PAT 1009 Product of Polynomials (25)
PAT 1009 Product of Polynomials (25)
week
发布于 2018-08-27 09:32:10
发布于 2018-08-27 09:32:10
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 0.5Sample Output
3 3 3.6 2 6.0 1 1.6#include<iostream>
#include<iomanip>
#include<math.h>
using namespace std;
struct Polynomials
{
int exponents;
double coefficients;
};
int main()
{
int i,j,n,m,exponents,cnt=0;
double coefficients;
Polynomials a[11],b[11],c[2001];
cin>>n;
for(i=0;i<n;i++)
{
cin>>a[i].exponents>>a[i].coefficients;
}
cin>>m;
for(i=0;i<m;i++)
{
cin>>b[i].exponents>>b[i].coefficients;
}
for(i=0;i<=2000;i++)
{
c[i].coefficients=0;
c[i].exponents=i;
}
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
//cout<<a[i].exponents<<" "<<a[i].coefficients<<endl;
//cout<<b[j].exponents<<" "<<b[j].coefficients<<endl;
exponents=a[i].exponents+b[j].exponents;//指数相加
coefficients=a[i].coefficients*b[j].coefficients;//系数相乘
c[exponents].coefficients+=coefficients;
//cout<<exponents<<" "<<coefficients<<endl;
}
}
for(i=2000;i>=0;i--)
{
if(fabs(c[i].coefficients)!=0)
{
cnt++;
}
}
cout<<cnt;
for(i=2000;i>=0;i--)
{
if(fabs(c[i].coefficients)!=0)
{
cout<<" "<<c[i].exponents<<" "<<fixed<<setprecision(1)<<c[i].coefficients;
}
}
return 0;
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原始发表：2016年03月07日，如有侵权请联系 cloudcommunity@tencent.com 删除
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