【CodeForces 628C 】Bear and String Distance （贪心）

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本文介绍了一种通过贪心算法解决字符串匹配问题的方法，给定一个字符串和一个整数k，寻找与原字符串各字符间ASCII值之差总和等于k的新字符串。

Description

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 use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/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 .

Sample Input

Input

4 26
bear

Output

roar

Input

2 7
af

Output

Input

3 1000
hey

Output

-1

题目大意

约定两个字母之间的差为两者ASCII值的差的绝对值，给定一串英文字符的字母个数n以及规定要达到的差值d，输入一个字符串s，求与这个字符串差为d的字符串

思路

用贪心从第一位开始，与’z’和’a’比较，如果用’z’或者’a’作为这一位的答案之后还是无法达到差值k，那么如果这一位与’z’的差值大就用’z’作为这一位的答案，反之用’a’，总之就是要是k减去这一位的差值要尽可能的小，然后后移一位继续贪心。如果到某一位（这一位必须小于等于n）后不满足上述条件，那么这是当前串的差值与输入的字符串的差值一定小于’z’-s[i]>=k或者s[i]-‘a’>=k,那么如果是第一种情况就用s[i]+k作为这一位的答案，反之用s[i]-k作为这一位的答案。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=1e5+5;

int n,k;
char str[maxn];
char ans[maxn];

int main()
{
    while(~scanf("%d %d",&n,&k))
    {
        scanf("%s",&str);
        int len=strlen(str);
        for(int i=0;i<len;i++)
        {
            if(str[i]+k>'z'&&str[i]-k<'a')
            {
                if('z'-str[i]>=str[i]-'a')
                {
                    ans[i]='z';
                    k-='z'-str[i];
                }
                else
                {
                    ans[i]='a';
                    k-=str[i]-'a';
                }
            }
            else if(str[i]+k<='z')
            {
                ans[i]=str[i]+k;
                k=0;
            }
            else if(str[i]-k>='a')
            {
                ans[i]=str[i]-k;
                k=0;
            }

        }
        if(!k) printf("%s\n",ans);
        else printf("-1\n");
    }
    return 0;
}