Codeforces Round #324 (Div. 2) 584C. Marina and Vasya

                             

               解题报告： Codeforces Round #324 (Div. 2)

C. Marina and Vasya

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.

More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print  - 1.

Input

The first line contains two integers n and t (1 ≤ n ≤ 105, 0 ≤ t ≤ n).

The second line contains string s1 of length n, consisting of lowercase English letters.

The third line contain string s2 of length n, consisting of lowercase English letters.

Output

Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1.

Sample test(s)

input

3 2
abc
xyc

output

ayd

input

1 0
c
b

output

-1

题意：长度为n的字符串s1,s2,s3。f(s1, s3) = f(s2, s3) = t，

s1,s3有t个字符不同，s2,s3有t个字符不同。给你 n,t,s1,s2求s3。

如果没有输出-1。

分析：s1,s2相同的sum个  如果 n-t-sum>(n-sum)/2  

不存在这样的s3 ，-1。

#include 
    
     
#include 
     
      
#include 
      
       
using namespace std;
const int N=100005;
char s1[N],s2[N],s3[N];
int b[N];
int main()
{
    int t,n,i,a,sum=0;
    scanf("%d%d",&n,&t);
    scanf("%s%s",s1,s2);
    for(i=0; i
       
        (n-sum)/2) printf("-1\n");
    else
    {
        if(sum>=n-t)
        {
            int k=n-t;
            for(i=0; i
        
         0)
                {
                    if(b[i]==0)
                    {
                        if((s1[i]-'a'+1)%26==s2[i]-'a') s3[i]=(s1[i]-'a'+2)%26+'a';
                        else s3[i]=(s1[i]-'a'+1)%26+'a';
                    }

                    else
                    {
                        s3[i]=s1[i];
                        k--;
                    }
                }
                else
                {
                    if((s1[i]-'a'+1)%26==s2[i]-'a') s3[i]=(s1[i]-'a'+2)%26+'a';
                    else s3[i]=(s1[i]-'a'+1)%26+'a';
                }
            }
        }
        else
        {
            int ss1=n-t-sum,ss2=n-t-sum;
            for(i=0; i
         
          0) { s3[i]=s1[i]; ss1--; } else if(ss2>0) { s3[i]=s2[i]; ss2--; } else { if((s1[i]-'a'+1)%26==s2[i]-'a') s3[i]=(s1[i]-'a'+2)%26+'a'; else s3[i]=(s1[i]-'a'+1)%26+'a'; } } } } printf("%s\n",s3); } return 0; } 
         
        
       
      
     
    

                             

               解题报告： Codeforces Round #324 (Div. 2)