Codeforces Round #789 (Div. 2) C- Tokitsukaze and Strange Inequality(二维前缀和)

思路：见代码注释

#pragma GCC optimize(2)
#pragma GCC optimize(1)
#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
const ull base=131;
#define MAX 5009
#define PI 3.141592653589793
using namespace std;
inline void solve(){
    int len;cin>>len;
    vector<int> a(len+1);
    for(int i=1;i<=len;i++) cin>>a[i];
    vector<vector<int>> dp(len+1,vector<int>(len+1,0));
    for (int i = 1; i <= len; ++i) {
      for (int j = 1; j <= len; ++j) if (a[j] < i) dp[i][j] = 1;//如果小于，设这点为一个逆序对
      for (int j = 1; j <= len; ++j) dp[i][j] += dp[i][j - 1];//前缀和求出在1-j内逆序对个数
  }
    ll ans=0;
    for(int b=1;b<=len;b++)
        for(int c=b+1;c<=len;c++){
            ans+=(dp[a[c]][b-1]*(dp[a[b]][len]-dp[a[b]][c]));//枚举c下标的数字在1-b(即a的范围数字)的逆序对*(枚举b下标状态下的数字在总长范围内-枚举b状态下的数字在1-c内(即所有不是b范围的)
        }
    cout<<ans<<"\n";
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);std::cout.tie(0);
    int sum;cin>>sum;
    while(sum--){
       solve(); 
    }
}