POJ 2527 polynomial remains polynomial operation

Polynomial remains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1236		Accepted: 700

Description

Given the polynomial
A (x) = an xn + ... + A1 X + a0,
Compute the remainder R (x) when a (x) was divided by xk+1.

Input

The input consists of a number of cases. The first line of all case specifies the 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.

Assume that the coefficients of the remainder can is represented by 32-bit integers.

Sample Input

5 26 3 3 2 0 15 20 0 3 2 0 14 11 4 1 1 16 32 3-3 4 1 0 11 05 10 073 51 2 3 4-1-1

Sample Output

3 2-3 -1-2-1 2-3001 2 3 4

Source

Alberta Collegiate Programming Contest 2003.10.18

AC Code

#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;}

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POJ 2527 polynomial remains polynomial operation