Codeforces Round #672 (Div. 2) A. Cubes Sorting (思维)_综合

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题意：有一个长度为n的数组，每次能够交换相邻俩数的位置，问能否在 $\frac{n*(n-1)}{2}-1$ 次操作中使得数组非递减

题解：只有当数组严格递减的时候，操作次数是 $\frac{n*(n-1)}{2}$ ，故判断数组是否严格递减即可

#include<bits/stdc++.h>#pragma GCC optimize(2)
#define rep(i,a,n) for (int i=a;i<=n;i++)//i为循环变量，a为初始值，n为界限值，递增
#define per(i,a,n) for (int i=a;i>=n;i--)//i为循环变量， a为初始值，n为界限值，递减。using namespace std;
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0)
#define ll long long
#define ull unsigned long long
#define ld long double
const int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<60;
const int mod=1e9+7;ll read()//快读
{
    ll w = 1, s = 0;char ch = getchar();while (ch < '0' || ch>'9') {
     if (ch == '-') w = -1; ch = getchar(); }while (ch >= '0' && ch <= '9') {
     s = s * 10 + ch - '0';    ch = getchar(); }return s * w;
}void solve(){
    int n,a[50010];cin>>n;rep(i,1,n) cin>>a[i];int f=0;rep(i,2,n){
    if(a[i]-a[i-1]<0){
    }else {
    f=1;break;}}if(f) cout<<"YES"<<endl;else cout<<"NO"<<endl;
}int main(){
    int _;cin>>_;while(_--){
    solve();}return 0;
}