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<1)<需要头文件#include 
    
   
 
   
 
   
 
   
评测结果
 
   
 
    
 
     
 
      
时间
 
      
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题目
 
      
语言
 
      
用时(ms)
 
      
内存(kB)
 
      
用户
 
     
 
    
 
    
 
     
 
      
7月17日 20:46
 
      
答案正确
 
      
25
 
      
1009
 
      
C++ (g++ 4.7.2)
 
      
1
 
      
360
 
      
datrilla
 
     
 
    
 
   
 
   
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测试点
 
      
结果
 
      
用时(ms)
 
      
内存(kB)
 
      
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0
 
      
答案正确
 
      
1
 
      
256
 
      
5/5
 
     
 
     
 
      
1
 
      
答案正确
 
      
1
 
      
360
 
      
5/5
 
     
 
     
 
      
2
 
      
答案正确
 
      
1
 
      
360
 
      
5/5
 
     
 
     
 
      
3
 
      
答案正确
 
      
1
 
      
232
 
      
5/5
 
     
 
     
 
      
4
 
      
答案正确
 
      
1
 
      
232
 
      
5/5
 
     
 
    
 
   
 
    
   
#include
#include
#include
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>index1[0][i]>>e[0][i];
 } 
 cin>>n2; 
 for(i=0;i>index1[1][i]>>e[1][i];
  {
   for(j=0;j0][j]+index1[1][i]]+=e[0][j]*e[1][i];
   }
  }
 }  
 i=2001;
 count=0;
 while(i--)
 {
  if(fabs(index[i])>esp)count++;
 }
 cout<2001;
 while(i--)
 {
  if(fabs(index[i])>esp){
            cout<<" "<" ";
   cout<1)<return 0;
}