1009. Product of Polynomials (25)

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

用到(小数点保留一位)cout<<setiosflags(ios::fixed)<<setprecision(1)<<index[i];需要头文件#include<iomanip>

评测结果

测试点

#include<iostream>
#include<iomanip>
#include<math.h>
using namespace std;
#define esp 0.05
int main()
{
 double index[2001]={0},e[2][11];
 int index1[2][11];
 int n1,n2,i,j,count;  
 cin>>n1; 
 for(i=0;i<n1;i++)
 {
  cin>>index1[0][i]>>e[0][i];
 } 
 cin>>n2; 
 for(i=0;i<n2;i++)
 {  
  cin>>index1[1][i]>>e[1][i];
  {
   for(j=0;j<n1;j++)
   {
    index[index1[0][j]+index1[1][i]]+=e[0][j]*e[1][i];
   }
  }
 }  
 i=2001;
 count=0;
 while(i--)
 {
  if(fabs(index[i])>esp)count++;
 }
 cout<<count;
 i=2001;
 while(i--)
 {
  if(fabs(index[i])>esp){
            cout<<" "<<i<<" ";
   cout<<setiosflags(ios::fixed)<<setprecision(1)<<index[i];}
 }
 cout<<endl;
 return 0;
}