Quotient Polynomial

最新推荐文章于 2020-03-05 18:06:59 发布

飞火恋月

最新推荐文章于 2020-03-05 18:06:59 发布

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本文介绍了一种解决多项式除法问题的简单算法，并通过示例详细解释了如何求解给定多项式和整数k下的商多项式q(x)及余数r。提供了完整的C++代码实现。

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A polynomial of degree n can be expressed as

If k is any integer then we can write:

Here q(x) is called the quotient polynomial of p(x) of degree (n-1) and r is any integer which is called the remainder.

For example, if p(x) = x³- 7x²+ 15x - 8 and k = 3 then q(x) = x²- 4x + 3 and r = 1. Again if p(x) = x³- 7x²+ 15x - 9 and k = 3 then q(x) = x²- 4x + 3 and r = 0.

In this problem you have to find the quotient polynomial q(x) and the remainder r. All the input and output data will fit in 32-bit signed integer.

Input
Your program should accept an even number of lines of text. Each pair of line will represent one test case. The first line will contain an integer value for k. The second line will contain a list of integers (a_n, a_{n-1, …} a₀), which represent the set of co-efficient of a polynomialp(x). Here 1 ≤ n ≤ 10000. Input is terminated by <EOF>.

Output
For each pair of lines, your program should print exactly two lines. The first line should contain the coefficients of the quotient polynomial. Print the reminder in second line. There is a blank space before and after the ‘=’ sign. Print a blank line after the output of each test case. For exact format, follow the given sample.

Sample Input	Output for Sample Input
3 1 –7 15 –8 3 1 –7 15 –9	q(x): 1 -4 3 r = 1 q(x): 1 -4 3 r = 0

这个题其实并不难一开始的那次错误是我想复杂了其实只要想通了算法很简单开始的时候扯得太远了水题一个上代码

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<ctype.h>
using namespace std;
const int maxn= 10001;
int main()
{
    char ch;
    int a[maxn],n,k,r,b[maxn];
    while(scanf("%d",&k)!=EOF)
    {
        int i=1,flag=1;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        while (scanf("%d%c",&a[i],&ch)) //不要直接输入整串字符串那样往外取数太麻烦了
        {
            if (ch=='\n') break;
            i++;
        }
        n=i;
        b[n-1]=a[1];
        for(i=n-2; i>=0; i--)
        {
            b[i]=k*b[i+1]+a[n-i];
        }
        printf("q(x):");
        for (i=n-1; i>0; i--)
            printf(" %d",b[i]);
        printf("\n");
        printf("r = %d\n",b[0]);
        printf("\n");
    }
    return 0;
}