1009 Product of Polynomials

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

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

product：乘积

• 代码

  #include
  #include
  #include
  #include
  #include
  #include
  using namespace std;
  
  int main(){
     int n1;
     cin >> n1;
     double arr[1001] = {0.0};
     double ans[2001] = {0.0};
     int zhishu;
     float xishu;
     for(int i = 0; i < n1; i ++)
     {
         cin >> zhishu >> xishu;
         arr[zhishu] = xishu;
     }
     int n2;
     cin >> n2;
     for(int i = 0; i < n2; i ++)
     {
          cin >> zhishu >> xishu;
          **for(int j = 0; j < 1001;j ++ )
          {
              ans[j + zhishu] += arr[j] * xishu;
          }**
     }
     int cnt = 0;
     for(int i = 0; i < 2001;i ++)
     {
         if(ans[i] != 0.0) cnt ++; //系数不等于0
     }
     cout << cnt ;
     for**( int i = 2000; i >= 0; i--)**
     {
         if(ans[i] != 0.0) printf(" %d %.1f", i, ans[i]); //输出指数和系数
     }
  
      return 0;
  }
  
  

  • 总结

    和那道1002做法原理差不多，姊妹题，不过这道需要注意是按指数由大到小进行输出。还有个英文问题，一开始不知道这product的意思是乘积（哭