codeforces 658D. Bear and Polynomials

D. Bear and Polynomials

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Input
The first line contains two integers n and
k (1 ≤ n ≤ 200 000, 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.

Output
Print the number of ways to change one coefficient to get a valid polynomial
Q that Q(2) = 0.

Examples

Input

3 1000000000
10 -9 -3 5

Output

3

Input

3 12
10 -9 -3 5

Output

2

Input

2 20
14 -7 19

Output

0

Note
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.

题意：改变一个系数使Q（2） = 0， 问有几种可能；

思路：直接求和肯定是不行的，太大了， 又因为它叫我的求的是Q（2），所以我们可以根据二进制吧所有值都放到最高位来计算。

如果进位是遇到奇数， 那么能改的系数位数只能在0 - 该位， 因为无论怎么改后面的，前面都不会为0；

遍历的时候从后往前，如果遇到值某一位的值很大，那么直接break即可，因为前面怎么改都不会为0；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<stack>
#include <queue>
#include <set>
using namespace std;
const int inf = 2147483647;
const double PI = acos(-1.0);
const int mod = 1000000007;
long long a[200005], b[200005];
int main()
{
long long n, k;
while (~scanf("%I64d%I64d", &n, &k))
{
for (long long i = 0; i <= n; ++i)
{
scanf("%I64d", &a[i]);
b[i] = a[i];
}
for (long long i = 0; i < n; ++i)
{
b[i + 1] += b[i] / 2;
b[i] %= 2;
}
long long x = 0;
for (long long i = 0; i <= n; ++i)
{
if (b[i])
{
x = i;
break;
}
}
long long ans = 0;
long long s = 0;

for (long long i = n; i >= 0; --i)
{
s = s * 2 + b[i];
if (abs(s) > 1e10)
break;
if (i <= x)
{
long long p = abs(s - a[i]);
if (p == 0 && i == n)
continue;
if (p <= k)
ans++;
}
}
printf("%I64d\n", ans);

}
}