PAT Advanced 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 (,) are the exponents and coefficients, respectively. It is given that 1, 0.
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
求多项式乘积,并输出每个系数非零项的系数和次数。ans[i]存储结果中次数为i的项的系数,src1[i]、src2[i]分别存储输入中次数为i的项的系数。遍历所有非零位相乘,结果存入ans中即可。
#include <bits/stdc++.h> using namespace std; float ans[2010],src1[2010],src2[2010]; int main() { int K,tn; float an; cin>>K; for(int i=0;i<K;i++) { cin>>tn>>an; src1[tn]=an; } cin>>K; for(int i=0;i<K;i++) { cin>>tn>>an; src2[tn]=an; } for(int i=0;i<2010;i++) { if(src1[i]!=0) { for(int j=0;j<2010;j++) { if(src2[j]!=0) { ans[i+j]+=src1[i]*src2[j]; } } } } int cnt=0; for(int i=0;i<2010;i++) { if(ans[i]!=0) cnt++; } cout<<cnt; for(int i=2009;i>=0;i--) if(ans[i]!=0) { cout<<' '<<i<<' '; cout<<fixed<<setprecision(1)<<ans[i]; } cout<<endl; return 0; }