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
10.25
6 2 10.245
10.3
3 100 9.2
9.2
Hint
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.
直接从最前面的一个大于5的数开始，模拟进位就好。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int main()
{
char str[200010];
memset(str,0,sizeof(str));
int n,t,i,j,jiayi=0;;
int dian =0,flag;
// n = read();
// t = read();
scanf("%d %d",&n,&t);
flag = n;
//getchar();
scanf("%s",str+1);
for(i=1;i<=n;i++)
if(str[i] == '.')
dian = i;
for(i=dian+1;i<=n;i++)
{
if(str[i]>='5')
{
flag = i;
break;
}
}
if(i==n+1)
{
puts(str+1);
return 0;
}
while(t--)//次数
{
str[flag] = '0';
for(i=flag-1;i>dian;i--)
{
str[i]++;
if(str[i]<':')
break;
if(str[i]==':')
str[i]='0';
}
if(i==dian)
{
jiayi = 1;
break;
}
if(str[i]>='5')
flag = i;
else
{
flag = i;
break;
}
}
if(jiayi)//整数进位
{
for(i=dian-1;i>=0;i--)
{
str[i]++;
if(str[i]<':')
break;
if(str[i]==':')
{
str[i]='0';
}
}
if(str[0])
cout << '1';
for(i=1;i<dian;i++)
cout << str[i];
return 0;
}
for(i=1;i<=flag;i++)
{
cout << str[i];
}
return 0;
}