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

  1. 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 ...

  2. 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 ...

  3. 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, ...

  4. 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/ ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. 阿里云centos7.3安装lamp环境

    参考文档:http://www.jb51.net/article/96649.htm http://m.blog.csdn.net/qq_33813365/article/details/766337 ...

  2. AngularJs -- ngMessages(1.3+)

    ngMessages(1.3+) 表单和验证是AngularJS中复杂的组件之一.用AngularJS默认的方式来写,不是特别好,不简洁. 在AngualrJS1.3发布前,表单验证必须以这种方式编写 ...

  3. 关于System.getProperty("java.io.tmpdir");的输出,及System.getProperty();参数

    1,首先来介绍下System.getProperty("java.io.tmpdir")输出因为这个输出有点特殊. 理论介绍:他是获取系统临时目录.可以是window的temp,l ...

  4. 200行代码实现RPC框架

    之前因为项目需要,基于zookeeper和thrift协议实现了一个简单易用的RPC框架,核心代码不超过200行. zookeeper主要作用是服务发现,thrift协议作为通信传输协议, 基于com ...

  5. 第12月第30天 love2d

    1. Linux On Linux, you can use one of these command lines: love /home/path/to/gamedir/ love /home/pa ...

  6. HTML5+CSS把footer固定在底部

    在刚开始给网页写footer的时候,我们会碰到一个让人烦躁的问题:当页面内容太少时,footer显示在了页面中间,这是我们不希望出现的,我们希望它能够永远呆在底部,不管网页的内容是多还是少.下面的代码 ...

  7. 微信小程序实现首页图片多种排版布局!

    先来个效果图: 使用技术主要是flex布局,绝对定位布局,小程序前端页面开发,以及一些样式! 直接贴代码,都有详细注释,熟悉一下,方便以后小程序开发! wxml: <view class='in ...

  8. 微信小程序调用接口返回数据或提交数据

    由于小程序发起网络请求需要通过 wx.request 文档地址 https://mp.weixin.qq.com/debug/wxadoc/dev/api/network-request.html 习 ...

  9. dataframe 差集

    >>>data_a={'state':[1,1,2],'pop':['a','b','c']}>>>data_b={'state':[1,2,3],'pop':[' ...

  10. vue里面使用Velocity.js

    英文文档:http://velocityjs.org/ https://github.com/julianshapiro/velocity 中文手册(教程):http://www.mrfront.co ...