HDU 5147 Sequence II 树状数组

点击打开链接

Sequence II

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 577    Accepted Submission(s): 226

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

151 3 2 4 5

 

Sample Output

4

 

Source

BestCoder Round #23

 

官方题解：

要统计四元组的数量我们可以通过枚举c，然后统计区间[1,c-1]有多少二元组(a,b)满足a<b且Aa<Ab，以及统计出区间[c+1,n]有多少d满足Ac<Ad，根据乘法原理，把这两项乘起来就可以统计到答案里了.然后我们来处理子问题：区间[1,c-1]内有多少二元组(a,b).那么我们可以枚举b，然后统计区间[1,b-1]内有多少a满足Aa<Ab，那么这个可以通过用树状数组询问前缀和来实现.时间复杂度是O(nlogn).

//1684MS1840K#include<stdio.h>#include<string.h>#define ll __int64#define N 50007int a[N],pre[N],suf[N],C[N];//pre记录前缀，suf记录后缀int lowbit(int x){    return x&(-x);}void add(int pos,int val){    while(pos<N)    {        C[pos]+=val;        pos+=lowbit(pos);    }}int getsum(int x){    int result=0;    while(x>0)    {        result+=C[x];        x-=lowbit(x);    }    return result;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        memset(C,0,sizeof(C));        for(int i=1;i<=n;i++)//计算前缀的个数        {            pre[i]=getsum(a[i]);            add(a[i],1);        }        memset(C,0,sizeof(C));        for(int i=n;i>=1;i--)//计算后缀的个数        {            suf[i]=n-i-getsum(a[i]);//倒着输入，第i个数后面有n-i个数，再看看这n-i个数中是不是存在比a[i]小的            add(a[i],1);        }        ll ans=0,p=0;//p代表前缀的总个数        for(int i=1;i<n;i++)//枚举c的位置        {            ans+=suf[i]*p;            p+=pre[i];        }        printf("%I64d\n",ans);    }}