链接:

https://codeforces.com/contest/1251/problem/B

题意:

A palindrome is a string t which reads the same backward as forward (formally, t[i]=t[|t|+1−i] for all i∈[1,|t|]). Here |t| denotes the length of a string t. For example, the strings 010, 1001 and 0 are palindromes.

You have n binary strings s1,s2,…,sn (each si consists of zeroes and/or ones). You can swap any pair of characters any number of times (possibly, zero). Characters can be either from the same string or from different strings — there are no restrictions.

Formally, in one move you:

choose four integer numbers x,a,y,b such that 1≤x,y≤n and 1≤a≤|sx| and 1≤b≤|sy| (where x and y are string indices and a and b are positions in strings sx and sy respectively),

swap (exchange) the characters sx[a] and sy[b].

What is the maximum number of strings you can make palindromic simultaneously?

思路:

考虑所有串的长度,当有奇数长度存在时,保证可以交换出全部,当没有时, 可能存在0或1为奇数,不能配对,有奇数可以将奇数减位,同时保证剩下的偶数相等。

代码:

#include<bits/stdc++.h>
using namespace std;
typedef long long LL; int n;
string s; int main()
{
int t;
scanf("%d", &t);
while(t--)
{
int one = 0, zer = 0;
int q = 0;
scanf("%d", &n);
for (int i = 1;i <= n;i++)
{
cin >> s;
int len = (int)s.length();
if (len%2)
q++;
for (int j = 0;j < len;j++)
{
if (s[j] == '0')
zer++;
else
one++;
}
}
if (q > 0)
printf("%d\n", n);
else
{
if (one%2 || zer%2)
printf("%d\n", n-1);
else
printf("%d\n", n);
}
}
return 0;
}

Educational Codeforces Round 75 (Rated for Div. 2) B. Binary Palindromes的更多相关文章

  1. Educational Codeforces Round 75 (Rated for Div. 2)

    知识普及: Educational使用拓展ACM赛制,没有现场hack,比赛后有12h的全网hack时间. rank按通过题数排名,若通过题数相等则按罚时排名. (罚时计算方式:第一次通过每题的时间之 ...

  2. Educational Codeforces Round 75 (Rated for Div. 2) D. Salary Changing

    链接: https://codeforces.com/contest/1251/problem/D 题意: You are the head of a large enterprise. n peop ...

  3. Educational Codeforces Round 75 (Rated for Div. 2) C. Minimize The Integer

    链接: https://codeforces.com/contest/1251/problem/C 题意: You are given a huge integer a consisting of n ...

  4. Educational Codeforces Round 75 (Rated for Div. 2) A. Broken Keyboard

    链接: https://codeforces.com/contest/1251/problem/A 题意: Recently Polycarp noticed that some of the but ...

  5. Educational Codeforces Round 75 (Rated for Div. 2)D(二分)

    #define HAVE_STRUCT_TIMESPEC#include<bits/stdc++.h>using namespace std;pair<int,int>a[20 ...

  6. Educational Codeforces Round 94 (Rated for Div. 2) C. Binary String Reconstruction (构造)

    题意:给你一个字符串\(s\),原字符串为\(w\),如果\(i>x\)且\(w_{i-x}=1\),那么\(s_{i}=1\),如果\(i+x\le n\)且\(w_{i+x}=1\),那么\ ...

  7. Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship

    Problem   Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec P ...

  8. Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems(动态规划+矩阵快速幂)

    Problem   Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec P ...

  9. Educational Codeforces Round 43 (Rated for Div. 2)

    Educational Codeforces Round 43 (Rated for Div. 2) https://codeforces.com/contest/976 A #include< ...

随机推荐

  1. 树莓派上跑.NET的segment fault错误

    答案:树莓派1和树莓派zero是不支持的,原因是.net需要arm v7 详情看这里 可以用 cat /proc/cpuinfo | grep 'model name' |uniq 看一下cpu

  2. 在CentOS下搭建GoLang开发环境

    下载安装包 [vagrant@dk ~]$ wget http://mirrors.ustc.edu.cn/golang/go1.13.linux-amd64.tar.gz 使用USTC 中科大开源镜 ...

  3. Java调用SqlLoader将大文本导入数据库

    Java调用SqlLoader将大文本导入数据库 业务场景:将一千万条数据,大约500M的文本文档的数据导入到数据库 分析:通过Java的IO流解析txt文本文档,拼接动态sql实现insert入库, ...

  4. 单源最短路——朴素Dijkstra&堆优化版

    朴素Dijkstra 是一种基于贪心的算法. 稠密图使用二维数组存储点和边,稀疏图使用邻接表存储点和边. 算法步骤: 1.将图上的初始点看作一个集合S,其它点看作另一个集合 2.根据初始点,求出其它点 ...

  5. 修改织梦DedeCMS投票漏洞

    织梦/dedecms系统我们都知道是有很多漏洞的,我在调试投票功能的时候正好要用到投票功能,这不就出现了漏洞,下面我就给大家展示如何修复这个织梦投票漏洞 首先我们打开//dedevote.class. ...

  6. X86驱动:恢复SSDT内核钩子

    SSDT 中文名称为系统服务描述符表,该表的作用是将Ring3应用层与Ring0内核层,两者的API函数连接起来,起到承上启下的作用,SSDT并不仅仅只包含一个庞大的地址索引表,它还包含着一些其它有用 ...

  7. IDEA 导入 NodeJS 项目部署启动

    1.导入项目 2.添加模块 3.配置启动项 4.启动 5.备注 如果不明白,新建一个项目查看配置详情 原文地址:https://blog.csdn.net/tiankongzhichenglyf/ar ...

  8. webpack 3.1 升级webpack 4.0

    webpack 3.1 升级webpack 4.0 为了提升打包速度以及跟上主流技术步伐,前段时间把项目的webpack 升级到4.0版本以上 webpack 官网:https://webpack.j ...

  9. Vivado cordic IP求模求角教程

    前言 当需要对复数求模的时候,用FPGA怎么求呢?怎么开根号? 方法1:先求幅值平方和,再使用cordic IP开根号.(蠢办法) 方法2:直接用cordic求取模值. 此处只介绍方法2,资源占用更少 ...

  10. BASIS小问题汇总1

    try to start SAP system but failed 2019-04-04 Symptom: when i tried to start SAP system, using the c ...