Codeforces 639C Bear and Polynomials 【思维】
题目链接：Codeforces 639C Bear and Polynomials
C. Bear and Polynomials
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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：给定一个2的n项式的系数。每次可以将任意一个系数修改为绝对值不超过k的值，问你有多少种方案使得该多项式结果为0。
处理多项式学会了新的技能，ORZ。。。
思路：对于多项式 ∑ni=0a[i]∗2i ，我们可以将 a[i] 的值加权到 2i+1 上，方法就是 a[i+1]=a[i+1]+a[i]2 且 a[i]=a[i]mod2 。这样处理后，多出了第n+1次幂，且前面所有幂次的系数为-1、0、1中的一个。之后我们可以倒着消去高次幂的系数，加到低次幂的系数上，具体是 a[i]=a[i]+a[i+1]∗2 且 a[i+1]=0 。
发现我们总是可以将后面项的系数合并到第i项上(0<=i<=n)，假设合并到第i项系数为xx，此时第i项系数只有取a[i]-xx且前面所有项系数均为0才可以使得整个多项式结果为0。因为前面所有项的值sum满足 −2i+1<=sum<=2i−1 ，二者的和不可能为0。若标记第一个非0系数项为F，只有当项数i <= F时才是合法的，判定下是否合法即可。这里有两个trick
1，我们倒着向低次幂添加系数时，每次累乘2，若不加限制乘下去会爆，所以当系数 > 2*k时就跳出；
2，当n <= F时，说明0 - n-1项的系数全为0，这时当系数与原来的系数一样时是不合法的方案。
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MOD = 1e9 + 7;
const int MAXN = 2*1e5 + 10;
void add(LL &x, LL y) { x += y; x %= MOD; }
LL x[MAXN], y[MAXN];
int main()
{
int n, k;
while(scanf("%d%d", &n, &k) != EOF) {
for(int i = 0; i <= n; i++) {
scanf("%lld", &x[i]);
y[i] = x[i];
}
y[n+1] = 0;
for(int i = 0; i <= n; i++) {
y[i+1] += y[i] / 2;
y[i] %= 2;
//printf("%lld\n", y[i+1]);
}
int F = n + 1;
for(int i = 0; i <= n; i++) {
if(y[i]) {
F = i; break;
}
}
// for(int i = 0; i <= n; i++) {
// cout << y[i] << endl;
// }
int ans = 0; LL xx = y[n+1];
for(int i = n; i >= 0; i--) {
xx = 2 * xx + y[i];
if(abs(xx) > 2 * k) break;
if(i > F) continue;
if(i == n && x[i] == xx) continue;
ans += (abs(x[i] - xx) <= k);
}
printf("%d\n", ans);
}
return 0;
}