题目

思路：

赛时代码（先求右起最长有序区间长度，再求左边最小值是否小于等于右边有序区间左端点的数）

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int maxn = 1e6 + 5;
int a[maxn];
int n;
map<int, int> pos;
void solve(){int i, j;cin >> n;bool mul = 0;for(i = 1; i <= n; i++){cin >> a[i];}int cnt = 0;for(i = n - 1; i >= 1; i--){if(a[i] <= a[i + 1]) cnt++;else break;}//cout << cnt << '\n';if(cnt == n - 1){cout << 0 << '\n';return;}int r = a[n - cnt];int l = 1e18;for(i = 1; i <= n - cnt - 1; i++){l = min(l, a[i]);}if(l <= r){cout << -1 << '\n';return;}cout << n - cnt - 1 << '\n';
}
signed main(){ios::sync_with_stdio(0);cin.tie(0);int T;cin >> T;while(T--){solve();}return 0;
}