Codeforces Round #379 (Div. 2) A. Anton and Danik 水题
A. Anton and Danik
题目连接:
http://codeforces.com/contest/734/problem/A
Description
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Sample Input
6
ADAAAA
Sample Output
Anton
Hint
题意
问你A多,还是D多,然后输出谁赢谁输
题解:
签到题
代码
#include<bits/stdc++.h>
using namespace std;
string s;
int main()
{
cin>>s;
cin>>s;
int a=0,b=0;
for(int i=0;i<s.size();i++)
if(s[i]=='A')a++;
else b++;
if(a>b)puts("Anton");
else if(a==b)puts("Friendship");
else puts("Danik");
}
Codeforces Round #379 (Div. 2) A. Anton and Danik 水题的更多相关文章
- Codeforces Round #379 (Div. 2) D. Anton and Chess 水题
D. Anton and Chess 题目连接: http://codeforces.com/contest/734/problem/D Description Anton likes to play ...
- Codeforces Round #379 (Div. 2) B. Anton and Digits 水题
B. Anton and Digits 题目连接: http://codeforces.com/contest/734/problem/B Description Recently Anton fou ...
- Codeforces Round #379 (Div. 2) D. Anton and Chess —— 基础题
题目链接:http://codeforces.com/contest/734/problem/D D. Anton and Chess time limit per test 4 seconds me ...
- Codeforces Round #404 (Div. 2) B. Anton and Classes 水题
B. Anton and Classes 题目连接: http://codeforces.com/contest/785/problem/B Description Anton likes to pl ...
- Codeforces Round #404 (Div. 2) A - Anton and Polyhedrons 水题
A - Anton and Polyhedrons 题目连接: http://codeforces.com/contest/785/problem/A Description Anton's favo ...
- Codeforces Round #297 (Div. 2)A. Vitaliy and Pie 水题
Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec Memory Limit: 256 MBSubmit: xxx ...
- Codeforces Round #379 (Div. 2) A B C D 水 二分 模拟
A. Anton and Danik time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforces Round #290 (Div. 2) A. Fox And Snake 水题
A. Fox And Snake 题目连接: http://codeforces.com/contest/510/problem/A Description Fox Ciel starts to le ...
- Codeforces Round #322 (Div. 2) A. Vasya the Hipster 水题
A. Vasya the Hipster Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/581/p ...
随机推荐
- java多条件不定条件查询
网站或各类管理系统都会用到搜索,会用到一个或多个不确定条件搜索,单条件搜索比较简单,有时候会有多个条件共同查询,如果系统中已经提供了相关的方法供你使用最好,像我做这老系统改版,需要添加搜索,就要自己写 ...
- jsp_注释
jsp支持两种注释的语法操作,一种是显示注释(在客户端允许看的见),另一种是隐式注释 显示注释:<!--注释内容--> 隐式注释: 格式一://单行注释 格式二:/*多行注释*/ 格式三: ...
- GitLab的Gravatar头像服务不可用
由于www.gravatar.com在国内不可正常使用,导致我们搭建的GitLab在网页上会阻塞大量时间,并最终无法显示头像.我们可以将其替换成"多说"的头像服务.我使用的是CE ...
- asp.net正则表达式学习例子
asp.net 获取网页Document时常会用到 edited by:曹永思-博客园 1.获取某个class的div内的标签 获取<div class="imgList2" ...
- jQuery学习总结(二)
简单选择器: 在使用jQuery 选择器时,我们首先必须使用“$()”函数来包装我们的CSS 规则. 而CSS 规则作为参数传递到jQuery 对象内部后,再返回包含页面中对应元素的jQuery 对象 ...
- Mine Number(搜索,暴力) ACM省赛第三届 G
Mine Number Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 Every one once played the gam ...
- 中国大学MOOC-陈越、何钦铭-数据结构-2016秋期中考试
判断题: 1-1 算法分析的两个主要方面是时间复杂度和空间复杂度的分析. (2分) 1-2 将N个数据按照从小到大顺序组织存放在一个单向链表中.如果采用二分查找,那么查找的平均时间复杂度是O(logN ...
- java环境配置笔记
1.使用Eclipse,要安装jdk,jdk现在可用1.7版本 2.打开Eclipse,配置maven,打开window-preferencess,在maven-user settings处,设置ma ...
- 【Win10 UWP】QQ SDK(一):SDK基本使用方法
每当开发一个应用需要社交分享的应用时,总是心里咯噔一下:到底什么时候分享能加上QQ和微信?除了WP8.0版本的微信SDK,官方似乎从未正面发布过适应时代发展的QQ SDK,就连后台,也没有一个可以创建 ...
- Javascript开发之工具归纳
写在前面 由于JS开发对我来说是全新的技术栈,开发过程中遇到了各种各样的框架.工具,同时也感叹一下相对于.Net的框架(工具框架以及测试框架等)JS框架真的是太丰富了.社区的力量果然强大---也是由此 ...