Codeforces Round #324 (Div. 2)C

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;}