B. Error Correct System
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ford Prefect got a job as a web developer for a small company that makes towels. His current work task is to create a search engine for the website of the company. During the development process, he needs to write a subroutine for comparing strings S and T of
equal length to be "similar". After a brief search on the Internet, he learned about the Hamming distance between two strings S and T of
the same length, which is defined as the number of positions in which S and T have
different characters. For example, the Hamming distance between words "permanent" and "pergament"
is two, as these words differ in the fourth and sixth letters.

Moreover, as he was searching for information, he also noticed that modern search engines have powerful mechanisms to correct errors in the request to improve the quality of search. Ford doesn't know much about human beings, so he assumed that the most common
mistake in a request is swapping two arbitrary letters of the string (not necessarily adjacent). Now he wants to write a function that determines which two letters should be swapped in string S,
so that the Hamming distance between a new string S and string T would
be as small as possible, or otherwise, determine that such a replacement cannot reduce the distance between the strings.

Help him do this!

Input

The first line contains integer n (1 ≤ n ≤ 200 000)
— the length of strings S and T.

The second line contains string S.

The third line contains string T.

Each of the lines only contains lowercase Latin letters.

Output

In the first line, print number x — the minimum possible Hamming distance between strings S and T if
you swap at most one pair of letters in S.

In the second line, either print the indexes i and j (1 ≤ i, j ≤ n, i ≠ j),
if reaching the minimum possible distance is possible by swapping letters on positions i and j,
or print "-1 -1", if it is not necessary to swap characters.

If there are multiple possible answers, print any of them.

Sample test(s)
input
9
pergament
permanent
output
1
4 6
input
6
wookie
cookie
output
1
-1 -1
input
4
petr
egor
output
2
1 2
input
6
double
bundle
output
2
4 1
Note

In the second test it is acceptable to print i = 2, j = 3.

题意:交换S或者T中的两个字符。使得两串的差异度最小。有三种情况,一种是交换后正好两个位置都相应同样,一种是交换后仅仅有一个位置同样,还有就是交换也不能满足相应同样。

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#define N 2222
using namespace std; char s[200009],t[200009];
int n;
int a[N][N]; int main()
{
while(~scanf("%d",&n))
{
scanf("%s %s",s,t); int num=0;
memset(a,0,sizeof a); for(int i=0;i<n;i++)
if(s[i]!=t[i])
{
int x=s[i]-'a';
int y=t[i]-'a';
num++;
a[x][y]=i+1;
} for(int i=0;i<26;i++)//交换后两位置都相应同样了
for(int j=0;j<26;j++)
if(a[i][j] && a[j][i])
{
cout<<num-2<<endl;
cout<<a[i][j]<<" "<<a[j][i]<<endl;
return 0;
} for(int i=0;i<26;i++)//交换后仅仅有一个位置相应同样
for(int j=0;j<26;j++)
if(a[i][j])
{
for(int k=0;k<26;k++)
if(a[j][k])
{
cout<<num-1<<endl;
cout<<a[i][j]<<" "<<a[j][k]<<endl;
return 0;
}
} cout<<num<<endl;
cout<<-1<<" "<<-1<<endl;//无法通过交换降低差异度 }
return 0;
}

Codeforces Round #296 (Div. 2) B. Error Correct System的更多相关文章

  1. 字符串处理 Codeforces Round #296 (Div. 2) B. Error Correct System

    题目传送门 /* 无算法 三种可能:1.交换一对后正好都相同,此时-2 2.上面的情况不可能,交换一对后只有一个相同,此时-1 3.以上都不符合,则不交换,-1 -1 */ #include < ...

  2. Codeforces Round #296 (Div. 1) C. Data Center Drama 欧拉回路

    Codeforces Round #296 (Div. 1)C. Data Center Drama Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xx ...

  3. CodeForces Round #296 Div.2

    A. Playing with Paper 如果a是b的整数倍,那么将得到a/b个正方形,否则的话还会另外得到一个(b, a%b)的长方形. 时间复杂度和欧几里得算法一样. #include < ...

  4. Codeforces Round #296 (Div. 2B. Error Correct System

    Ford Prefect got a job as a web developer for a small company that makes towels. His current work ta ...

