CodeForces 657B Bear and Polynomials
题目链接：http://codeforces.com/problemset/problem/657/B
Bear and Polynomials
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
3 1000000000
10 -9 -3 5
3
3 12
10 -9 -3 5
2
2 20
14 -7 19
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.
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 <bits/stdc++.h>
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){<span style="white-space:pre"> </span>// 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)<span style="white-space:pre"> </span>// 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];<span style="white-space:pre"> </span>// 按权求和
if (abs(sum) > 2e9)<span style="white-space:pre"> </span>// 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)<span style="white-space:pre"> </span>// 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;
}