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 + ;
 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 = ;
     while(x > )
     {
         ret += c[x];
         x -= lowbit(x);
     }
     return ret;
 }

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

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

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