codeforces 628C Bear and String Distance 水

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

Description
定义两个小写字母之间的距离为这两个字母在字母表中的距离，如dis(a,z)=25,dis(a,c)=2，两个长度相同串的距离为这两个串对应位置字母距离之和。现给出一个长度为n的数字串s和一个距离k，问是否存在一个长度为n的串ss，使得dis(s,ss)=k，如果存在任意输出一解，如果不存在则输出-1
Input
第一行为两个整数n,k(1<=n<=10^5,1<=k<=10^6)，第二行为一长度为n的数字串
Output
如果存在一个长度为n的数字串ss，使得dis(s,ss)=k则输出ss，如果不存在则输出-1
Sample Input
4 26
bear
Sample Output
roar

#include <bits/stdc++.h>using namespace std;int main(){    int n;    int k;    cin>>n>>k;    string s;    cin>>s;    int len=s.length();    int ans=0;    for(int i=0;i<len;i++)    {        ans+=max((int)(s[i]-'a'),(int)('z'-s[i]));    }    if(ans<k)    {        printf("-1\n");        return 0;    }    else    {        for(int i=0;i<len;i++)        {            int res1='z'-s[i];            int res2=abs('a'-s[i]);            int flag=0;            int res=res1;            if(res1<res2)            {                res=res2;                flag=1;            }            if(res>k&&k!=0)            {                if(!flag)                printf("%c",s[i]+k);                else printf("%c",s[i]-k);                k=0;            }            else if(k==0)            {                printf("%c",s[i]);            }            else            {                if(!flag)                printf("z");                else printf("a");                k-=res;            }        }        cout<<endl;    }}