等于 //然后,要使最小的要和最大的一样大,就要移动(大-小)步, //要使较小的要和较大的一 ...

  8. codeforces C. Devu and Partitioning of the Array

    题意:给你n个数,然后分成k部分,每一个部分的和为偶数的有p个,奇数的有k-p个,如果可以划分,输出其中的一种,不可以输出NO; 思路:先输出k-p-1个奇数,再输出p-1个偶数,剩余的在进行构造.  ...

  9. 【Henu ACM Round#20 D】 Devu and Partitioning of the Array

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 一开始所有的数字单独成一个集合. 然后用v[0]和v[1]记录集合的和为偶数和奇数的集合它们的根节点(并查集 然后先让v[0]的大小 ...

随机推荐

  1. iOS开发中地图开发的简单应用

    iOS上使用地图比Android要方便,只需要新建一个MKMapView,addSubView即可.这次要实现的效果如下: 有标注(大头针),定位,地图. 1.添加地图 1.1 新一个Single V ...

  2. Android百度地图之显示地图

    添加地图显示 一.在百度官网下载相关的SDK (网址:http://developer.baidu.com/map/sdkandev-download.htm) 解压下载好的BaiduMap_Andr ...

  3. Android存储之SQLiteDatbase

    SQLiteDatabase的方式会生成一个数据库文件,每个应用最多只对应一个数据库文件,即.db文件. 可以使用很多第三方工具进行打开,查看数据库里的内容. 昨晚试了好几种工具,如navicat,s ...

  4. js window.onload事件

    1.最简单的调用方式 直接写到html的body标签里面,如: ? 1 2 3 4     <html>       <body onload="func()"& ...

  5. HDU 5012 Dice DFS

    简单DFS //#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h ...

  6. (Problem 2)Even Fibonacci numbers

    Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting w ...

  7. 用命令行方式关闭linux防火墙

    #/sbin/iptables -I INPUT -p tcp --dport 80 -j ACCEPT #/sbin/iptables -I INPUT -p tcp --dport 22 -j A ...

  8. IOS的处理touch事件处理(按照手指的移动移动一个圆,开发环境用的ios7,storyboard)

    先看下页面的效果图: 首先定义这个ball它有两个属性和两个方法: @property(nonatomic) CGPoint location; @property(nonatomic) CGFloa ...

  9. Search a 2D Matrix【python】

    class Solution: # @param matrix, a list of lists of integers # @param target, an integer # @return a ...

  10. 网页制作之JavaScript部分 2 - DOM操作

    1.DOM的基本概念  htmlDOM是一种面向对象的树的模型,它包含html中的所有元素:通过html可以找到所有包含在dom中的元素. DOM是文档对象模型,这种模型为树模型:文档是指标签文档:对 ...