1002 A+B for 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 sum 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 to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 2 1.5 1 2.9 0 3.2

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

polynomial 多项式

exponent 指数

coefficient 系数

• 代码

  #include
  #include
  #include
  #include
  #include
  #include
  using namespace std;
  
  int main(){
     int n1;
     cin >> n1;
     float c[1001]={0};
     int zhishu;
     float xishu;
     **for(int i = 0; i < n1; i ++)
     {
         cin >> zhishu >> xishu;
         c[zhishu] += xishu;
     }**
     int n2;
     cin >> n2;
     for(int i = 0; i < n2; i ++)
     {
          cin >> zhishu >> xishu;
          c[zhishu] += xishu;
     }
     int cnt = 0;
     for(int i = 0; i < 1001;i ++)
     {
         if(c[i] != 0) cnt ++; //系数不等于0
     }
     cout << cnt ;
     for( int i = 1000; i >= 0; i--)
     {
         if(c[i] != 0.0) printf(" %d %.1f", i, c[i]); //输出指数和系数
     }
  
      return 0;
  }
  
  

• 总结

  注意这题要按照系数的由大到小排序，保留一位小数。通过这题掌握利用数组计数的方法。