【Codeforces 817B. Makes And The Product】

B. Makes And The Product
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

题意 ： 找出最小的三个数有多少可能结果

a[3] != a[2],
a[3] == a[2] && a[2] != a[1]
a[3] == a[2] == a[1]
三种情况

AC代码：

#include<cstdio>#include<map>#include<algorithm>using namespace std;const int MAX = 1e5 + 10;typedef long long LL;LL a[MAX];map <LL,LL> m;int main(){    int n;    scanf("%d",&n);    for(int i = 1; i <= n; i++)        scanf("%lld",&a[i]),m[a[i]]++;    sort(a + 1,a + 1 + n);    if(a[3] != a[2]) printf("%lld\n",m[a[3]]);    else if(a[3] == a[2] && a[2] != a[1]) printf("%lld\n",m[a[3]] * (m[a[3]] - 1) / 2);    else printf("%lld\n",m[a[1]] * (m[a[1]] - 1) * (m[a[1]] - 2) / 6);    return 0;}