codeforces817b Makes And The Product

B. Makes And The Product

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

After returning from the army Makes received a gift — an array a consisting of n positive
integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (i,  j,  k) (i < j < k),
such that ai·aj·ak is
minimum possible, are there in the array? Help him with it!

Input

The first line of input contains a positive integer number n (3 ≤ n ≤ 105) —
the number of elements in array a. The second line contains n positive
integer numbers ai (1 ≤ ai ≤ 109) —
the elements of a given array.

Output

Print one number — the quantity of triples (i,  j,  k) such that i,  j and k are
pairwise distinct and ai·aj·ak is
minimum possible.

Examples

input

4
1 1 1 1

output

4

input

5
1 3 2 3 4

output

2

input

6
1 3 3 1 3 2

output

1

Note

In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.

In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3,
then the answer is 2.

In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices.

最小乘积必然是前3小的元素乘积

考虑第i个点的贡献 如果ai∈前三小的元素，那么他的贡献就是1-(i-1)中第二元素个数*(i+1)-n中第三元素个数+1-(i-1)第三元素个数*(i+1)-n第二元素个数

注意long long与特判第二元素==第三元素

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=300000+10;
int has[maxn];
int s[maxn],t[maxn];
int A[maxn],B[maxn];
int tot=0;
inline int bs(int x){
int l=1,r=tot;
while(l+1<r){
int mid=(l+r)>>1;
if(has[mid]>x)
r=mid-1;
else l=mid;
}
if(has[r]==x)
return r;
return l;
}
int main(){
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
int n;
scanf("%d",&n);
//n=100000;
for(int i=1;i<=n;i++){
scanf("%d",&A[i]);
//A[i]=1;
B[i]=A[i];
}
int x,y,z;
sort(B+1,B+n+1);
long long ans=0;
for(int i=1;i<=n;i++)
if(B[i]!=B[i-1])
has[++tot]=B[i];
x=bs(B[1]),y=bs(B[2]),z=bs(B[3]);
for(int i=1;i<=n;i++){
int u=bs(A[i]);
t[u]++;
}
for(int i=1;i<=n;i++){
int u=bs(A[i]);
t[u]--;
if(u==x){
if(y==z)
ans+=(long long)s[y]*t[z];
else ans+=(long long)s[y]*t[z]+(long long)s[z]*t[y];
}
else if(u==y){
if(x==z)
ans+=(long long)s[x]*t[z];
else ans+=(long long)s[x]*t[z]+(long long)s[z]*t[x];
}
else if(u==z){
if(y==x)
ans+=(long long)s[x]*t[y];
else ans+=(long long)s[x]*t[y]+(long long)s[y]*t[x];
}
s[u]++;
}
cout<<ans<<endl;
return 0;
}