Educational Codeforces Round 8 C. Bear and String Distance 贪心

C. Bear and String Distance

题目连接：

http://www.codeforces.com/contest/628/problem/C

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

4 26

bear

Sample Output

roar

Hint

题意

现在定义了一个dis(a,b)，dis(a,b) = abs(a-b)，等于这两个数的ASCII码的距离

然后现在给你一个串，让你构造另外一个串，使得这两个串之间的距离和恰好等于k

题解：

贪心，我们暴力向最远的地方靠就好了

代码

#include<bits/stdc++.h>
using namespace std;

string s,s1;
int main()
{
    int n,k;
    cin>>n>>k;
    cin>>s;
    for(int i=0;i<s.size();i++)
    {
        int dis1 = 'z'-s[i];
        int dis2 = s[i]-'a';
        if(dis1>dis2)
        {
            int ddd = min(dis1,k);
            k-=ddd;
            s1+=s[i]+ddd;
        }
        else
        {
            int ddd = min(dis2,k);
            k-=ddd;
            s1+=s[i]-ddd;
        }
    }
    if(k)return puts("-1");
    else cout<<s1<<endl;
}