B. Odd Sum Segments CF(分割数组)
题目地址
http://codeforces.com/contest/1196/problem/B
3 seconds
256 megabytes
standard input
standard output
You are given an array aa consisting of nn integers a1,a2,…,ana1,a2,…,an. You want to split it into exactly kk non-empty non-intersecting subsegments such that each subsegment has odd sum (i. e. for each subsegment, the sum of all elements that belong to this subsegment is odd). It is impossible to rearrange (shuffle) the elements of a given array. Each of the nn elements of the array aa must belong to exactly one of the kk subsegments.
Let's see some examples of dividing the array of length 55 into 33 subsegments (not necessarily with odd sums): [1,2,3,4,5][1,2,3,4,5] is the initial array, then all possible ways to divide it into 33 non-empty non-intersecting subsegments are described below:
- [1],[2],[3,4,5][1],[2],[3,4,5];
- [1],[2,3],[4,5][1],[2,3],[4,5];
- [1],[2,3,4],[5][1],[2,3,4],[5];
- [1,2],[3],[4,5][1,2],[3],[4,5];
- [1,2],[3,4],[5][1,2],[3,4],[5];
- [1,2,3],[4],[5][1,2,3],[4],[5].
Of course, it can be impossible to divide the initial array into exactly kk subsegments in such a way that each of them will have odd sum of elements. In this case print "NO". Otherwise, print "YES" and any possible division of the array. See the output format for the detailed explanation.
You have to answer qq independent queries.
The first line contains one integer qq (1≤q≤2⋅1051≤q≤2⋅105) — the number of queries. Then qq queries follow.
The first line of the query contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of elements in the array and the number of subsegments, respectively.
The second line of the query contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the ii-th element of aa.
It is guaranteed that the sum of nn over all queries does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105).
For each query, print the answer to it. If it is impossible to divide the initial array into exactly kk subsegments in such a way that each of them will have odd sum of elements, print "NO" in the first line. Otherwise, print "YES" in the first line and any possible division of the array in the second line. The division can be represented as kk integers r1r1, r2r2, ..., rkrk such that 1≤r1<r2<⋯<rk=n1≤r1<r2<⋯<rk=n, where rjrj is the right border of the jj-th segment (the index of the last element that belongs to the jj-th segment), so the array is divided into subsegments [1;r1],[r1+1;r2],[r2+1,r3],…,[rk−1+1,n][1;r1],[r1+1;r2],[r2+1,r3],…,[rk−1+1,n]. Note that rkrk is always nn but you should print it anyway.
3
5 3
7 18 3 14 1
5 4
1 2 3 4 5
6 2
1 2 8 4 10 2
YES
1 3 5
NO
NO 题意: 给定一个n,m。下一行有n个数。问是否可以分割成m组,使每组的和为奇数。如果可以,打印YES,并且输出分割出来的每组的最右边坐标(可能有多种,打印一组即可);否则输出“NO”;
解析: 首先明确一点:偶数+奇数==奇数 奇数+奇数=偶数;
所以呢,在给定的一组数中可以忽略偶数的存在,因为偶数与奇数的运算并不能改变结果的奇偶性。首先统计奇数的个数ob。要分成m组,假设每一组1个奇数,则至少需要m个奇数,因为偶数的存在改变不了结局。
如果奇数个数ob比m还少,肯定分不出来m组。所以要ob>=m; 根据贪心的思想,从前往后分割。在前面,只要for到一个奇数,就算分割到一组,直到还剩最后一组。需要看最后一组奇数个数,ob-(k-1)即为剩余奇数个数
如果是偶数个,肯定不行因为加起来是偶数,输出“NO”;奇数个的话,输出YES,打印最后一位的坐标,就行了。
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
const int maxn=;
typedef long long ll;
ll a[maxn];
int main()
{
ll t;
cin>>t;
while(t--)
{
ll n,m;
cin>>n>>m;
ll od=;
for(int i=;i<n;i++)
{
cin>>a[i];
if(a[i]%!=)
od++; //统计奇数个数
}
if(od>=m&&(od-(m-))%!=)
{
cout<<"YES"<<endl;
ll k=;
for(int i=;i<n;i++)
{
if(a[i]%!=)
{
k++; //当前所分的组数
if(k==m)
{
cout<<n<<endl;break;
}
else
{
cout<<i+<<" ";
}
}
}
// cout<<endl;
}
else
cout<<"NO"<<endl;
}
return ;
}
B. Odd Sum Segments CF(分割数组)的更多相关文章
- Codeforces Round #575 (Div. 3) B. Odd Sum Segments (构造,数学)
B. Odd Sum Segments time limit per test3 seconds memory limit per test256 megabytes inputstandard in ...
- lintcode 容易题:Partition Array by Odd and Even 奇偶分割数组
题目: 奇偶分割数组 分割一个整数数组,使得奇数在前偶数在后. 样例 给定 [1, 2, 3, 4],返回 [1, 3, 2, 4]. 挑战 在原数组中完成,不使用额外空间. 解题: 一次快速排序就可 ...
