1009. Product of Polynomials (25)

本是古典何须时尚 2022-05-31 07:08 101阅读 0赞

This time, you are supposed to find A\*B where A and B are two polynomials.

**Input Specification:**

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:K N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.

**Output Specification:**

For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

**Sample Input**

2 1 2.4 0 3.2
    2 2 1.5 1 0.5

**Sample Output**

3 3 3.6 2 6.0 1 1.6

--------------------

题目大意：

这次，你应该找到A\*B，其中A和B是两个多项式。  
输入规格：  
每一个输入文件包含一个测试用例。每个案占2行，每一行包含一个多项式的信息：K N1 aN1 N2 aN2 ... NK aNK, K是多项式中非零项的个数，Ni和aNi (i= 1,2，…K)分别是指数和系数。给定1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.  
输出规范：  
对于每个测试用例你应该在一行中输出A和B的乘积，与输入格式相同。注意，在每一行的末尾没有多余的空格。请精确到小数点后一位。

代码：

#include<stdio.h>
    double a[1001],b[1001],c[2001];
    int main()
    {
        int i,j,n,m,k,t,count=0;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",&k);
            scanf("%lf",&a[k]);
        }
        scanf("%d",&m);
        for(i=0;i<m;i++)
        {
            scanf("%d",&t);
            scanf("%lf",&b[t]);
        }
        for(i=0;i<1001;i++)
        {
            if(a[i]!=0)
            {
                for(j=0;j<1001;j++)
                {
                    if(b[j]!=0)
                    {
                        c[i+j]+=a[i]*b[j];
                    }
                }
            }
        }
        for(i=2000;i>=0;i--)
        {
            if(c[i]!=0)
            {
                count++;
            }
        }
        printf("%d",count);
        for(i=2000;i>=0;i--)
        {
            if(c[i]!=0)
            {
                printf(" %d %.1lf",i,c[i]);
            }
        }
        return 0;
    }