Monocarp has got two strings ss and tt having equal length. Both strings consist of lowercase Latin letters "a" and "b".
Monocarp wants to make these two strings ss and tt equal to each other. He can do the following operation any number of times: choose an index pos1pos1 in the string ss, choose an index pos2pos2 in the string tt, and swap spos1spos1 with tpos2tpos2.
You have to determine the minimum number of operations Monocarp has to perform to make ss and tt equal, and print any optimal sequence of operations — or say that it is impossible to make these strings equal.
The first line contains one integer nn (1≤n≤2⋅105)(1≤n≤2⋅105) — the length of ss and tt.
The second line contains one string ss consisting of nn characters "a" and "b".
The third line contains one string tt consisting of nn characters "a" and "b".
If it is impossible to make these strings equal, print −1−1.
Otherwise, in the first line print kk — the minimum number of operations required to make the strings equal. In each of the next kk lines print two integers — the index in the string ss and the index in the string tt that should be used in the corresponding swap operation.
4 abab aabb
2 3 3 3 2
1 a b
-1
8 babbaabb abababaa
3 2 6 1 3 7 8
In the first example two operations are enough. For example, you can swap the third letter in ss with the third letter in tt. Then s=s= "abbb", t=t= "aaab". Then swap the third letter in ss and the second letter in tt. Then both ss and tt are equal to "abab".
In the second example it's impossible to make two strings equal.
题意：
给出s和t两个长度相等的字符串,问需要交换多少次s和t才能相等
输出次数和需要交换的下标
思路：
分两种情况进行讨论
ababab
babbab
ab=2,ba=1
所以ab+ba=3
即偶+奇=奇（直接输出-1即可）
可以交换成功的,进行分类讨论：
只有ab+ba的和为偶数才可以交换成功，
而出现偶数的可能只有偶+偶=偶或者奇+奇=偶
故此时可以进行下一步判断哪个奇数哪个偶数
abababb
babbaba
ab=2,ba=2
所以ab+ba=4
即偶+偶=偶
所以ab和ab之间交换,ba和ba之间进行交换
abababbab
babbababa
ab=3,ba=3
所以ab+ba=6
即奇+奇=偶
把多余的那一项拿出来单独进行交换即可
1 #include<stdio.h>
2 #include<string.h>
3 #include<algorithm>
4 #include<iostream>
5 using namespace std;
6 typedef long long ll;
7 const int N=2e5+20;
8 char s[N],t[N];
9 int ab[N],ba[N];//记录下标
10
11 int main()
12 {
13 int n;
14 while(~scanf("%d",&n))
15 {
16 scanf("%s%s",s,t);
17 // int ab=0,ba=0;
18 // for(int i=0;i<n;i++)
19 // {
20 // if(s[i]=='a'&&t[i]=='b')
21 // ab++;
22 // else if(s[i]=='b'&&t[i]=='a')
23 // ba++;
24 // else
25 // continue;
26 // }
27 // if(ab!=ba)
28 // printf("-1\n");//s=ab、t=ab这样子不对了就
29 int p=1,q=1;
30 for(int i=0; i<n; i++)
31 {
32 if(s[i]=='a'&&t[i]=='b')
33 ab[p++]=i+1;
34 else if(s[i]=='b'&&t[i]=='a')
35 ba[q++]=i+1;
36 else
37 continue;
38 }
39 p--,q--;
40 int sum=p+q;
41 if(sum%2)//如果和是奇数
42 {
43 printf("-1\n");
44 continue;
45 }
46 if(p%2&&q%2)
47 //即奇+奇=偶
48 //把多余的那一项拿出来单独进行交换即可
49 {
50 int kk=(p-1)/2+(q-1)/2+1;
51 printf("%d\n",kk+1);
52 for(int i=1;i<=p-1;i+=2)
53 printf("%d %d\n",ab[i],ab[i+1]);
54 for(int i=1;i<=q-1;i+=2)
55 printf("%d %d\n",ba[i],ba[i+1]);
56 printf("%d %d\n%d %d\n",ba[q],ba[q],ba[q],ab[p]);
57 }
58 else if(p%2==0&&q%2==0)
59 //即偶+偶=偶
60 //所以ab和ab之间交换,ba和ba之间进行交换
61 {
62 printf("%d\n",(p+q)/2);
63 for(int i=1;i<=p;i+=2)
64 printf("%d %d\n",ab[i],ab[i+1]);
65 for(int i=1;i<=q;i+=2)
66 printf("%d %d\n",ba[i],ba[i+1]);
67 }
68 }
69 return 0;
70 }