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
Submit:
#include <iostream> #include <cmath> using namespace std; //目标:求多项式的乘积 int main() { int a,b,i,j,exp,k=0; scanf("%d",&a); float coe,arr[1001]={0.0},temp[2001]={0.0};//相乘后指数最大为2000 for (i=0;i<a;i++) { scanf("%d %f",&exp,&coe); arr[exp] = coe; } scanf("%d",&b); for (i=0; i<b; i++) { scanf("%d %f",&exp,&coe); for (j=0;j<1001;j++) //遍历第一行所有系数 if (arr[j] != 0.0) temp[exp+j] += coe*arr[j];//把第二行输入的每一个系数与第一行每一个相乘 结果保存到temp数组 +=保存相同系数的和 } for (i=0;i<2001;i++) if (temp[i] != 0.0) k++; printf("%d",k); for (i=2001;i>=0;i--) if (temp[i] != 0.0) printf(" %d %.1f",i,temp[i]); return 0; }