#include <iostream> #include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int num; cin >> num; // Reading input from STDIN while(num) { int n; cin>>n; int A[n]; int ma=0; long long sum=0; for(int i=0;i<n;++i) { cin>>A[i]; if(ma<A[i]) ma=A[i]; sum=sum+A[i]; } sort(A,A+n); long long res=0; for(int i=0;i<n1;++i) { for(int j=i+1;j<n;++j) { res=res+abs(A[i]A[j]); } } res=res*ma; cout<<res%1000000007<<"\n"; } }
Problem Description
Andrew is very fond of Maths.He has N boxes with him,in each box there is some value which represents the Strength of the Box.The ith box has strength A[i]. He wants to calculate the Overall Power of the all N Boxes.
Overall Power here means Sum of Absolute Difference of the strengths of the boxes(between each pair of boxes) multiplied by the Maximum strength among N boxes. Since the Overall Power could be a very large number,output the number modulus 10^9+7(1000000007).
Input
First line of the input contains the number of test cases T. It is followed by T test cases. Each test case has 2 lines. First line contains the number of boxes N. It is followed by a line containing N elements where ith element is the strength of Andrew’s ith box.
Output
For each test case, output a single number, which is the Overall Power for that testcase.
Constraints
1<=T<= 10
2<=N<=10^5
0<=A[i]<=10^9

Test Case 1
Input (stdin)
1 3 3 1 2
Expected Output
12

Test Case 2
Input (stdin)
2 2 1 2 5 4 5 3 1 2
Expected Output
2 100