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

C. Marina and Vasya

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/584/problem/C

Description

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 Input

3 2
abc
xyc

Sample Output

ayd

HINT

题意

给你两个字符串，要求你构造出第三个字符串，使得第三个字符串和第一个字符串和第二个字符串的不同个数，都是k个

题解：

难点就是重叠的时候

只要过了下面数据就差不多了吧

3 2

abc

def

这个只要交替染色就好了

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <functional>
#include <queue>
#define N 100000+100
#define pf(x) ((x)*(x))
#define D(x) (1/x)
#define LL long long
using namespace std;
const double PI=3.141592653589793;char s1[N],s2[N],s3[N];
bool v[N];
int l,t;
int cnt;
bool s;void build()
{cnt=0;if(cnt==t)return;for(int i=0;i<l;i++)if(s1[i]==s2[i]){v[i]=true;s3[i]=s1[i];cnt++;if(cnt==t) return;}s=true;for(int i=0;i<l;i++)if(s1[i]!=s2[i]){v[i]=true;if(s) s3[i]=s1[i];else s3[i]=s2[i];s=!s;if(s) cnt++;if(cnt==t) return;}
}int main()
{cin>>l>>t;cin>>s1>>s2;
//    l=strlen(s1);t=l-t;build();s3[l]='\0';if(cnt<t) {cout<<"-1"<<endl;return 0;}
//    cout<<i<<endl;for(int i=0;i<l;i++)if(!v[i]){for(int j='a';j<='z';j++)if(s1[i]!=j && s2[i]!=j){s3[i]=j;break;}}cout<<s3<<endl;return 0;
}