Codeforces Round #197 (Div. 2) A. Helpful Maths【字符串/给一个连加计算式,只包含数字 1、2、3,要求重新排序,使得连加的数字从小到大】
2 seconds
256 megabytes
standard input
standard output
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
The first line contains a non-empty string s — the sum Xenia needs to count. String s contains no spaces. It only contains digits and characters "+". Besides, string s is a correct sum of numbers 1, 2 and 3. String s is at most 100 characters long.
Print the new sum that Xenia can count.
3+2+1
1+2+3
1+1+3+1+3
1+1+1+3+3
2
2
【代码】:
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm> using namespace std; const int N=100005;
int k;
char s[110];
int a[110]; int main()
{
scanf("%s",s);
for(int i=0;i<strlen(s);i+=2)//+=2
{
a[k++]=s[i]-'0';
}
sort(a,a+k); printf("%d",a[0]);//第一个前面无+,单独输出
for(int i=1;i<k;i++)
printf("+%d",a[i]);
printf("\n");
}
Codeforces Round #197 (Div. 2) A. Helpful Maths【字符串/给一个连加计算式,只包含数字 1、2、3,要求重新排序,使得连加的数字从小到大】的更多相关文章
- 线段树 Codeforces Round #197 (Div. 2) D. Xenia and Bit Operations
题目传送门 /* 线段树的单点更新:有一个交叉更新,若rank=1,or:rank=0,xor 详细解释:http://www.xuebuyuan.com/1154895.html */ #inclu ...
- Codeforces Round #197 (Div. 2) (A、B、C、D、E五题合集)
A. Helpful Maths 题目大意 给一个连加计算式,只包含数字 1.2.3,要求重新排序,使得连加的数字从小到大 做法分析 把所有的数字记录下来,从小到大排序输出即可 参考代码 #inclu ...
- Codeforces Round #197 (Div. 2)
A.Helpful Maths 分析:将读入的字符转化为数字,直接排个序就可以了. #include <cstdlib> #include <cstring> #include ...
- Codeforces Round #197 (Div. 2) : E
看了codeforces上的大神写的题解之后,才知道这道题水的根本! 不过相对前面两题来说,这道题的思维要难一点: 不过想到了水的根本,这题也真心不难: 方法嘛,就像剥洋葱一样,从外面往里面剥: 所以 ...
- [置顶] Codeforces Round #197 (Div. 2)(完全)
http://codeforces.com/contest/339/ 这场正是水题大放送,在家晚上限制,赛后做了虚拟比赛 A,B 乱搞水题 C 我是贪心过的,枚举一下第一个拿的,然后选使差值最小的那个 ...
- Codeforces Round #287 (Div. 2) D. The Maths Lecture [数位dp]
传送门 D. The Maths Lecture time limit per test 1 second memory limit per test 256 megabytes input stan ...
- Codeforces Round #197 (Div. 2) C,D两题
开了个小号做,C题一开始看错范围,D题看了半小时才看懂,居然也升到了div1,囧. C - Xenia and Weights 给出一串字符串,第i位如果是1的话,表示有重量为i的砝码,如果有该种砝码 ...
- Codeforces Round #197 (Div. 2) : C
哎....这次的比赛被安叔骂的好惨! 不行呢,要虐回来: 这道搜索,老是写错,蛋疼啊! 果然是基础没打好! #include<cstdio> using namespace std; ], ...
- Codeforces Round #197 (Div. 2) : D
这题也是一个线段树的水题: 不过开始题目没看明白,害得我敲了一个好复杂的程序.蛋疼啊.... 最后十几分钟的时候突然领悟到了题意,但是还是漏掉一个细节,老是过不去... 以后比赛的时候不喝啤酒了,再也 ...
随机推荐
- vue 路由跳转记住滚动位置,返回时回到上次滚动位置
参考:https://blog.csdn.net/qq_40204835/article/details/79853685 方法一: 利用Keep-Alive和监听器 1.首先在路由中引入需要的模块 ...
- jdom xpath定位带xmlns命名空间的节点(转)
jdom xpath定位带xmlns命名空间的节点 2013-06-29 0个评论 作者:baozhengw 收藏 我要投稿 关键词:jdom xpath xmlns 命名 ...
- Django项目:CMDB(服务器硬件资产自动采集系统)--01--01CMDB获取服务器基本信息
AutoClient #settings.py # ————————01CMDB获取服务器基本信息———————— import os BASEDIR = os.path.dirname(os.pat ...
- [转]10 Tips for Learning a New Technology
We live in a very exciting time. Never before has education been so cheaply available to the masses ...
- css 始终显示滚动条,内容超出显示有滑块的滚动条,内容没有超出显示空的滚动条
1.内容没有超出显示空的滚动条 <div class="div1"> 前端开发者前端开发者前端开发者前端开发者前端开发者 </div> css代码: .di ...
- phpSpider 单页测试_模拟登陆
<?php require './vendor/autoload.php'; use phpspider\core\phpspider; use phpspider\core\requests; ...
- MySQL系列(一)--基础知识(转载)
安装就不说了,网上多得是,我的MySQL是8.0版本,可以参考:CentOS7安装MySQL8.0图文教程和MySQL8.0本地访问设置为远程访问权限 我的MySQL安装在阿里云上面,阿里云向外暴露端 ...
- TZ_01MyBatis_log4j.propertiies
# Set root category priority to INFO and its only appender to CONSOLE. #log4j.rootCategory=INFO, CON ...
- 实体类No default constructor found 找不到默认构造函数;
root cause org.springframework.beans.BeanInstantiationException: Could not instantiate bean class [c ...
- XML解析器之JAXP与DOM4J
XML是一种数据格式,那么需要对XML文件进行操作就需要用到XML解析器---------针对dom方式和sax方式提供了不同的解析技术-----需要不同的XML解析器 dom方式:会把文档中所有元素 ...