POJ-2527 Polynomial Remains-多項式相除

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.

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.

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.

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

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

Alberta Collegiate Programming Contest 2003.10.18

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 10010;
int main()
{
    int n,k;
    int val[maxn];
    while(scanf("%d%d",&n,&k)!=EOF,n!=-1 || k !=-1)
    {
        int i;
        for(i = 0 ; i <= n ; ++i)
        {
            scanf("%d",&val[i]);
        }
        //進行除法運算
        for(i = n ; i >= k ; --i)
        {
            if(val[i] == 0)
            {
                continue;
            }

            val[i-k] = val[i-k] - val[i];
            val[i] = 0;
        }
        //調整陣列長度,即高位的0不用輸出
        int t = n;
        while(val[t] == 0 && t > 0)
        {
            --t;
        }
        for(i = 0 ; i < t ; ++i)
        {
            printf("%d ",val[i]);
        }
        printf("%d\n",val[t]);
    }
    return 0;
}

因為輸入輸出資料太多，所以用cin、cout會超時。。