Title source :  HackerRank
Base time limit :1  second Space restriction :131072 KB The score is : 5  difficulty :1 Level algorithm problem
  Collection
  Focus on
There is such a program ,fun I'll do an array of integers A Is evaluated , among Floor Indicates rounding down :
 
fun(A)
    sum = 0
    for i = 1 to A.length
        for j = i+1 to A.length
            sum = sum + Floor((A[i]+A[j])/(A[i]*A[j])) 
    return sum
 
Give the array A, It's up to you to calculate fun(A) Result . for example :A = {1, 4, 1},fun(A) = [5/4] + [2/1] + [5/4] = 1 + 2 + 1 = 4.
Input
 The first 1 That's ok :1 Number N, Represents an array A The length of (1 <= N <= 100000).
The first 2 - N + 1 That's ok : Each row 1 Number A[i](1 <= A[i] <= 10^9).
Output
 Output fun(A) Calculated results of .
Input Example
3
1 4 1
Output Example
4

 The laws of mathematics , Only 1 and 2 The amount of money has an impact on the final result /
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
typedef long long LL;
#define M 1000000007
/*
1 The effect of ,[(1+n)/n]
n=1 sum+=2 n>1 sum+=1
2 n =1,sum+=1,n=2,sum+=1,n>2,sum+=0;
3 [(3+n)/3*n] 1_1 2_0 3_0 4_0
4 [(4+n)/4*n] 1_1 2_0 3_0 4_0
Only 1
----
set up n1,n2,n-n1-n1 Respectively 1 The number of ,2 The number of , Other numbers
The final result is equal to
2*(n1*(n1-1)/2) //1 and 1 Between
+ n2*(n2-1)/2
+ n1*(n-n1-n2)
*/
int main()
{
vector<LL> v;
LL n,t,cnt=,num=;
scanf("%lld",&n);
for(LL i=;i<n;i++)
{
scanf("%lld",&t);
if(t==)
cnt+=n-;
else if(t==)
num++;
}
printf("%lld\n",cnt+num*(num-)/);
return ;
}

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