1009 Product of Polynomials (25分) 多项式乘法
1009 Product of Polynomials (25分)
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
注意系数为0的情况
#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<stack>
#include<algorithm>
#include<map>
#include<string.h>
#include<string>
#define MAX 1000000
#define ll long long
using namespace std;
double a[],b[],c[];
double y;
int x;
int main()
{
int k;
cin>>k;
for(int i=;i<;i++)
a[i]=b[i]=c[i]=;
for(int i=;i<k;i++)
{
cin>>x>>y;
a[x]=y;
}
cin>>k;
for(int i=;i<k;i++)
{
cin>>x>>y;
b[x]=y;
}
for(int i=;i<;i++)
{
for(int j=;j<;j++)
c[i+j]=c[i+j]+a[i]*b[j];//合并同类项
} int cnt=;
for(int i=;i<;i++)
{
if(c[i]!=)//去除多项式系数为0的项
cnt++;
}
cout<<cnt;
for(int i=;i>=;i--)
{
if(c[i]!=)
printf(" %d %.1lf",i,c[i] );
}
cout<<endl;
return ;
}
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