Codeforces 628 C. Bear and String Distance【贪心】

C. Bear and String Distance time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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 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 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 .

Examples input 4 26 bear output roar input 2 7 af output db input 3 1000 hey output -1

题目大意：

给你一个长度为N的字符串，让你找到另外一个字符串长度也是N，使得各个位子上的字符差的和是k.

如果不存在输出-1.

思路：

每个位子进行贪心，从第一个位子向后扫，肯定我们希望前边的字符差尽可能的差的多一些，那么之后的字符是可以和原字符串相等的。

所以我们对每个位子进行枚举，只要k>0.那么我们此时就尽量将字符差拉大，并且控制不超过范围即可。

Ac代码：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; char a[500000]; char ans[500000]; int main() { int n,k; while(~scanf("%d%d",&n,&k)) { scanf("%s",a); for(int i=0;i<n;i++) { if(k>0) { int tmp=0; for(int j=1;j<=26;j++) { if(abs(j-(a[i]-'a'+1))>tmp) { tmp=abs(j-(a[i]-'a'+1)); ans[i]=j-1+'a'; } } if(k>=tmp)k-=tmp; else { for(int j=1;j<=26;j++) { if(abs(j-(a[i]-'a'+1))==k) { ans[i]=j-1+'a'; } } k=0; } } else ans[i]=a[i]; } if(k>0)printf("-1\n"); else { for(int i=0;i<n;i++)printf("%c",ans[i]); printf("\n"); } } }