POJ 2527 Polynomial Remains 多项式运算

Polynomial Remains

Description

Given the polynomial

a(x) = a_n xⁿ + ... + a₁ x + a₀,

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

3 2
-3 -1
-2
-1 2 -3
0
0
1 2 3 4

Source

Alberta Collegiate Programming Contest
2003.10.18

AC代码

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int main(){
    int n,k;
    int my[10000+10];
    while(cin>>n>>k){
        if(n==-1&&k==-1)break;
        for(int i=0;i<=n;i++)cin>>my[i];
        ///for(int i=n;i>0;--i)cout<<my[i]<<"x^"<<i<<'+';
        ///cout<<my[0]<<'\12';
        for(int i=n;i>=k;--i){
            if(my[i]==0)continue;
            my[i-k]=my[i-k]-my[i];
            my[i]=0;
        }
        int len=n;
        while(len>=0&&!my[len])len--;
        for(int i=0;i<len;++i)cout<<my[i]<<' ';
        cout<<my[len]<<'\12';
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。