CF 719C Efim and Strange Grade

CF 719C C. Efim and Strange Grade

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
inputCopy
6 1
10.245
outputCopy
10.25
inputCopy
6 2
10.245
outputCopy
10.3
inputCopy
3 100
9.2
outputCopy
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**　就是给你一个字符串让你遵循四舍五入的准则，看看四舍五入后数最大是多少(注意只能对小数位进行四舍五入)字符串长度为你可以进行次四舍五入.
** 思路**　就是模拟尽情的模拟，注意卡点吧。

你需要注意他这个四舍五入不一定要从最后一位开始，可以从中间开始。

注意进位

注意小数点后没有多余的后缀０

注意要早据尾端最远的大于等于５的位置这样才是最优解。

找到据尾端最远的大于等于５的位置后往前找就是找４，找不到４就书名无法在进位了就退出就行了

注意无法进位，(四舍五入后没原数大)就输出数

#include <stack>
#include <map>
#include <queue>
#include <string>
#include<iostream>
#include<stdio.h>
#include<string.h>
#include <algorithm>
#include <math.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
typedef pair<ll, ll> pii;
#define
#define
#define
#define
#define
const int maxn = 1e6 + 10;
#define
#define
#define
const int mod = 1e9 + 7;
ll n, m, flag;
char str[maxn], s;
int main() {
cin >> n >> m;
cin >> str + 1;
for(int i = 1; i <= n; i++) {
if(str[i] == '.') {
flag = i;
break;
}
}
str[0] = '0';
if(flag == 0) {
cout << str + 1 << endl;
} else {
ll l = flag - 1;
for(int i = n; i > flag; i--) {
if(str[i] >= '5') {
l = i;
}
}
if(l == flag - 1) {
for(int i = 1; i <= n; i++) {
cout << str[i];
}
cout << endl;
return 0;
}
ll sum = l - flag;
m = min(m, sum);///最多那么多位
for(int i = l, j = 1; j <= m; i--, j++) {
if(str[i] >= '5' && str[i] <= '9') {
if(i > flag + 1 && str[i - 1] == '4') {
str[i]='0';
str[i - 1] += 1;
l = i - 1;
}else if(i>flag+1){
str[i-1]+=1;
l=i-1;
break;
}
else if(i==flag+1){
ll q=flag-1;
l=flag-1;
while(str[q]=='9'){
str[q]='0';
q--;
}
str[q]+=1;
break;
}
}else {
break;
}
}
for(int i=0;i<=l;i++){
if(i==0){
if( str[0]!='0')
cout<<str[i];
}
else {
cout<<str[i];
}
}
cout<<endl;
}
return 0;
}