Digit Queries#12312

Digit Queries (Анализа на решението)

Problem Author:	Hasan Saleh (MathModel)
Prepared by:	Hasan Saleh (MathModel)
Edited by:	Hasan Saleh (MathModel)

Solution

Observe the pattern will be :

n - 1 times 999..9 1 time 8 n - 1 times 000...0 1 time 1

\underbrace{999..9}_{n-1 \text{ times}} \: \underbrace{8}_{1 \text{ time}} \: \underbrace{000...0}_{n-1 \text{ times}} \: \underbrace{1}_{1 \text{ time}}

This can be proven via expansion of $(1 0^{n} - 1)^{2}$ (10^n-1)^2 (Exercise left).

The index handling is easy from the pattern

$(1 \leq i < n)$ (1 \le i < n) then answer is $9$ .
$(i = n)$ (i=n) then answer is $8$ .
$(n < i < 2 n)$ (n<i<2n) then answer is $0$ .
$(i = 2 n)$ (i=2n) then answer is $1$ .
$(i > 2 n)$ (i>2n) then answer is $- 1$ .

for _ in range(int(input())):
    n,i=map(int,input().split())
    if(1<=i and i<=n-1):
        print(9)
    elif(i==n):
        print(8)
    elif(n+1<=i and i<2*n):
        print(0)
    elif(i==2*n):
        print(1)
    else:
        print(-1)

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