CodeForces 628C

 CodeForces 628C  

Description

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

The distance between two letters is defined as the difference between their positions in the alphabet. For example, dist(c,e)=dist(e,c)=2;, and .dist(a,z)=dist(z,a)=25

Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, dist(af,db)=dist(a,d)+dist(f,b)=3+4=7, and dist(bear,roar)=16+10+0+0=26.

Limak gives you a nice string s and an integerk. He challenges you to find any nice strings' that dist(s,s’)=k . 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 andk (1 ≤ n ≤ 10⁵,0 ≤ k ≤ 10⁶).

The second line contains a string s of lengthn, 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 dist(s,s’)=k .

Sample Input

Input

4 26
bear

Output

roar

Input

2 7
af

Output

db

Input

3 1000
hey

Output

-1

题意：两个字符串比对，两个字符串相同的位置上的字符的Ascall码值的差值的绝对值的和。现在已知其中的一个字符串和差值总和m，求另一个字符串（有多个输出其中任意一个字符串）。

思路：从第一个字符开始找到与他差值最大的字符’a‘或'z,并记录下来，存在ans数组中,然后从总和m中减去当前字符的最大差值path，依次往下进行。当m<path时，让当前字符加上m即可得到当前位待求字符，剩下的待求字符和已知字符串剩下的字符相同即可。若字符串全部找到后，m仍然大于0，则说明不可能找到待求字符串与已知字符串的差值和等于m.

My  solution：

/*2016.4.22*/

#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
char c[100100],ans[100100];
int path,n,m;
int solve(char b)//计算当前字符所能得到的最大差值 
{
	int x,y;
	x=b-'a';//与字符‘a’比较得到差值 
	y='z'-b;//与字符‘z’比较得到差值 
	if(y<0)
	y=-1*y;
	if(y>x)//取最大差值的字符（标记）进行返回，path则记录最大差值 
	{
		path=y;
		return 1;
	}
	else
	{
		path=x;
		return 0;
	}
}
int main()
{
	int i,j,k,h;
	while(scanf("%d%d",&n,&m)==2)
	{
		scanf("%s",c);
		for(i=0;i<n;i++)
		{
			path=0;
			k=solve(c[i]);//k用来标识最大差值字符是a,还是z
			if(path<=m)//当前差值小于差值和 
			{
				if(k==1)//最大差值字符是z
				{
					ans[i]='z';//记录该位的字符 
					m-=path;//更新剩余差值和 
				}
				else//最大差值字符是a
				{
					ans[i]='a';//记录该位的字符
					m-=path;
				}
			}
			else//当前最大差值小于差值和
			{
				if(k==1)//最大差值是由字符z得到，现在差值用不完 
				{
					ans[i]=c[i]+m;//即可得待求字符 
					m=0;//更新m值 
				} 
				else//最大差值是由字符a得到
				{
					ans[i]=c[i]-m;
					m=0;
				}
				i++;
				break;
			}	
		}
		for(;i<n;i++)//当i<n时,差值和已经匹配完了 ,剩余差值和为0，即剩余的待求字符与剩余的已知字符相同 
		ans[i]=c[i];
		if(m>0)//无法找到合适的字符串 
		printf("-1\n"); 
		else
		{
			ans[n]='\0';//字符串最后一位应为'\0' 
		     printf("%s\n",ans);
		}
		
	}
	return 0;
}