題目：

Patrick likes to play baseball, but sometimes he will spend so many hours hitting home runs that his mind starts to get foggy! Patrick is sure that his scores acrossnsessions follow the identity permutation (ie. in the first game he scores1point, in the second game he scores2 points and so on). However, when he checks back to his record, he sees that all the numbers are mixed up!

Define a special exchange as the following: choose any subarray of the scores and permute elements such that no element of subarray gets to the same position as it was before the exchange. For example, performing a special exchange on[1,2,3]can yield[3,1,2]but it cannot yield[3,2,1]since the2is in the same position.

Given a permutation ofnintegers, please help Patrick find the minimum number of special exchanges needed to make the permutation sorted! It can be proved that under given constraints this number doesn't exceed10^18.

An arrayaais a subarray of an arraybbif acan be obtained from bby deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

思路：

①全部都是正確位置 0

②從前後跑到第一個不是a[i] = i，從後往前跑到第一個不是a[i] = i，判斷中間的是不是都是a[i] != i，如果是就是1，不是就是2

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <string>
 6 #include <vector>
 7 #include <cmath>
 8  
 9 using namespace std;
10  
11 #define ll long long
12 #define pb push_back
13 #define fi first
14 #define se second
15  
16 const int N = 2e5 + 10;
17 int a[N];
18  
19 void solve()
20 {      
21     int T;
22     cin >> T;
23     while(T--){
24         int n;
25         cin >> n;
26         for(int i = 1; i <= n; ++i) cin >> a[i];
27         int same = 0;
28         for(int i = 1; i <= n; ++i){
29             if(a[i] == i) same++;
30         }
31         if(same == n) cout << 0 << endl;
32         else{
33             int l, r;
34             for(int i = 1; i <= n; ++i){
35                 if(a[i] == i) continue;
36                 l = i - 1;
37                 break;
38             }
39             for(int i = n; i >= 1; --i){
40                 if(a[i] == i) continue;
41                 r = i + 1;
42                 break;
43             }
44             same = 0;
45             //cout << "l = " << l << "  " << " r = " << r << endl;
46             for(int i = l + 1; i <= r - 1; ++i){
47                 if(a[i] == i) same++;
48             }
49             if(same == 0) cout << 1 << endl;
50             else cout << 2 << endl;
51         }
52     }
53 }
54  
55 int main()
56 {
57     ios::sync_with_stdio(false);
58     cin.tie(0);
59     cout.tie(0); 
60     solve();
61  
62     return 0;
63 }