CodeForces 657B Bear and Polynomials

题目链接：http://codeforces.com/problemset/problem/657/B

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 exceedingk by the absolute value. More formally:

Let a₀, a₁, ..., a_n denote the coefficients, so. Then, a polynomialP(x) is valid if all the following conditions are satisfied:

• a_i is integer for everyi;
• |a_i| ≤ k for everyi;
• a_n ≠ 0.

Limak has recently got a valid polynomial P with coefficients a₀, a₁, a₂, ..., a_n. He noticed that P(2) ≠ 0 and he wants to change it. He is going to change one coefficient to get avalid polynomial Q of degreen 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 integersa₀, a₁, ..., a_n (|a_i| ≤ k, a_n ≠ 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 polynomialQ 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:

1. He can set a₀ =  - 10. Then he would getQ(x) =  - 10 - 9x - 3x² + 5x³ and indeedQ(2) =  - 10 - 18 - 12 + 40 = 0.
2. Or he can set a₂ =  - 8. ThenQ(x) = 10 - 9x - 8x² + 5x³ and indeedQ(2) = 10 - 18 - 32 + 40 = 0.
3. Or he can set a₁ =  - 19. ThenQ(x) = 10 - 19x - 3x² + 5x³ and indeedQ(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 of10⁹. 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 is2 this time.

思路：既然让x等于2，那就好办了。像处理二进制一样，将所有系数变换到最高项系数那里。然后从最高项系数开始，按权求和，一直到不大于第一个系数不为0的项再进行判断。为什么呢？因为如果在其之后，将它的系数往最高项移动，对低项没影响，而低项有不为0的项，这样是无解的。

直接附上AC代码：

#include 
using namespace std;
typedef long long ll;
const int maxn = 200005;
ll num[maxn], t[maxn];
int n, k;

int main(){
    #ifdef LOCAL
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> k;
    for (int i=0; i<=n; ++i){
        cin >> num[i];
        t[i] = num[i];
    }
    for (int i=1; i<=n; ++i){           // 低项系数高项移
        num[i] += num[i-1]/2ll;
        num[i-1] %= 2ll;
    }
    int flag = 0;
    for (int i=0; i<=n; ++i)            // 找移动之后的第一个不为零的项
        if (num[i] != 0ll){
            flag = i;
            break;
        }
    ll sum = 0ll;
    int ans = 0;
    for (int i=n; i>=0; --i){
        sum = sum*2ll+num[i];           // 按权求和
        if (abs(sum) > 2e9)             // 已经不满足系数绝对值条件，再往前更加不满足
            break;
        if (i <= flag){
            ll temp = abs(sum-t[i]);
            if (i==n && temp==0ll)      // 满足此条件，说明要使最高项系数为0，不可取
                continue;
            if (temp <= k)
                ++ans;
        }
    }
    cout << ans << endl;
    return 0;
}