Acm Marina and Vasya的题解

最新推荐文章于 2019-03-24 23:08:00 发布

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最新推荐文章于 2019-03-24 23:08:00 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/irrelevant/p/5514483.html

本文介绍了一个字符串匹配的问题，任务是找出两个长度相同的字符串中，存在第三个字符串，在恰好t个字符上与前两者不同。通过分析给出的字符串，文章提供了一种算法实现思路及代码示例。

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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

output

-1

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

题意不难理解

也没有什么坑点就是容易搞错求得不同的位置字母转换以下就成了求n-t个相同的位置字母了首先求出两行字母串本来就相同的位置字母的数量根据这个和要求分为三种情况（其实我不知道是不是可以少分点当时就是这么想的我也难得去想怎么简化）下面代码都有哈

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;

const int MAXN=100010;
char ans[MAXN],bns[MAXN];

int main(){
    
    int num,count;
    int n,t,k1,k2;
    scanf("%d%d",&n,&t);
    scanf("%s%s",ans,bns);
    num=0;
    count=n-t;
    for(int i=0;i<n;i++){
        
        if(ans[i]==bns[i]) num++;
        
    }
    if(num==count){
        
        for(int i=0;i<n;i++){
            
            if(ans[i]==bns[i]) printf("%c",ans[i]);
            else{
                
                if(ans[i]!='a'&&bns[i]!='a') printf("a");
                else if(ans[i]!='b'&&bns[i]!='b') printf("b");
                else if(ans[i]!='c'&&bns[i]!='c') printf("c");
                
            }
            
        }
        printf("\n");
        
    }
    else if(num<count){
        
        if(count-num>(n-num)/2) printf("-1\n");
        else{
            
            k1=k2=0;
            for(int i=0;i<n;i++){
                
                if(ans[i]==bns[i]) printf("%c",ans[i]);
                else{
                    
                    if(k1!=count-num){
                        
                        printf("%c",ans[i]);
                        k1++;
                        
                    }
                    else{
                        
                        if(k2!=count-num){
                            
                            printf("%c",bns[i]);
                            k2++;
                            
                        }
                        else{
                            
                            if(ans[i]!='a'&&bns[i]!='a') printf("a");
                            else if(ans[i]!='b'&&bns[i]!='b') printf("b");
                            else if(ans[i]!='c'&&bns[i]!='c') printf("c");
                            
                        }
                        
                    }
                
                }
                
            }
            printf("\n");
            
        }
        
    }
    else if(num>count){
        
        k1=0;
        for(int i=0;i<n;i++){
            
            
            if(ans[i]==bns[i]){
                
                if(k1!=num-count){
                    
                    if(ans[i]!='a'&&bns[i]!='a') printf("a");
                    else if(ans[i]!='b'&&bns[i]!='b') printf("b");
                    else if(ans[i]!='c'&&bns[i]!='c') printf("c");
                    k1++;
                    
                }
                else{
                    
                    printf("%c",ans[i]);
                    
                }
                
            }
            else{
                
                if(ans[i]!='a'&&bns[i]!='a') printf("a");
                else if(ans[i]!='b'&&bns[i]!='b') printf("b");
                else if(ans[i]!='c'&&bns[i]!='c') printf("c");
                
            }
            
        }
        printf("\n");
        
    }
    
}