2018_1_30_Polynomial Remains_多项式除法_数组模拟

最新推荐文章于 2022-07-31 23:52:26 发布

ujn20161222

最新推荐文章于 2022-07-31 23:52:26 发布

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本文介绍了一种计算多项式 a(x) 除以 x^k+1 的余数 r(x) 的方法。通过输入多项式的系数和 k 的值，可以计算出余数的系数。示例展示了如何处理不同多项式的情况。

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Polynomial Remains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1366		Accepted: 764

Description

Given the polynomial
a(x) = a _n x ⁿ + ... + a ₁ x + a ₀,
compute the remainder r(x) when a(x) is divided by x ^k+1.

Input

The input consists of a number of cases. The first line of each case specifies the two integers n and k (0 <= n, k <= 10000). The next n+1 integers give the coefficients of a(x), starting from a0 and ending with an. The input is terminated if n = k = -1.

Output

For each case, output the coefficients of the remainder on one line, starting from the constant coefficient r0. If the remainder is 0, print only the constant coefficient. Otherwise, print only the first d+1 coefficients for a remainder of degree d. Separate the coefficients by a single space.

You may assume that the coefficients of the remainder can be represented by 32-bit integers.

Sample Input

5 2
6 3 3 2 0 1
5 2
0 0 3 2 0 1
4 1
1 4 1 1 1
6 3
2 3 -3 4 1 0 1
1 0
5 1
0 0
7
3 5
1 2 3 4
-1 -1

Sample Output

Source

Alberta Collegiate Programming Contest 2003.10.18

#include<iostream>
#include<algorithm>
#include<string> 
#include<cstring>
#include<vector>
#include<cmath>
#include<fstream>
#include<iomanip>
using namespace std;
const int N=10000+5;
int a[N];
int main(){
	int n,k;
	while(scanf("%d%d",&n,&k)){
		if(n==-1&&k==-1)return 0;
		for(int i=0;i<=n;i++)
		scanf("%d",&a[i]);
		for(int i=n;i>=k;i--){
			if(a[i]!=0){
				a[i-k]-=a[i];
				a[i]=0;
			}
		}
		while(n>=0&&!a[n])n--;
		for(int i=0;i<n;i++)
		printf("%d ",a[i]);
		printf("%d\n",a[n]);
	}
	return 0;
}