HDU 5147 Sequence II ( 树状数组 )

Sequence II

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30    Accepted Submission(s): 16

Problem Description

Long long ago, there is a sequence A with length n. All numbers in this sequence is no smaller than 1 and no bigger than n, and all numbers are different in this sequence.
Please calculate how many quad (a,b,c,d) satisfy:
1. 1≤a<b<c<d≤n
2. Aa<Ab
3. Ac<Ad

 

 

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case begins with a line contains an integer n.
The next line follows n integers A1,A2,…,An.

[Technical Specification]
1 <= T <= 100
1 <= n <= 50000
1 <= Ai <= n

 

 

Output

For each case output one line contains a integer,the number of quad.

 

 

Sample Input

1

5

1 3 2 4 5

 

 

Sample Output

4

 

 

BC#23的一条算是简单题。

一开始看错题 ， 以为是  （ a < b < c < d ）Aa < Ac , Ab < Ad 

一直想不出来。

看清题之后 ， 变简单了很多 。

枚举（1~n）作为c位置。

求前面 a < b < c 符合 Aa < Ab 的个数再乘上前面大于Ac的数的个数（即未插入BIT数）。 

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
using namespace std;
const int N = 50010;
const double PI = acos(-1.0);
const double eps = 1e-6;
typedef long long LL;
typedef pair<double,double> pii;
#define X first
#define Y second

int n , A[N] ;
LL cnt[N] , c[N];

void init() { memset( c , 0 ,sizeof c ); }
int lowbit( int x ) { return x&-x; }
void update(  int pos , int key ) { while( pos < n + 10 ) { c[pos] += key ;pos += lowbit(pos);}}
LL query( int pos ) { LL res = 0 ; while( pos > 0 ) {res += c[pos]; pos -= lowbit(pos); } return res ; }

void Run() {
    init();
    scanf("%d",&n);
    for( int i = 1 ; i <=n ; ++i ){
        scanf("%d",&A[i]);
    }
    LL res = 0 ;
    for( int i = 1 ; i <= n ; ++i ) {
        LL low = query( A[i] - 1 ) , up = i - 1 - low;
        cnt[i] = cnt[i-1] + low ;
        update( A[i] , 1 ) ;
        res += cnt[i-1] * ( n - A[i] - up ) ;
    }
    printf("%I64d\n",res);
}
int main() {
//    freopen("in.txt","r",stdin);
    int _ , cas = 1 ;
    scanf("%d",&_); while( _-- ) {
        Run();
    }
}