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!
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.
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.
4 1 1 1 1
4
5 1 3 2 3 4
2
6 1 3 3 1 3 2
1
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.
就是給定N個數,然後按大小順序排序後,找前三個數,問有多少種不同的組合。
如果是 6
1 1 2 2 3 3
的話前三個數,是1,1,2然後2有兩個可以替換,所以是2種情況。以此類推
如果都是一個數的話,那麼就是排列組合。
要用LLD才行
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
int number[100005];
int main()
{
int n;
long long int all = 1;
scanf("%d",&n);
for(int i = 0; i < n; i++)
{
scanf("%d",&number[i]);
}
sort(number,number+n);
int count = 1,index = 1;
while(number[index] == number[0] && index < n)
{
index++;
count++;
}
if(count >= 3)
{
all=all*index;
all=all*(index-1);
all=all*(index-2);
all=all/6;
}
else if(count == 2)
{
for(int i = 3; i < n; i++)
if(number[i] == number[2])
all++;
}
else
{
int k = 3;
while(k<n&&(number[k]==number[2]))
k++;
k--;
if(number[1] == number[2])
{
all=k;
all=all*(k-1);
all=all/2;
}
else
{
all=all+(k-2);
}
}
printf("%lld\n",all);
return 0;
}