HDU5792 World is Exploding（树状数组）2016 Multi-University Training Contest 5

World is Exploding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 183 Accepted Submission(s): 80

Problem Description

Given a sequence A with length n,count how many quadruple (a,b,c,d) satisfies: a≠b≠c≠d,1≤a< b≤n,1≤c< d≤n,Aa< Ab,Ac>Ad.

Input

The input consists of multiple test cases.

Each test case begin with an integer n in a single line.

The next line contains n integers A1,A2⋯An.

1≤n≤50000

0≤Ai≤1e9

Output

For each test case,output a line contains an integer.

Sample Input

4

2 4 1 3

4

1 2 3 4

Sample Output

1

0

Author

ZSTU

Source

2016 Multi-University Training Contest 5

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统计一个数列中某一个数左边有多少个比他大，比他小，右边有多少个比他大，比他小。然后就是一个排列组合的问题了，先求出总的对数，再扣除不符合的（可能有中间两个数相等的情况）。另外统计的时候用树状数组优化一下。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <vector>
#define MAX 50005
using namespace std;
int n;
struct BIT {
long long sum[MAX];
void init() {
memset(sum, 0, sizeof sum);
}
int lowbit(int x) {
return (x & (-x));
}
void update(int x, long long val) {
for(int i = x; i <= n; i += lowbit(i)) {
sum[i] += val;
}
}
long long query(long long x) {
long long ans = 0;
for(int i = x; i; i -= lowbit(i)) {
ans += sum[i];
}
return ans;
}
}bit1,bit2,bit3,bit4;

int main()
{
while(~scanf("%d",&n))
{
bit1.init();
bit2.init();
bit3.init();
bit4.init();
vector<int> list;
list.clear();
int a[MAX];
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
list.push_back(a[i]);
}
sort(list.begin(),list.end());
list.erase(unique(list.begin(),list.end()),list.end());
long long l1[MAX],l2[MAX],r1[MAX],r2[MAX];
long long ans1=0,ans2=0;
for(int i=1;i<=n;i++)
{
a[i]=lower_bound(list.begin(),list.end(),a[i])-list.begin()+1;
l1[i]=bit1.query(a[i]-1);
l2[i]=bit2.query(n-a[i]);
ans1+=l1[i];
ans2+=l2[i];
bit1.update(a[i], 1);
bit2.update(n-a[i]+1,1);
}
for(int i=n;i>=1;i--)
{
r1[i]=bit3.query(a[i]-1);
r2[i]=bit4.query(n-a[i]);
bit3.update(a[i], 1);
bit4.update(n-a[i]+1,1);
}
long long ans=ans1*ans2;
for(int i=1;i<=n;i++)
{
ans-=l1[i]*l2[i];
ans-=l1[i]*r1[i];
ans-=r2[i]*r1[i];
ans-=l2[i]*r2[i];
}
printf("%lld\n",ans);
}
return 0;
}