Codeforces CF#628 Education 8 C. Bear and String Distance

C. Bear and String Distance
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.

The distance between two letters is defined as the difference between their positions in the alphabet. For example, , and .

Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, , and .

Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that . Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to usegets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead ofScanner/System.out in Java.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).

The second line contains a string s of length n, consisting of lowercase English letters.

Output

If there is no string satisfying the given conditions then print "-1" (without the quotes).

Otherwise, print any nice string s' that .

Examples
input
4 26
bear
output
roar
input
2 7
af
output
db
input
3 1000
hey
output
-1

 

题意:
    给出一个长度为n的字符串,定义两个字符串的距离就是对应位的字母在字母表上的距离之和,也就是ae跟ea距离为4+4=8。
    给出k,要求构造一个等长的字母串,使得两串距离恰好为k。

  

题解:
    很容易想到每一位都尽量拉长距离,尽快达到k,然后全部一样即可。

 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <string>
 4 #include <iostream>
 5 #include <algorithm>
 6 using namespace std;
 7 
 8 const int N = 100010;
 9 int n, k;
10 char str[N];
11 char ans[N];
12 
13 int main() {
14     scanf("%d%d", &n, &k);
15     scanf("%s", str);
16     memset(ans, 0, sizeof(ans));
17     for(int i = 0; i < n; ++i) {
18         int dis = min(k, max(str[i] - 'a', 'z' - str[i]));
19         if(str[i] - 'a' >= dis) ans[i] = str[i] - dis;
20         else ans[i] = str[i] + dis;
21         k -= dis;
22     }
23     if(k) puts("-1");
24     else printf("%s\n", ans);
25     return 0;
26 }
View Code

 

posted @ 2016-12-02 12:07  yanzx6  阅读(127)  评论(0编辑  收藏