2040: Bear and Polynomials
Time Limit: 2 Sec Memory Limit: 256 MBSubmit: 0 Solved: 0
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Description
Limak is a little polar bear. He doesn't have many toys and thus he often plays with polynomials.
He considers a polynomial valid if its degree is n and its coefficients are integers not exceeding k by the absolute value. More formally:
Let a0,a1,...,an denote the coefficients, so 
. Then, a polynomial P(x) is valid if all the following conditions are satisfied:
- ai is integer for every i;
- |ai|≤k for every i;
- an≠0.
Limak has recently got a valid polynomial P with coefficients a0,a1,a2,...,an. He noticed that P(2)≠0 and he wants to change it. He is going to change one coefficient to get a valid polynomial Q of degree n that Q(2)=0. Count the number of ways to do so. You should count two ways as a distinct if coefficients of target polynoms differ.
The first line contains two integers n and k (1≤n≤200000,1≤k≤109)− the degree of the polynomial and the limit for absolute values of coefficients.
The second line contains n+1 integers a0,a1,...,an (|ai|≤k,an≠0)− describing a valid polynomial . It's guaranteed that P(2)≠0.
Print the number of ways to change one coefficient to get a valid polynomial Q that Q(2)=0.
3 1000000000
10 -9 -3 5
3
3 12
10 -9 -3 5
2
2 20
14 -7 19
0
In the first sample, we are given a polynomial P(x)=10-9x-3x2+5x3.
Limak can change one coefficient in three ways:
- He can set a0=-10. Then he would get Q(x)=-10-9x-3x2+5x3 and indeed Q(2)=-10-18-12+40=0.
- Or he can set a2=-8. Then Q(x)=10-9x-8x2+5x3 and indeed Q(2)=10-18-32+40=0.
- Or he can set a1=-19. Then Q(x)=10-19x-3x2+5x3 and indeed Q(2)=10-38-12+40=0.
In the second sample, we are given the same polynomial. This time though, k is equal to 12 instead of 109. Two first of ways listed above are still valid but in the third way we would get |a1|>k what is not allowed. Thus, the answer is 2 this time.