Codeforces 639C Bear and Polynomials 【思维】
2016-04-08 17:01
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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:
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; }
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