C. Marina and Vasya
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.
Examples
input
3 2
abc
xyc
output
ayd
input
1 0
c
b
output
-1
思路比较简单，先把不同的换成相同的有几个
然后两个相等的直接落下来并且m–
其他的就把q串落下来m个，w也一样
如果个数不够就-1
思路好像
疯狂wa
模拟题神烦
#include<iostream>
#include<cmath>
#include<string>
#include<algorithm>
using namespace std;
string q, w, e;
int main()
{
int n, m;
cin >> n >> m;
m = n - m;
cin >> q >> w;
e = q;
int yy = 0;
for (int a = 0;a < n;a++)
{
if (q[a] == w[a] && m>0)
{
m--;
e[a] = q[a];
yy++;
}
else e[a] = '.';
}
if (yy + m + m > n)
{
cout << -1;
return 0;
}
int jj = 0;
int ee = m;
for (int a = 0;a < n;a++)
{
if (q[a] == w[a])continue;
jj++;
if (jj > m + ee)
{
for (int b = 'a';b <= 'z';b++)
{
if (b != q[a] && b != w[a])
{
e[a] = b;
break;
}
}
}
else if (jj > m)e[a] = w[a];
else e[a] = q[a];
}
for (int a = 0;a < n;a++)
{
for (int b = 'a';b <= 'z';b++)
{
if (e[a] == '.'&&b!=q[a]&&b!=w[a])
{
e[a] = b;
break;
}
}
}
cout << e;
}