The Magical Stone Solution Codechef

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The Magical Stone Solution Codechef

Initially, there is a magical stone of mass 2N lying at the origin of the number line. For the next N seconds, the following event happens:

Let us define the decomposition of a magical stone as follows: If there is a magical stone of mass M>1 lying at coordinate X, then it decomposes into two magical stones, each of mass M2 lying at the coordinates X−1 and X+1 respectively. The original stone of mass M gets destroyed in the process.
Each second, all the magical stones undergo decomposition simultaneously.
Note that there can be more than one stone at any coordinate X.

Given a range [L,R], find out the number of stones present at each of the coordinates in the range [L,R]. As the number of stones can be very large, output them modulo (109+7).

Input Format
The first line contains a single integer T – the number of test cases. Then the test cases follow.
The first and only line of each test case contains three integers N, L and R, as described in the problem statement.

Output Format
For each testcase, output in a single line a total of (R−L+1) space-separated integers. The ith integer will denote the number of stones present at X=(L+i−1) coordinate. As the number of stones can be very large, output them modulo (109+7).

Constraints
1≤T≤100
1≤N≤106
−N≤L≤R≤N
Sum of (R−L+1) over all the test cases doesn’t exceed 105.

Sample Input 1
3
2 -2 2
2 0 2
150000 48 48

Sample Output 1
1 0 2 0 1
2 0 1
122830846

Explanation

Test case 1: Let us look at the number of stones for x=−2 to x=2 as the time progresses:

t=0: {0,0,1,0,0}
t=1: {0,1,0,1,0}
t=2: {1,0,2,0,1}
We have to output the number of stones at x=−2 to x=2, which is {1,0,2,0,1}.

Test case 2: Similar to first test case, We have to output the number of stones at x=0 to x=2, which is {2,0,1}.

SOLUTION

Program: The Magical Stone Solution in Python

x = int(1e9 + 7)
factorial = [1]
for i in range(1, 1000001):
    factorial.append((factorial[-1]*i)%x)

for _ in range(int(input())):
    n, l, r = map(int, input().split())
    while l <= r:
        if (l+n)%2 == 0:
            c = (l+n)//2
            val = factorial[n] * pow(factorial[c], x-2, x) * pow(factorial[n-c], x-2, x)
            print(val%x, end=' ')
        else:
            print(0, end=' ')
        l += 1
    print()

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