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CodeForces - 628C Bear and String Distance （模拟）

2016-04-22 21:54 330 查看

CodeForces
- 628C

Bear and String Distance

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

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 useBufferedReader/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

db

Input

3 1000
hey

Output

-1

//题意：

定义dist(字母1， 字母2) = 两字母ascll之差。

给定一个串a和一个值k，找到另一个串b(长度相等)使得两串各个位置的dist()之和 == k。若得不到k输出-1#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN = 1e5 + 100;
char s[MAXN];
int main()
{
int n, k;
while(~scanf("%d%d", &n, &k))
{
scanf("%s", s);
for(int i = 0; i < n; i++)
{
int t = s[i] - 'a';
int d = 25 - t;
if(t >= d)
{
if(k >= t)
{
s[i] = 'a';
k -= t;
}
else
{
s[i] = s[i] - k;
k -= k;
break;
}
}
else
{
if(k >= d)
{
s[i] = 'z';
k -= d;
}
else
{
s[i] += k;
k -= k;
break;
}
}
}
if(k == 0)
printf("%s\n", s);
else
puts("-1");
}
return 0;
}

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