  5. Codeforces Round #296 (Div. 1) E. Triangles 3000

    http://codeforces.com/contest/528/problem/E 先来吐槽一下,一直没机会进div 1, 马力不如当年, 这场题目都不是非常难,div 2 四道题都是水题! 题目 ...

  6. 水题 Codeforces Round #296 (Div. 2) A. Playing with Paper

    题目传送门 /* 水题 a或b成倍的减 */ #include <cstdio> #include <iostream> #include <algorithm> ...

  7. Codeforces Round #296 (Div. 2) A. Playing with Paper

    A. Playing with Paper One day Vasya was sitting on a not so interesting Maths lesson and making an o ...

  8. Codeforces Round #296 (Div. 2) A B C D

    A:模拟辗转相除法时记录答案 B:3种情况:能降低2,能降低1.不能降低分别考虑清楚 C:利用一个set和一个multiset,把行列分开考虑.利用set自带的排序和查询.每次把对应的块拿出来分成两块 ...

  9. Codeforces Round #296 (Div. 1) B - Clique Problem

    B - Clique Problem 题目大意:给你坐标轴上n个点,每个点的权值为wi,两个点之间有边当且仅当 |xi - xj| >= wi + wj, 问你两两之间都有边的最大点集的大小. ...

随机推荐

  1. easyui框架Date日期类型以json形式显示到前台datagrid时,显示为[object Object]

    如下图,easyui当后台把时间数据返回转换成json然后加载在easyui的datagrid里面,显示为[object Object]      需要对时间格式添加格式的显示方法 /** * 时间格 ...

  2. Python-字符编码详解

    1. 字符编码简介 1.1. ASCII ASCII(American Standard Code for Information Interchange),是一种单字节的编码.计算机世界里一开始只有 ...

  3. OAuth2.0官方文档中文翻译

    http://page.renren.com/699032478/note/708597990 (一)背景知识 OAuth 2.0很可能是下一代的“用户验证和授权”标准,目前在国内还没有很靠谱的技术资 ...

  4. 基于CentOS的MySQL学习补充四--使用Shell批量从CSV文件里插入数据到数据表

    本文出处:http://blog.csdn.net/u012377333/article/details/47022699 从上面的几篇文章中,能够知道怎样使用Shell创建数据库.使用Shell创建 ...

  5. Oracle 10g 数据库的备份和还原

    一.备份数据库 1.在图形工具中,如sqldeveloper,pl/sqldeveloper用以下这句查找空表 select 'alter table '||table_name||' allocat ...

  6. ui-router(三)controller与template

    这篇就是在以前的基础上,把客户端angular.js 负责的部分整体串起来演示一下. 我们按照angular执行顺序来做前提准备: (1)Client 根目录下 index.html 首先加载angu ...

  7. Graphics View框架

    Qt4.2开始引入了Graphics View框架用来取代Qt3中的Canvas模块,并在很多地方作了改进,Graphics View框架实现了模型-视图结构的图形管理,能对大量图元进行管理,支持碰撞 ...

  8. Makefile学习之路5——通过函数增强功能

    通过函数能显著增强Makefile的功能.对于simple项目的Makefile,尽管使用了模式规则,但还是有一件比较麻烦的事情,就是要在Makefile中指明每一个项目源文件.下面介绍几个后期会使用 ...

  9. Codeforces461A Appleman and Toastman 贪心

    题目大意是Appleman每次将Toastman给他的Ni个数拆分成两部分后再还给Toastman,若Ni == 1则直接丢弃不拆分.而Toastman将每次获得的Mi个数累加起来作为分数,初始时To ...

  10. Linux kernel 之 uart 驱动解析

    uart 是一种非常之常见的总线,比如DEBUG信息输出,小数据量数据传输,485,以及蓝牙的控制,GPS,很多都是通过uart 进行数据传输并进行控制. 在Linux kernel 内部,uart ...