pat甲级1009. Product of Polynomials (25)

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本文介绍了一种解决多项式乘法问题的算法，并通过一个具体示例展示了如何使用桶排序的方法来高效地求解两个多项式的乘积。代码采用C++编写，详细解释了输入输出格式及注意事项。

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1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

题意分析：

每一个测试样例包含两行，第一个数k,随后输出k对的数，N1代表指数，a_{N1代表系数，对于每一个样例输出A*B的结果，同时注意输出格式哟，利用桶排序，}

注意定义的类型，定义数组为Double型的哟，详解见代码

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string.h>
#include<cstdio>
#include<stdio.h>
#include<iomanip>
using namespace std;
int main()
{
        int k;
        double a[1001],b[1001],c[2002];
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        cin>>k;
        int e;
        double c1;
        for(int i=1;i<=k;i++)
        {
                cin>>e>>c1;
                a[e]+=c1;
        }
        cin>>k;
        for(int i=1;i<=k;i++)
        {
                cin>>e>>c1;
                b[e]+=c1;
        }
        for(int i=0;i<=1000;i++)
        {
                for(int j=0;j<=1000;j++)
                {
                        c[i+j]+=a[i]*b[j];
                }
        }
        int count=0;
        for(int i=0;i<=2000;i++)
        {
                if(c[i]!=0.0)
                        count++;
        }
        cout<<count;
        for(int i=2000;i>=0;i--)
        {
                if(c[i]!=0.0)
                       //cout<<" "<<i<<" "<<c[i];
                {
                        cout<<" "<<i;
                        cout<<fixed<<setprecision(1);
                        cout<<" "<<c[i];
                }
        }
        cout<<endl;
        return 0;
}