CF 1372C Omkar and Baseball

题目：

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 across n sessions follow the identity permutation (ie. in the first game he scores 1 point, in the second game he scores 2 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 the 2 is in the same position.

Given a permutation of n integers, 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 exceed 10^18.

An array aa is a subarray of an array bb if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

 思路：

①全部都是正确位置   0

②从前后跑到第一个不是a[i] = i，从后往前跑到第一个不是a[i] = i，判断中间的是不是都是a[i] != i，如果是就是1，不是就是2

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cmath>
 
using namespace std;
 
#define ll long long
#define pb push_back
#define fi first
#define se second
 
const int N = 2e5 + 10;
int a[N];
 
void solve()
{      
    int T;
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        for(int i = 1; i <= n; ++i) cin >> a[i];
        int same = 0;
        for(int i = 1; i <= n; ++i){
            if(a[i] == i) same++;
        }
        if(same == n) cout << 0 << endl;
        else{
            int l, r;
            for(int i = 1; i <= n; ++i){
                if(a[i] == i) continue;
                l = i - 1;
                break;
            }
            for(int i = n; i >= 1; --i){
                if(a[i] == i) continue;
                r = i + 1;
                break;
            }
            same = 0;
            //cout << "l = " << l << "  " << " r = " << r << endl;
            for(int i = l + 1; i <= r - 1; ++i){
                if(a[i] == i) same++;
            }
            if(same == 0) cout << 1 << endl;
            else cout << 2 << endl;
        }
    }
}
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0); 
    solve();
 
    return 0;
}