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 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

6 1
10.245

output

10.25

input

6 2
10.245

output

10.3

input

3 100
9.2

output

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.

题解：

1.整段数字被分成两部分， 小数部分和正数部分。 先对小数部分进行进位。

2.首先找到小数点后第一个大于等于5的位置，对这个位置进行四舍五入。进位时，先用个while循环处理完'999……'的进位，然后再对当前位+1。如果当前位<5，就可以马上停止了， 因为前面的数都不可能>=5。 如果当前位>=5, 则继续下一次四舍五入（一定是当前位，所以不必每次都遍历找第一个>=5的数），最多t次。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <sstream>
#include <algorithm>
using namespace std;
#define ms(a, b)  memset((a), (b), sizeof(a))
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 2e5+10;

char s[maxn];
int n, t, dot;

int main()
{
    scanf("%d%d%s",&n, &t, s+1);
    for(dot = 1; dot<=n; dot++)
        if(s[dot]=='.') break;

    int i;
    for(i = dot+1; i<=n; i++)
        if(s[i]>='5') break;

    if(i>n)
    {
        puts(s+1);
        return 0;
    }

    while(t--)
    {
        s[i--] = 0;
        while(s[i]!='.' && s[i]=='9')
            s[i--] = 0;
        if(s[i]!='.') s[i]++;

        if(s[i]=='.' || s[i]<'5') break;
    }

    if(s[dot+1]==0)
    {
        s[dot] = 0;
        i = dot-1;

        while(i>=1 && s[i]=='9')
            s[i--] = '0';
        if(i>=1) s[i]++;
        else putchar('1');
    }
    puts(s+1);
}