Educational Codeforces Round 23 B. Makes And The Product
2 seconds
256 megabytes
standard input
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!
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.
题意:
给一串数字,在其中选三个数字,使得这三个数字的乘积最小,问有多少个这样的组合
思路:
直接排序乘积最小那么这三个数字必定是最小的三个,又因为数字可能相同,会产成不同结果的情况一共有三种:
1.三个数字都相同 。
2.第一个和第二个不同,第二个与第三个相同。
3.第二个和第三个不同。
实现代码:
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
#define ll long long
map<ll,ll>mp;
int main()
{
ll m,i,a[];
cin>>m;
for(i=;i<m;i++){
cin>>a[i];
mp[a[i]]++;}
sort(a,a+m);
ll sum = ;
if(a[]==a[]&&a[]==a[]){
ll num = mp[a[]];
sum = (num*(num-)*(num-))/;
}
if(a[]!=a[]){
ll num = mp[a[]];
sum = num;
}
if(a[]!=a[]&&a[]==a[]){
ll num = mp[a[]];
// num = 99999;
sum = (num*(num-))/;
} cout<<sum<<endl;
}
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