Codeforces Round #296 (Div. 二) B(思维题)

Codeforces Round #296 (Div. 2) B(思维题)

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：给你两个长度为n的字符串S,T，交换S中的两个字符使两个字符串在相同位置上相同的字符最多，输出交换后仍不相同的字符的个数，以及交换的字符的位置，如果无论怎么交换都不行，则输出-1，-1.

思路：用一个二维数组a[i][j]来存储第一个字符串i和第二个字符串j发生不匹配的字母的位置，然后遍历询问是否存在a[][]。

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
char s[200010];
char t[200010];
int a[30][30];
int main()
{
    int n,i,j;
    scanf("%d",&n);
        scanf("%s %s",s,t);
        int cnt=0;
        memset(a,0,sizeof(a));
        for(i=0; i<n; i++) {
            if(s[i]!=t[i]) {
                a[s[i]-'a'][t[i]-'a']=i+1;
                cnt++;
            }
        }
        for(i=0;i<n;i++){
                if(s[i]!=t[i]&&a[t[i]-'a'][s[i]-'a']){
                    printf("%d\n%d %d\n",cnt-2,a[s[i]-'a'][t[i]-'a'],a[t[i]-'a'][s[i]-'a']);
                    return 0;
                }
        }
        for(i=0;i<n;i++){
            if(s[i]!=t[i]){
                for(j=0;j<26;j++){
                    if(a[t[i]-'a'][j]){
                        printf("%d\n%d %d\n",cnt-1,a[s[i]-'a'][t[i]-'a'],a[t[i]-'a'][j]);
                        return 0;
                    }
                }
            }
        }
     printf("%d\n-1 -1\n",cnt);
     return 0;
}

Codeforces Round #296 (Div. 二) B(思维题)

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