- Codeforces Round #575 (Div. 3) B. Odd Sum Segments 、C Robot Breakout
传送门 B题题意: 给你n个数,让你把这n个数分成k个段(不能随意调动元素位置).你需要保证这k个段里面所有元素加起来的和是一个奇数.问可不可以这样划分成功.如果可以打印YES,之后打印出来是从哪里开 ...
- LeetCode 548. Split Array with Equal Sum (分割数组使得子数组的和都相同)$
Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies fol ...
- lintcode373 奇偶分割数组
奇偶分割数组 分割一个整数数组,使得奇数在前偶数在后. 您在真实的面试中是否遇到过这个题? Yes 样例 给定 [1, 2, 3, 4],返回 [1, 3, 2, 4]. 我的方法:设定两个数组,分别 ...
- LeetCode 410——分割数组的最大值
1. 题目 2. 解答 此题目为 今日头条 2018 AI Camp 5 月 26 日在线笔试编程题第二道--最小分割分数. class Solution { public: // 若分割数组的最大值 ...
- Leetcode 410.分割数组的最大值
分割数组的最大值 给定一个非负整数数组和一个整数 m,你需要将这个数组分成 m 个非空的连续子数组.设计一个算法使得这 m 个子数组各自和的最大值最小. 注意:数组长度 n 满足以下条件: 1 ≤ n ...
- leetcode 410. 分割数组的最大值(二分法)
1. 题目描述 给定一个非负整数数组和一个整数 m,你需要将这个数组分成 m 个非空的连续子数组.设计一个算法使得这 m 个子数组各自和的最大值最小. 注意: 数组长度 n 满足以下条件: 1 ≤ n ...
- Odd sum (对本菜鸡来说是个极坑题)
https://codeforces.com/problemset/problem/797/B time limit per test 1 second memory limit per test 2 ...
随机推荐
- NIFI
Apache nifi 第一篇(概述) Apache nifi 第二篇(小白初试) nifi数据对接流程初次尝试 NIFI ExecuteSQL配置教程(1.8) Processor(处理器)之配置 ...
- WEB前段(HTML+JS),后端(MYSQL+PHP)开发基础
一.HTML HTML:超文本标记语言,可以加载JS/CSS/图片/链接等非文字的内容 一切的网页开发技术都需要建立在HTML的基础之上 HTML的结构和语法 HTML元素 注释: <!-- ...
- flask邮箱注册问题
app/models.py self.confirmed = True db.session.add(self) db.session.commit() 这里的数据修改完后必须commit提交上去,不 ...
- 【LeetCode】合并两个有序数组
[问题] 给定两个有序整数数组 nums1 和 nums2,将 nums2 合并到 nums1 中,使得 num1 成为一个有序数组. 说明:初始化 nums1 和 nums2 的元素数量分别为 m ...
- [YOLO]《YOLOv3: An Incremental Improvement》笔记
相比较于前两篇论文,个人感觉YOLO3作者有点来搞笑的!!!虽然加了一些新的点子进来,但是,论文的开头是这样的: 简单理解就是作者花了很多时间玩Twitter去了,所以没有做啥研究!!!! 然后: 你 ...
- No 'Access-Control-Allow-Origin' header is present on the requested resource——Web Api跨域问题
最近使用C#写了一个简单的web api项目,在使用项目中的.cshtml文档测试的时候没有任何问题,但是在外部HBuilder上面编写.html通过Ajax调用web api路径时报错: No 'A ...
- 剑指offer_2.3_Day_6
大家都知道斐波那契数列,现在要求输入一个整数n,请你输出斐波那契数列的第n项(从0开始,第0项为0). n<=39 public class Solution { public int Fibo ...
- 098-PHP二维数组的元素输出
<?php $stu=array(array(76,87,68), array(65,89,95), array(90,80,66), array(90,95,65)); //定义一个二维数组 ...
- ubuntu12.04安装JDK8
系统里已经有jdk1.6,下载并解压jdk后,按照网上的教程(bash.bashrc)没成功. sudo update-alternatives --install /usr/bin/java jav ...
- qvector 转为数组
在 qt 中想要把 qvector 转化为原始数据构成的数组,有几种方法: 直接使用循环读取 double *bytes = new double[vec.size()]; for (int i = ...