Codeforces 584C - Marina and Vasya (简单字符串构造

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 aand 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.

Examples

input

Copy

3 2
abc
xyc

output

Copy

ayd

input

Copy

1 0
c
b

output

Copy

-1

 题意:给出长度为n 和不同字符个数t 以及两个长度为n的字符串.f(a,b) 表示ai!=bi的和

求一个字符串f(s1,s3)==f(s2,s3)==t

思路：我们先构造出一个s3 它每个位置都不同于s1,s2，然后d1=d2=n-t表示需要相同的字符个数,然后直接替换s3的字符即可

最后判断 d1 与 d2是否为0即可输出答案

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pll;
#define eps 1e-6
#define pb push_back
const int INF = 0x3f3f3f3f;
const int maxn = 100000+5;
const int MOD = 1e9+7;
int f[maxn];
int main()
{
    int n,t;
    string a,b,s;
    cin>>n>>t;
    cin>>a>>b;
    for(int i=0; i<n; i++)
    {
        for(char c='a'; c<='z'; c++)
        {
            if(a[i]==c||b[i]==c) continue;
            s.push_back(c);
            break;
        }
    }
    int d1=n-t,d2=n-t;
    for(int i=0; i<n; i++)
    {
        if(d1==0) break;
        if(a[i]==b[i])
        {
            s[i]=a[i];
            d1--;
            d2--;
            f[i]=1;
        }
    }
    for(int i=0; i<n; i++)
    {
        if(f[i]) continue ;
        if(d1)
        {
            d1--;
            s[i]=a[i];
            f[i]=1;
        }
        else if(d2)
        {
            d2--;
            s[i]=b[i];
            f[i]=1;
        }
    }
    if(!d1&&!d2)
        cout<<s<<endl;
    else cout<<-1<<endl;
}

 总结：字符串构造题写得太少emm

 

 

 PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~