Codeforces Round #373 (Div. 2) -- C - Efim and Strange Grade （贪心）

大体题意：

给你一个最多长度为20W的小数，你有t次进位的机会，每一次 你可以在小数点后任意一个位置进位，求进位后小数的最大值？

思路：

贪心思路，没想到比赛想的不够好，超时了= =

其实还是很简单的！

直接找到小数点后第一个可以进位的数>=5的位置，然后从这个位置开始往前进，最多进t次，注意处理，小数点前的进位和加上前导1的情况即可！

详细见代码：

#include <bits/stdc++.h>using namespace std;const int maxn = 200000 + 10;char s[maxn];int main(){    int n,t;    scanf("%d%d%s",&n, &t, s);    int len = strlen(s);    int pos;    for (int i = 0; i < len; ++i){        if (s[i] == '.'){            pos = i;            break;        }    }    int i;    for (i = pos+1; i < len; ++i){        if (s[i] >= '5')break;    }    bool qian0 = 0;    int flag = 1;    if (i == len)flag = 0;    len = i;    for (--i ; i > pos && t > 0 && flag; --i,--t){        s[i] += flag;        if (s[i] >= '5' && t > 1)flag = 1,len = i;        else flag = 0;    }    if (i == pos && flag){        for (i = pos-1; i >= 0 && flag; --i){            s[i] += 1;            if (s[i] > '9')s[i] = 48,flag = 1;            else flag = 0;        }    }    if (flag == 1 && i == -1)qian0 = 1;    if (qian0)printf("1");    s[len] = 0;    if (s[len-1] == '.')s[len-1] = 0;    puts(s);    return 0;}

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 thant seconds. Note, that he can choose to not use allt 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 then-th digit one has to take a look at the digitn + 1. If it is less than 5 than the n-th digit remain unchanged while all subsequent digits are replaced with0. Otherwise, if the n + 1 digit is greater or equal to5, the digit at the position n is increased by 1 (this might also change some other digits, if this one was equal to9) 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 is1.1, while if we round 1.5 to the nearest integer, the result is2. Rounding number 1.299996121 in the fifth decimal place will result in number1.3.

Input

The first line of the input contains two integers n andt (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 with0.

Output

Print the maximum grade that Efim can get in t seconds. Do not print trailing zeroes.

Examples

Input

6 110.245

Output

10.25

Input

6 210.245

Output

10.3

Input

3 1009.2

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 then10.3 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.