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:
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 ≤ 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.
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:
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.
题目大意:给你一个n次多项式,问你能不能通过改变其中一项的系数来使整个多项式值为0,其中系数的绝对值不能超过k。
分析:我们通过把整个多项式转换成二进制形式,然后解只能通过第一个先导非零项和它前边的几项产生,因为一个2的整数次幂无法表示比它还小的项,倒着枚举并统计前边几项的和并统计答案。
#include <cstdio> #include <iostream> #define MAXN 2147483647 using namespace std; long long n,k,tot,num,now,a[200001],f[200001]; int main() { cin.sync_with_stdio(false); cin>>n>>k; for(int i = 0;i <= n;i++) cin>>a[i]; now = -1; for(int i = 0;i <= n;i++) { f[i] += a[i]; if(i != n) { f[i+1] += f[i]/2; f[i] %= 2; } if(now == -1 && f[i] != 0) now = i; } for(int i = n;i >= 0;i--) { tot *= 2; tot += f[i]; if(tot > MAXN || tot < -MAXN) break; if(i <= now) { long long ans = tot - a[i]; if(ans <= k && ans >= -k && !(i == n && ans == 0)) num++; } } cout<<num<<endl; }