1 second

256 megabytes

standard input

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.

The first line of the input contains two integers *n* and *t* (1 ≤ *n* ≤ 200 000, 1 ≤ *t* ≤ 10^{9}) — 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.

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

1 2 |
6 1 10.245 |

1 |
10.25 |

1 2 |
6 2 10.245 |

1 |
10.3 |

1 2 |
3 100 9.2 |

1 |
9.2 |

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.

显然可以看出在离小数点更近的地方开始取整最终得到的值最大，所以可以用动态规划的方式来确定从哪一位开始向上取整，确定以后再从确定的地方逐步加一然后得到最终结果。

令\(dp_{i}\) 为向上取整需要消耗的机会次数，对于当前字符串\(a\)，可以得到：

\(dp_{i}=\begin{cases}

1 & a_{i}\ge5\\

INF & a_{i}\lt4\\

dp_{i+1}+1 & a_{i}=4

\end{cases}

\)

然后从左至右找到第一个\(dp_{i}\le t\)即从获取的索引开始向上取整，每取一次整\(t = t-dp_{i}\)，最后去掉结尾的0，输出结果即可。

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/* * File Name: 719C.cpp * Author: razrLeLe * Mail: razrlele@gmail.com */ #include <bits/stdc++.h> #define int64 long long using namespace std; #define INF 0x3f3f3f3f #define LOCAL const int maxn = 200007; string in_str; int dp[maxn]; int dot; int solve(int i) { if(dp[i] != -1) return dp[i]; else if(in_str[i]=='.' || in_str[i] < '4' || i < dot) dp[i] = INF; else if( in_str[i] >= '5') dp[i] = 1; else dp[i] = 1 + solve(i+1); return dp[i]; } int main(int argc, char ** argv) { #ifdef LOCAL freopen("/Users/yuyue/build/data.txt", "r", stdin); // freopen("/Users/yuyue/build/out.txt", "w", stdout); #endif int n, t; while( cin >> n >> t) { memset(dp, -1, maxn*sizeof(int)); cin >> in_str; dot = in_str.find("."); for(int i = 0; i < n; i++) solve(i); int pos = n-1; for( int i = 0; i < n ; i++) if(dp[i] <= t) { pos = i; break; } in_str = in_str.substr(0, pos+1); int end; for( int i=pos; i>0 ;i--) { if( in_str[i] > '9') { in_str[i] = '0'; if( in_str[i-1] != '.') in_str[i-1]++; else in_str[i-2]++; } else if( t >= dp[i]) { in_str[i]= '0'; if( in_str[i-1] != '.') in_str[i-1]++; else in_str[i-2]++; t -= dp[i]; } } if( in_str[0] > '9') { in_str[0] = '0'; in_str = "1"+in_str; } dot = in_str.find("."); end = in_str.length()-1; for( ; end>dot;) { if(in_str[end] == '0') end--; else break; } if( end == dot) end--; cout << in_str.substr(0, end+1)<<endl; } return 0; } |