PROBLEM LINK:

Practice
Contest

Author:Dmytro Berezin
Tester:Shiplu Hawlader and Mahbubul Hasan
Editorialist:Lalit Kundu

DIFFICULTY:

SIMPLE-EASY

PREREQUISITES:

Precomputation

PROBLEM:

Given N(<=10^5 ) digits (0<=each digit<=9), and M(<=10^5) queries, each consisting of one x(1<=x<=n). Print B1 - B2.
B1 = sum of all ( a_x - a_y ) such that x > y and a_x> a_y.
B2 = sum of all ( a_x - a_y ) such that x > y and a_x< a_y.

EXPLANATION:

We define a new array A where A[i][j] is the number of times digit i occurs in a₁,a₂,...a_j.
Code:

for i=0 to 9:    // calculate the row A[i] for each i=0 to 9.    
    for j=1 to n:   
        A[i][j]=A[i][j-1];   
        if a[j]==i: A[i][j]++;   // if a[j]==i, current value one more than previous.

For each input query p, let q=a_p, we do:

B1=0    
for i=0 to q-1:  // sum of those where ax > ay    
    B1 += A[i][p-1]*(q-i)   // difference is (x-i)

B2=0    
for i=q+1 to 9:  // sum of those where ax < ay    
    B2 += A[i][p-1]*(q-i)    // difference is (x-i)

print B1-B2

Time Complexity: O( N * 10 ) [ Computation of A ] + M * 10 ( For M queries.)

AUTHOR'S AND TESTER'S SOLUTIONS:

To be updated soon.