Codeforces Round #373 (Div. 2) C. Efim and Strange Grade 水题
C. Efim and Strange Grade
题目连接:
http://codeforces.com/contest/719/problem/C
Description
Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. First he got disappointed, as he expected a way more pleasant result. Then, he developed a tricky plan. Each second, he can ask his teacher to round the grade at any place after the decimal point (also, he can ask to round to the nearest integer).
There are t seconds left till the end of the break, so Efim has to act fast. Help him find what is the maximum grade he can get in no more than t seconds. Note, that he can choose to not use all t seconds. Moreover, he can even choose to not round the grade at all.
In this problem, classic rounding rules are used: while rounding number to the n-th digit one has to take a look at the digit n + 1. If it is less than 5 than the n-th digit remain unchanged while all subsequent digits are replaced with 0. Otherwise, if the n + 1 digit is greater or equal to 5, the digit at the position n is increased by 1 (this might also change some other digits, if this one was equal to 9) and all subsequent digits are replaced with 0. At the end, all trailing zeroes are thrown away.
For example, if the number 1.14 is rounded to the first decimal place, the result is 1.1, while if we round 1.5 to the nearest integer, the result is 2. Rounding number 1.299996121 in the fifth decimal place will result in number 1.3.
Input
The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 109) — the length of Efim's grade and the number of seconds till the end of the break respectively.
The second line contains the grade itself. It's guaranteed that the grade is a positive number, containing at least one digit after the decimal points, and it's representation doesn't finish with 0.
Output
Print the maximum grade that Efim can get in t seconds. Do not print trailing zeroes.
Sample Input
6 1
10.245
Sample Output
10.25
Hint
题意
给你一个数,你最多做T次四舍五入,问你最大能够是多少
题解:
找到第一个能够进位的数,然后贪心的去四舍五入就好了。
水题
代码
#include<bits/stdc++.h>
using namespace std;
string s;
int main()
{
int n,t;
scanf("%d%d",&n,&t);
cin>>s;
int i=0;
while(s[i]!='.')i++;
while(i<n&&s[i]<'5')i++;
if(i==n)
{
cout<<s<<endl;
return 0;
}
i--;int len=0;
while(t>0)
{
if(s[i]!='.')s[i]++;
else{
i--;len=i;
while(i>=0&&s[i]=='9')s[i--]='0';
if(i==-1)cout<<'1';
else s[i]++;
break;
}
if(s[i]<'5')
{
len=i;
break;
}
else
{
len=i;
i--;
}
t--;
}
for(int k=0;k<=len;k++)
cout<<s[k];
cout<<endl;
}
Codeforces Round #373 (Div. 2) C. Efim and Strange Grade 水题的更多相关文章
- Codeforces Round #373 (Div. 2) C. Efim and Strange Grade —— 贪心 + 字符串处理
题目链接:http://codeforces.com/problemset/problem/719/C C. Efim and Strange Grade time limit per test 1 ...
- Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 水题
A. Vitya in the Countryside 题目连接: http://codeforces.com/contest/719/problem/A Description Every summ ...
- Codeforces Round #368 (Div. 2) A. Brain's Photos (水题)
Brain's Photos 题目链接: http://codeforces.com/contest/707/problem/A Description Small, but very brave, ...
- Codeforces Round #185 (Div. 2) A. Whose sentence is it? 水题
A. Whose sentence is it? Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/ ...
- Codeforces Round #371 (Div. 2) A. Meeting of Old Friends 水题
A. Meeting of Old Friends 题目连接: http://codeforces.com/contest/714/problem/A Description Today an out ...
- Codeforces Round #355 (Div. 2) B. Vanya and Food Processor 水题
B. Vanya and Food Processor 题目连接: http://www.codeforces.com/contest/677/problem/B Description Vanya ...
- Codeforces Round #310 (Div. 2) B. Case of Fake Numbers 水题
B. Case of Fake Numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...
- Codeforces Round #309 (Div. 2) B. Ohana Cleans Up 字符串水题
B. Ohana Cleans Up Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/554/pr ...
- Codeforces Round #309 (Div. 2) A. Kyoya and Photobooks 字符串水题
A. Kyoya and Photobooks Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...
随机推荐
- CF #442 div2
A 判断下5个名字出现了几次.pre数据巨弱,就这么一水题在std测刷掉了非常多的人.. /** @Date : 2017-10-24 16:04:41 * @FileName: A.cpp * @P ...
- 消息队列之RabbitMQ的.Net客户端EasyNetQ
https://www.cnblogs.com/CoderAyu/p/9072408.html http://www.cnblogs.com/panzi/p/6337568.html http://w ...
- Mogodb 学习一
0.MongoDB和关系型数据的几个重要对象对比 MongoDB中的数据库.集合.文档 类似于关系型数据库中的数据库.表.行 MongoDB中的集合是没有模式的,所以可以存储各种各样的文档 1.启动M ...
- 一个简单的二叉搜索树(C++实现)
参考:http://www.cnblogs.com/skywang12345/p/3576373.html 这里主要就是自己实现的代码,删除动作有点不一样: #ifndef __BSTREE_H__ ...
- HDU 4627 The Unsolvable Problem 杭电多校联赛第三场1009 数学题
题意描述:给出一个n,要求在所有满足n = a+b的a和b里面求a和b的最小公倍数最大的两个数的最小公倍数. 解题报告:比赛的时候看到这个题的第一反应就是寻找这两个数一定是在a和b比较接近的地方找,这 ...
- 使用 scm-manager 搭建 git/svn 代码管理仓库(二)
主要介绍scm的配置. 1.配置为在Windows服务中启动scm-manager的启动方式有多种,可以在DOS(即命令行CMD模式)中启动,也可以在Windows服务中启动. 下面我们采用Windo ...
- CodeForces Contest #1110: Global Round 1
比赛传送门:CF #1110. 比赛记录:点我. 涨了挺多分,希望下次还能涨. [A]Parity 题意简述: 问 \(k\) 位 \(b\) 进制数 \(\overline{a_1a_2\cdots ...
- MySQL5.7 GTID在线开启与关闭【转】
当前场景 当前某些业务还有未开启GTID服务组,升级5.7后,如何检测是否符合开启GTID条件,如何在线修改切换使用GTID:已经升级5.7后,已经开启GTID,如何快速回滚后退: 线上gtid如 ...
- python格式化输出【转】
今天写代码时,需要统一化输出格式进行,一时想不起具体细节,用了最笨的方法,现在讲常见的方法进行一个总结. 一.格式化输出 1.整数的输出 直接使用'%d'代替可输入十进制数字: >>> ...
- 秘钥登录服务器执行shell脚本
做自动化的时候,有时候避免不了要和服务器有互动,刚巧碰上一个项目,需要执行命令才能完成本次测试. 昨天遇到的是秘钥形式的,只有秘钥和用户名,百度找了许久也没有思路,(能账号密码登录服务器的还简单些), ...