mirror of
https://github.com/Kiritow/OJ-Problems-Source.git
synced 2024-03-22 13:11:29 +08:00
118 lines
2.4 KiB
C++
118 lines
2.4 KiB
C++
#include <set>
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#include <map>
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#include <list>
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#include <cmath>
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#include <ctime>
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#include <deque>
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#include <queue>
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#include <stack>
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#include <cstdio>
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#include <string>
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#include <vector>
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#include <cctype>
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#include <cstring>
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#include <sstream>
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#include <fstream>
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#include <cstdlib>
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#include <cassert>
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#include <iostream>
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#include <algorithm>
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using namespace std;
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//Constant Declaration
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/*--------------------------*/
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//#define LL long long
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#define LL __int64
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const int M=100001;
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const int INF=1<<30;
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const double EPS = 1e-11;
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const double PI = acos(-1.0);
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/*--------------------------*/
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// some essential funtion
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/*----------------------------------*/
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void Swap(int &a,int &b){ int t=a;a=b;b=t; }
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int Max(int a,int b){ return a>b?a:b; }
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int Min(int a,int b){ return a<b?a:b; }
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int Gcd(int a,int b){ while(b){b ^= a ^=b ^= a %= b;} return a; }
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/*----------------------------------*/
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//for (i = 0; i < n; i++)
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/*----------------------------------*/
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LL c[M];
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int a[M];
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int left_lower[M], right_lower[M];//分别是,在第i个数左右边,比第i个数小的个数
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int n = 100000;
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int LowBit(int x)
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{
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return x&(-x);
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}
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int Sum(int k)
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{
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int sum = 0;
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while (k > 0)
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{
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sum += c[k];
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k -= LowBit(k);
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}
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return sum;
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}
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void Update(int k, int sc)
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{
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while (k <= 100000)
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{
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c[k] += sc;
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k += LowBit(k);
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}
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}
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int main()
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{
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//freopen("in.txt","r",stdin);
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//freopen("out.txt","w",stdout);
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int t, case1 = 0;
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scanf("%d", &t);
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int m;
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int i, j;
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int num;
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//scanf("%d%d", &n, &m);
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while (t--)
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{
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scanf("%d", &num);
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memset(c, 0, sizeof(c));
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memset(left_lower, 0, sizeof(left_lower));
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memset(right_lower, 0, sizeof(right_lower));
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for (i = 1; i <= num; i++)
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{
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scanf("%d", &a[i]);
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left_lower[i] += Sum(a[i] - 1);///Sum(a[i] - 1):插入该数前,区间1到a[i]-1的总个数
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Update(a[i], 1);
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}
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memset(c, 0, sizeof(c));
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for (i = num; i > 0; i--)//顺序插入
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{
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right_lower[i] += Sum(a[i] - 1);
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Update(a[i], 1);
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}
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LL ans = 0;
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for (i = 1; i <= num; i++)//逆序插入
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{
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ans += left_lower[i]*(num - i - right_lower[i]);//由于只能求比其小的个数,可以用i右边的总个数见减比他小的来求比他大的个数。
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ans += (i - 1 - left_lower[i])*right_lower[i];//
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}
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printf("%I64d\n", ans);
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}
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return 0;
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}
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