Bear and String Distance( 贪心)

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

.
Example Input

4 26
bear

Output

roar

Input

2 7
af

Output

db

Input

3 1000
hey

Output

-1

题意：
给一个长度为n 字符串，求一个距离此字符串为k 的字符串
字符串距离：两个字符串对应字符距离的和
思路：优先转换成'a'或'z'#include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
int main(){
int n,k;
scanf("%d%d",&n,&k);
string str;
cin>>str;
for(int i=0;i<str.size();i++){
int x='z'-str[i];
int y=str[i]-'a';
if(x>y){
if(k>x){
str[i]='z';
k-=x;
}
else{
str[i]+=k;
k=0;
}
}
else{
if(k>y){
str[i]='a';
k-=y;
}
else{
str[i]-=k;
k=0;
}
}
if(k==0) break;
}
if(k==0) cout<<str<<endl;
else cout<<"-1"<<endl;
return 0;
}