hdu 5147 Sequence II (树状数组 求逆序数)

题目链接

Sequence II

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 331    Accepted Submission(s): 151

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

题意：

很久很久以前，有一个长度为n的数列A，数列中的每个数都不小于1且不大于n，且数列中不存在两个相同的数.
请统计有多少四元组(a,b,c,d)满足：
1. 1≤a<b<c<d≤n
2. Aa<Ab
3. Ac<Ad

分析：
我的做法是把四元组分解成二元组来处理，分解的方法就是枚举把数组依次分为两部分，然后对每部分都用树状数组求逆序数，结果相乘就是满足条件的四元组的个数。
树状数组求逆序数的做法是，因为知道数列里的数是1-n，所以可以 以个数为c[]数组的元素，值为下标，通过求和来 求大于当前数 或者 小于当前数的个数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <algorithm>
#define LL __int64
const int maxn = 1e5 + 10;
using namespace std;
LL a[maxn], c[maxn], n, f[maxn];

int lowbit(int x)
{
    return x&(-x);
}
void add(int x,int d)
{
    while(x <= n)
    {
        c[x] += d;
        x +=lowbit(x);
    }
}
LL sum(int x)
{
    LL ret = 0;
    while(x > 0)
    {
        ret += c[x];
        x -= lowbit(x);
    }
    return ret;
}

int main()
{
    int t;
    LL i, ans, tmp;
    scanf("%d", &t);
    while(t--)
    {
        ans = 0;
        scanf("%I64d", &n);

        memset(f, 0, sizeof(f));
        memset(c, 0, sizeof(c));
        for(i = 1; i <= n; i++)
        {
            scanf("%I64d", &a[i]);
            add(a[i], 1);
            f[i] = f[i-1] + sum(a[i]-1);
        }

        memset(c, 0, sizeof(c));
        tmp = 0;
        for(i = n; i >= 1; i--)
        {
            tmp = sum(n) - sum(a[i]);
            add(a[i], 1);
            ans += tmp*f[i-1];
        }
        printf("%I64d
", ans);
    }
    return 0;
}