codeforces 719C. Efim and Strange Grade

C. Efim and Strange Grade

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

Copy

6 1
10.245

output

Copy

10.25

input

Copy

6 2
10.245

output

Copy

10.3

input

Copy

3 100
9.2

output

Copy

9.2

Note

In the first two samples Efim initially has grade 10.245.

During the first second Efim can obtain grade 10.25, and then 10.3 during the next second. Note, that the answer 10.30 will be considered incorrect.

In the third sample the optimal strategy is to not perform any rounding at all.

题意：你有n次操作，可以让给定的这个数的小数位四舍五入，求这个数经过t次操作后最大是多少

题解：小数位如果第一个小数位是>=5的情况下，整数位的个位就要+1，如果整数位是999这种情况就要变成1000

为了使得经过变换后的数最大，我们先从小数位的最前面开始进位

代码如下：

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
char str[];
char a[];
char b[];
int main(){
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n,t;
    scanf("%d%d",&n,&t);
    cin>>str;
    int pos=;
    //整数
    for(pos = ;; pos++) {
        if(str[pos] == '.') break;
        else a[pos] = str[pos];
    }
    pos++;
    for(int i=;str[i];i++,pos++){
        b[i]=str[pos];
    }
    pos=-;
    for(int i=;b[i];i++){
        if(b[i]>=''){
            pos=i;
            break;
        }
    }
    int flag=;
    for(int i=pos;i>=&&t>;i--){
        if(i!=&&b[i]>=''){
            b[i-]++;
            b[i]=;
        }else if(i==&&b[i]>=''){
            flag=;
        }else break;
        t--;
    }
    if(!flag) printf("%s.%s\n",a,b);
    else{
        int len=strlen(a);
        int num=;
        int i;
        for(i=len-;i>=;i--){
            if(a[i]!=''){
                a[i]++;
                break;
            }else{
                a[i]='';
                num++;
            }
        }
        if(num==len) cout<<"";
        cout<<a<<endl;
    }
    return ;
}

#include<bits/stdc++.h>
using namespace std;
string s;
int main(){
    int n,t;
    scanf("%d%d",&n,&t);
    cin>>s;
    int i=;
    while(s[i]!='.') i++;  //整数部分的长度
    while(i<n&&s[i]<'') i++;  //不能进位的长度
    if(i==n){
        //如果全部不能四舍五入就直接输出
        cout<<s<<endl;
        return ;
    }
    i--;
    int len=;
    while(t>){
        if(s[i]!='.') s[i]++;
        else{

            i--;
            len=i;
            while(i>=&&s[i]=='') s[i--]='';  //如果当前位是9，那么进位时注意将当前位改为0
            if(i==-) cout<<'';           //如果是9999的情况，就变成10000;
            else s[i]++;
            break;
        }
        if(s[i]<''){
            len=i;
            break;
        }else{
            len=i;
            i--;
        }
        t--;
    }
    for(int i=;i<=len;i++){
        cout<<s[i];
    }
    cout<<endl;
}