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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.
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.
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.
3 2
abc
xyc
ayd
1 0
c
b
1 #include<cstdio> 2 #include<vector> 3 #include<cmath> 4 #include<queue> 5 #include<map> 6 #include<cstring> 7 #include<algorithm> 8 using namespace std; 9 typedef long long ll; 10 typedef unsigned long long ull; 11 const int maxn=100005; 12 int main() 13 { 14 int t,n; 15 char s1[maxn],s2[maxn],s3[maxn]; 16 scanf("%d%d",&n,&t); 17 scanf("%s%s",s1,s2); 18 t=n-t; 19 int k1=0,k2=0; 20 for(int i=0;i<n;i++) 21 { 22 if(s1[i]==s2[i]&&(k1<t&&k2<t))s3[i]=s1[i],k1++,k2++; 23 else if(s1[i]!=s2[i]||k1==t) 24 for(int j=0;j<26;j++) 25 if(s1[i]!=‘a‘+j&&s2[i]!=‘a‘+j) 26 { 27 s3[i]=(char)(‘a‘+j); 28 break; 29 } 30 } 31 if(k1==t) 32 { 33 puts(s3); 34 return 0; 35 } 36 for(int i=0;i<n;i++) 37 { 38 if(s1[i]==s2[i])continue; 39 if(k1<t) 40 s3[i]=s1[i],k1++; 41 else if(k2<t) 42 s3[i]=s2[i],k2++; 43 else 44 for(int j=0;j<26;j++) 45 if(s1[i]!=‘a‘+j&&s2[i]!=‘a‘+j) 46 { 47 s3[i]=(char)(‘a‘+j); 48 break; 49 } 50 } 51 if(k1<t||k2<t)puts("-1"); 52 else puts(s3); 53 return 0; 54 }
-1
codeforces 584C Marina and Vasya
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原文地址：http://www.cnblogs.com/homura/p/4999324.html
