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

代码

#include<iostream>
#include<vector>
#include<string>
#include<math.h>
#include<algorithm>
#include<stack>
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

psogos