Codeforces Round #585 (Div. 2) D. Ticket Game
链接:
https://codeforces.com/contest/1215/problem/D
题意:
Monocarp and Bicarp live in Berland, where every bus ticket consists of n digits (n is an even number). During the evening walk Monocarp and Bicarp found a ticket where some of the digits have been erased. The number of digits that have been erased is even.
Monocarp and Bicarp have decided to play a game with this ticket. Monocarp hates happy tickets, while Bicarp collects them. A ticket is considered happy if the sum of the first n2 digits of this ticket is equal to the sum of the last n2 digits.
Monocarp and Bicarp take turns (and Monocarp performs the first of them). During each turn, the current player must replace any erased digit with any digit from 0 to 9. The game ends when there are no erased digits in the ticket.
If the ticket is happy after all erased digits are replaced with decimal digits, then Bicarp wins. Otherwise, Monocarp wins. You have to determine who will win if both players play optimally.
思路:
考虑两半, 如果和相等, 则?树也要相等.
如果l>r, 则l的问号要小于r,同时两边的问号差为sub, 考虑为了补充右边等于左边, 如果左边较小则填较大使其不成立, 较大则填较小.所有只有差值为sub/2*9时成立.
这样两人一起的贡献和为9.正好填充相等.
代码:
#include <bits/stdc++.h>
using namespace std;
string s;
int n;
int main()
{
cin >> n;
cin >> s;
int sum1 = 0, sum2 = 0, num1 = 0, num2 = 0;
for (int i = 0;i < n;i++)
{
if (i < n/2)
{
if (s[i] == '?')
num1++;
else
sum1 += s[i]-'0';
}
else
{
if (s[i] == '?')
num2++;
else
sum2 += s[i]-'0';
}
}
if (sum1 < sum2)
swap(sum1, sum2), swap(num1,num2);
if (sum1 == sum2)
{
if (num1 == num2)
puts("Bicarp");
else
puts("Monocarp");
return 0;
}
if (num1 > num2)
{
puts("Monocarp");
return 0;
}
int need = sum1-sum2;
int own = ((num2-num1)/2)*9;
if (need != own)
puts("Monocarp");
else
puts("Bicarp");
return 0;
}
Codeforces Round #585 (Div. 2) D. Ticket Game的更多相关文章
- Codeforces Round #585 (Div. 2)
https://www.cnblogs.com/31415926535x/p/11553164.html 感觉很硬核啊这场,,越往后越做不动,,,emmmm,,,(这场是奔着最后一题 2sat 来的, ...
- Codeforces Round #585 (Div. 2) C. Swap Letters
链接: https://codeforces.com/contest/1215/problem/C 题意: Monocarp has got two strings s and t having eq ...
- Codeforces Round #585 (Div. 2) B. The Number of Products(DP)
链接: https://codeforces.com/contest/1215/problem/B 题意: You are given a sequence a1,a2,-,an consisting ...
- Codeforces Round #585 (Div. 2) A. Yellow Cards(数学)
链接: https://codeforces.com/contest/1215/problem/A 题意: The final match of the Berland Football Cup ha ...
- Codeforces Round #585 (Div. 2) [补题]
前言 2019.9.16 昨天下午就看了看D题,没有写对,因为要补作业,快点下机了,这周争取把题补完. 2019.9.17 这篇文章或者其他文章难免有错别字不被察觉,请读者还是要根据意思来读,不要纠结 ...
- B. The Number of Products(Codeforces Round #585 (Div. 2))
本题地址: https://codeforces.com/contest/1215/problem/B 本场比赛A题题解:https://www.cnblogs.com/liyexin/p/11535 ...
- A. Yellow Cards ( Codeforces Round #585 (Div. 2) 思维水题
---恢复内容开始--- output standard output The final match of the Berland Football Cup has been held recent ...
- Codeforces Round #585 (Div. 2) E. Marbles (状压DP)
题目:https://codeforc.es/contest/1215/problem/E 题意:给你一个序列,你可以交换相邻的两个数,要达到一个要求,所有相同的数都相邻,问你交换次数最少是多少 思路 ...
- Codeforces Round #585 (Div. 2)E(状态压缩DP,思维)
#define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h>using namespace std;long long n,x;long lon ...
随机推荐
- 笔记-7:mysql视图
1.视图概述 2.创建视图 CREATE [OR REPLACE] VIEW view_name [(column_list)] AS SELECT_statement [WITH { CASCADE ...
- TZOJ1294吃糖果
#include<stdio.h> int main() { ],mi,i,max,s; scanf("%d",&t); while(t--) { scanf( ...
- 15_IO流
IO流 流 流的概念 流(stream)是指一连串流动字节/字符,按照先进先出的方式发送的信息的通道中. 数据源:流入通道中的数据的来源 目的地:流出通道的数据的目的地 输入流和输出流 数据源的数 ...
- 运维MES的日子里
可以看下<工业软件国内与国外差距,越来越小还是越来越大>.<MES技术国内现状与未来发展趋势>及<国际主要MES厂商>,总结来说中国工业是进步的,工业软件也是进步的 ...
- 利用贝叶斯算法实现手写体识别(Python)
在开始介绍之前,先了解贝叶斯理论知识 https://www.cnblogs.com/zhoulujun/p/8893393.html 简单来说就是:贝叶斯分类是一类分类算法的总称,这类算法均以贝叶斯 ...
- C#使用HtmlAgilityPack快速爬虫
HtmlAgilityPack真是一把网抓利器,可以迅速地从网页抓到想要的文本或数据,使用起来十分方便,引用时在NuGet安装添加并在头部引用using HtmlAgilityPack;即可. 针对网 ...
- Down State Flush Feature
Down State Flush Feature ========================================================== Citrix NetScaler ...
- Java基础加强-类加载器
/*类加载器*/ 把.class文件从硬盘上加载出来,将类的字节码(二进制)加载到内存中 /*类加载器及其委托机制*/ Java虚拟机中可以安装多个类加载器(可以自己编写),系统默认三个主要类加载器, ...
- IDEA 导入jar包
项IDEA的项目中导入下载好的jar包: 在intelij IDEA 中,点击File-Project Structure,出现界面的左侧点击Modules,然后点击“+”. 然后找到你要导入的jar ...
- Postman --> YApi
初始 Postman,才知其如此强大,慢慢接触学习吧~ “Modern software is built on APIs,Postman helps you develop APIs faster” ...